Displaying 1-10 of 15 results found.
Triangle T(n, k) read by rows: row n gives the coefficients of the numerator polynomials of the o.g.f. of the (n+1)-th diagonal of the Sheffer triangle A154537 (S2[2,1] generalized Stirling2), for n >= 0.
+20
1
1, 1, 2, 1, 16, 12, 1, 66, 284, 120, 1, 224, 2872, 5952, 1680, 1, 706, 21080, 116336, 146064, 30240, 1, 2160, 132228, 1531072, 4804656, 4130304, 665280, 1, 6530, 760500, 16271080, 101422640, 208791648, 132557760, 17297280, 1, 19648, 4155120, 151922560, 1661273440, 6556459008, 9657333504, 4766423040, 518918400, 1, 59010, 21993776, 1304454880, 23155279200, 155184721088, 427142449920, 477104352768, 189945688320, 17643225600
COMMENTS
The ordinary generating function (o.g.f.) of the (n+1)-th diagonal sequence of the Sheffer triangle A154537 = (e^x, e^(2*x) - 1), called S2[2,1], is GS2(2,1;n,x) = P(n, x)/(1 - 2*x)^(2*n+1), with the row polynomials P(n, x) = Sum_{k=0..n} T(n, k)*x^k, n >= 0. In the general case of Sheffer S2[d,a] = (e^(a*x), e^(d*x) - 1) (with gcd(d,a) = 1, d >= 0, a >= 0, and for d = 1 one takes a = 0) the o.g.f. of the (n+1)-th diagonal sequence is G(d,a;n,x) = P(d,a;n,x)/(1 - d*x)^(2*n + 1) with the numerator polynomial P and coefficient table T(d,a;n,k).
For the computation of the exponential generating function (e.g.f.) of the o.g.f.s of the diagonal sequences of a Sheffer triangle (lower triangular matrix) via Lagrange's theorem see a comment in A290311.
FORMULA
T(n, k) = [x^k] P(n, x) with the numerator polynomial in the o.g.f. of the (n+1)-th diagonal sequence of the triangle A154537. See a comment above.
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 ...
0: 1
1: 1 2
2: 1 16 12
3: 1 66 284 120
4: 1 224 2872 5952 1680
5: 1 706 21080 116336 146064 30240
6: 1 2160 132228 1531072 4804656 4130304 665280
7: 1 6530 760500 16271080 101422640 208791648 132557760 17297280
...
n = 8: 1 19648 4155120 151922560 1661273440 6556459008 9657333504 4766423040 518918400,
n = 9: 1 59010 21993776 1304454880 23155279200 155184721088 427142449920 477104352768 189945688320 17643225600.
...
n=3: The o.g.f. of the 4th diagonal sequence of A154537, [1, 80, 1320, ...], is P(3, x) = (1 + 66*x + 284*x^2 + 120*x^3)/(1 - 2*x)^7.
A symmetrical triangle sequence made from A154537:q(x,n)= Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]); p(x,n)=q(x,n)+x^n*q(1/x,n); t(n,m)=coefficients(p(x,n)).
+20
0
2, 3, 3, 5, 16, 5, 9, 62, 62, 9, 17, 208, 464, 208, 17, 33, 642, 2680, 2680, 642, 33, 65, 1880, 13404, 24320, 13404, 1880, 65, 129, 5322, 62188, 180488, 180488, 62188, 5322, 129, 257, 14752, 280144, 1209600, 1858752, 1209600, 280144, 14752, 257, 513
COMMENTS
Row sums are:
{2, 6, 26, 142, 914, 6710, 55018, 496254, 4868258, 51483878, 582795578,...}
FORMULA
q(x,n)= Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
p(x,n)=q(x,n)+x^n*q(1/x,n);
t(n,m)=coefficients(p(x,n)).
EXAMPLE
{2},
{3, 3},
{5, 16, 5},
{9, 62, 62, 9},
{17, 208, 464, 208, 17},
{33, 642, 2680, 2680, 642, 33},
{65, 1880, 13404, 24320, 13404, 1880, 65},
{129, 5322, 62188, 180488, 180488, 62188, 5322, 129},
{257, 14752, 280144, 1209600, 1858752, 1209600, 280144, 14752, 257},
{513, 40418, 1262544, 7828640, 16609824, 16609824, 7828640, 1262544, 40418, 513},
{1025, 110248, 5787604, 50950400, 140957728, 187181568, 140957728, 50950400, 5787604, 110248, 1025}
MATHEMATICA
p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]'
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]
+ Reverse[ CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 0, 10}]'
Flatten[%]
Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.
+10
638
1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
COMMENTS
Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
With regard to the preceding comment: For arbitrary (including non-disjoint) covers of an n-set by k nonempty subsets see A055154. - Manfred Boergens, May 20 2024 Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938. Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005 Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005 Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008 For tables of restricted Stirling numbers of the second kind see A143494 - A143496. S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009 S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009 Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010 S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011 If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014 S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015 The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017 With offsets 0 for n and k this is the Sheffer product matrix A007318* A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017 Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019 Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020 k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
J. Stirling, The Differential Method, London, 1749; see p. 7.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. T. Copeland's Shadows of Simplicity, A Class of Differential Operators and the Stirling Numbers,2015; Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015; The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions, 2011; Mathemagical Forests, 2008; and Addendum to "Mathemagical Forests", 2010.
FORMULA
S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0. S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019. For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n ( A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005 Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers ( A000110). - Emeric Deutsch, Nov 01 2006 Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007 Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007 Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)). (Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007 The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011 O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012 n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012 G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013 Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x). With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes. (End)
As a matrix equivalent of some inversions mentioned above, A008277* A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014 O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014 Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015 Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind. For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End) O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022 S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022 S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
EXAMPLE
The triangle S2(n, k) begins:
\ k 1 2 3 4 5 6 7 8 9
n \ 10 11 12 13 14 15 ...
----------------------------------------------------------------------------------
1 | 1
2 | 1 1
3 | 1 3 1
4 | 1 7 6 1
5 | 1 15 25 10 1
6 | 1 31 90 65 15 1
7 | 1 63 301 350 140 21 1
8 | 1 127 966 1701 1050 266 28 1
9 | 1 255 3025 7770 6951 2646 462 36 1
10 | 1 511 9330 34105 42525 22827 5880 750 45
1
11 | 1 1023 28501 145750 246730 179487 63987 11880 1155
55 1
12 | 1 2047 86526 611501 1379400 1323652 627396 159027 22275
1705 66 1
13 | 1 4095 261625 2532530 7508501 9321312 5715424 1899612 359502
39325 2431 78 1
14 | 1 8191 788970 10391745 40075035 63436373 49329280 20912320 5135130
752752 66066 3367 91 1
15 | 1 16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478 106470 4550 105 1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - Daniel Forgues, Jan 16 2016
MAPLE
seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007 stirling_2 := (n, k) -> (1/k!) * add((-1)^(k-i)*binomial(k, i)*i^n, i=0..k);
MATHEMATICA
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
B = BellMatrix[1&, rows];
a[n_, n_] := 1; a[n_, 1] := 1;
a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *) With[{m = 11},
Flatten@MapIndexed[Take[#, #2[[1]]] &,
Transpose@
Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
PROG
(PARI) for(n=1, 22, for(k=1, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013 (PARI) Stirling2(n, k)=sum(i=0, k, (-1)^i*binomial(k, i)*i^n)*(-1)^k/k! \\ M. F. Hasler, Mar 06 2012 (Haskell)
a008277 n k = a008277_tabl !! (n-1) !! (k-1)
a008277_row n = a008277_tabl !! (n-1)
(Maxima) create_list(stirling2(n+1, k+1), n, 0, 30, k, 0, n); /* Emanuele Munarini, Jun 01 2012 */ (J) n ((] (1 % !)) * +/@((^~ * (] (_1 ^ |.)) * (! {:)@]) i.@>:)) k NB. Stephen Makdisi, Apr 06 2016 (Magma) [[StirlingSecond(n, k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
CROSSREFS
Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle). Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810- A039813, A048994, A087107- A087111, A087127, A094262, A127701.
KEYWORD
nonn, easy, tabl, nice, core, changed
Triangle of B-analogs of Stirling numbers of the second kind.
+10
34
1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 40, 58, 16, 1, 1, 121, 330, 170, 25, 1, 1, 364, 1771, 1520, 395, 36, 1, 1, 1093, 9219, 12411, 5075, 791, 49, 1, 1, 3280, 47188, 96096, 58086, 13776, 1428, 64, 1, 1, 9841, 239220, 719860, 618870, 209622, 32340, 2388, 81, 1, 1
COMMENTS
Let M be an infinite lower triangular bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the subdiagonal. n-th row = M^n * [1,0,0,0,...]. - Gary W. Adamson, Apr 13 2009 A type B_n set partition is a partition P of the set {1, 2, ..., n, -1, -2, ..., -n} such that for any block B of P, -B is also a block of P, and there is at most one block, called a zero-block, satisfying B = -B. We call (B, -B) a block pair of P if B is not a zero-block. Then T(n,k) is the number of type B_n set partitions with k block pairs. See [Wang].
For example, T(2,1) = 4 since the B_2 set partitions with 1 block pair are {1,2}{-1,-2}, {1,-2}{-1,2}, {1,-1}{2}{-2} and {2,-2}{1}{-1} (the last two partitions contain a zero block).
(End)
Exponential Riordan array [exp(x), (1/2)*(exp(2*x) - 1)]. Triangle of connection constants for expressing the monomial polynomials x^n as a linear combination of the basis polynomials (x-1)*(x-3)*...*(x-(2*k-1)) of A039757. An example is given below. Inverse array is A039757. Equals matrix product A008277 * A122848. - Peter Bala, Jun 23 2014 T(n, k) also gives the (dimensionless) volume of the multichoose(k+1, n-k) = binomial(n, k) polytopes of dimension n-k with side lengths from the set {1, 3, ..., 1+2*k}. See the column g.f.s and the complete homogeneous symmetric function formula for T(n, k) below. - Wolfdieter Lang, May 26 2017 T(n, k) is the number of k-dimensional subspaces (i.e., sets of fixed points like rotation axes and symmetry planes) of the n-cube. See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019
LINKS
T. Piesk, Sets of fixed points of permutations of the n-cube: n = 3 and 4.
FORMULA
E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).
T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019 T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.
The row polynomials R(n,x) satisfy the Dobinski-type identity:
R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011 The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details. An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).
(End)
T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts. The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2). The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].
(End)
Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016 G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.
T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)
With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds: (1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k); (2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024
EXAMPLE
Triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 1
2: 1 4 1
3: 1 13 9 1
4: 1 40 58 16 1
5: 1 121 330 170 25 1
6: 1 364 1771 1520 395 36 1
7: 1 1093 9219 12411 5075 791 49 1
8: 1 3280 47188 96096 58086 13776 1428 64 1
9: 1 9841 239220 719860 618870 209622 32340 2388 81 1
10: 1 29524 1205941 5278240 6289690 2924712 630042 68160 3765 100 1
The sequence of row polynomials of A214406 begins [1, 1+x, 1+8*x+3*x^2, ...]. The o.g.f.'s for the diagonals of this triangle thus begin 1/(1-x) = 1 + x + x^2 + x^3 + ...
(1+x)/(1-x)^3 = 1 + 4*x + 9*x^2 + 16*x^3 + ...
(1+8*x+3*x^2)/(1-x)^5 = 1 + 13*x + 58*x^2 + 170*x^3 + ... . - Peter Bala, Jul 20 2012 Connection constants: x^3 = 1 + 13*(x-1) + 9*(x-1)*(x-3) + (x-1)*(x-3)*(x-5). Hence row 3 = [1,13,9,1]. - Peter Bala, Jun 23 2014 Complete homogeneous symmetric functions: T(3, 1) = h^{(2)}_2 = 1^2 + 3^2 + 1^1*3^1 = 13. The three 2D polytopes are two squares and a rectangle. T(3, 2) = h^{(3)}_1 = 1^1 + 3^1 + 5^1 = 9. The 1D polytopes are three lines. - Wolfdieter Lang, May 26 2017 T(4, 3) = 16 is the number of 3-dimensional subspaces (mirror hyperplanes) of the 4-cube. (These are 4 cubes and 12 cuboids.) See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019
MAPLE
A039755 := proc(n, k) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1, k-1)+(2*k+1)*procname(n-1, k) ; end if; end proc:
MATHEMATICA
t[n_, k_] = Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!); Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[1 ;; 56]]
PROG
(PARI) T(n, k)=if(k<0 || k>n, 0, n!*polcoeff(polcoeff(exp(x+y/2*(exp(2*x+x*O(x^n))-1)), n), k))
(Magma) [[(&+[(-1)^(k-j)*(2*j+1)^n*Binomial(k, j): j in [0..k]])/( 2^k*Factorial(k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 14 2019 (Sage) [[sum((-1)^(k-j)*(2*j+1)^n*binomial(k, j) for j in (0..k))/( 2^k*factorial(k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 14 2019
AUTHOR
Ruedi Suter (suter(AT)math.ethz.ch)
Triangle read by rows: 2-Stirling numbers of the second kind.
+10
31
1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
COMMENTS
This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4). The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277. For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder]. Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End) T(n,k)=S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)
LINKS
Andrei Z. Broder, The r-Stirling numbers, Report Number: CS-TR-82-949, 1982, Stanford University, Department of Computer Science.
FORMULA
T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0. T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n = k..oo} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n = 0..oo} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n = 0..oo} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i = 0..oo} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i = 0..oo} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n* A074051(n-2). This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]). Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse. Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012
EXAMPLE
Triangle begins
n\k|...2....3....4....5....6....7
=================================
2..|...1
3..|...2....1
4..|...4....5....1
5..|...8...19....9....1
6..|..16...65...55...14....1
7..|..32..211..285..125...20....1
...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
MAPLE
with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2, i)*(i+2)^(n-2), i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
MATHEMATICA
t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
PROG
(Sage)
@CachedFunction
def stirling2r(n, k, r) :
if n < r: return 0
if n == r: return 1 if k == r else 0
return stirling2r(n-1, k-1, r) + k*stirling2r(n-1, k, r)
A143494 = lambda n, k: stirling2r(n, k, 2)
for n in (2..6):
CROSSREFS
Cf. A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums), A143491, A143495, A143496, A143497.
Odd cubes: a(n) = (2*n + 1)^3.
+10
29
1, 27, 125, 343, 729, 1331, 2197, 3375, 4913, 6859, 9261, 12167, 15625, 19683, 24389, 29791, 35937, 42875, 50653, 59319, 68921, 79507, 91125, 103823, 117649, 132651, 148877, 166375, 185193, 205379, 226981, 250047, 274625, 300763, 328509, 357911, 389017, 421875
COMMENTS
Terms end in the repeating sequence 1, 7, 5, 3, 9, ... - Melvin Peralta, Jul 08 2015
REFERENCES
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3.
FORMULA
Sum_{n >= 0} 1/a(n) = 7 * zeta(3) / 8.
G.f.: (1+23*x+23*x^2+x^3)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 02 2012 Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/12)*(1 + sqrt(2)*cosh(sqrt(3)*Pi/4)) (Chamberland and Straub, 2013). - Amiram Eldar, Jan 26 2024
PROG
(Python)
def a(n): return (2*n+1)**3
Triangle of f-vectors of the simplicial complexes dual to the permutohedra of type B_n.
+10
21
1, 1, 2, 1, 8, 8, 1, 26, 72, 48, 1, 80, 464, 768, 384, 1, 242, 2640, 8160, 9600, 3840, 1, 728, 14168, 72960, 151680, 138240, 46080, 1, 2186, 73752, 595728, 1948800, 3037440, 2257920, 645120, 1, 6560, 377504, 4612608, 22305024, 52899840
COMMENTS
The Coxeter group of type B_n may be realized as the group of n X n matrices with exactly one nonzero entry in each row and column, that entry being either +1 or -1. The order of the group is 2^n*n!. The orbit of the point (1,2,...,n) (or any sufficiently generic point (x_1,...,x_n)) under the action of this group is a set of 2^n*n! distinct points whose convex hull is defined to be the permutohedron of type B_n. The rows of this table are the f-vectors of the simplicial complexes dual to these type B permutohedra. Some examples are given in the Example section below. See A060187 for the corresponding table of h-vectors of type B permutohedra. This is the (unsigned) triangle of connection constants between the polynomial sequences (2*x + 1)^n, n >= 0, and binomial(x+k,k), k >= 0. For example, (2*x + 1)^2 = 8*binomial(x+2,2) - 8*binomial(x+1,1) + 1 and (2*x + 1)^3 = 48*binomial(x+3,3) - 72*binomial(x+2,2) + 26*binomial(x+1,1) - 1. Cf. A163626. - Peter Bala, Jun 06 2019
FORMULA
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(2*i+1)^n.
Recurrence relation: T(n,k) = (2*k + 1)*T(n-1,k) + 2*k*T(n-1,k-1) with T(0,0) = 1 and T(0,k) = 0 for k >= 1.
Relation with type B Stirling numbers of the second kind: T(n,k) = 2^k*k!* A039755(n,k). E.g.f.: exp(t)/(1 + x - x* exp(2*t)) = 1 + (1 + 2*x)*t + (1 + 8*x + 8*x^2 )*t^2/2! + ... .
The polynomials in the first column of the array ((1+t)*P^(-1)-t*P)^(-1), P Pascal's triangle and I the identity, are the row polynomials of this table.
The polynomials in the first column of the array ((1+t)*I-t* A062715)^(-1) are, apart from the initial 1, the row polynomials of this table with an extra factor of t. Cf. A060187. (End) Integrating the above e.g.f. with respect to x from x = 0 to x = 1 gives Sum_{k = 0..n} (-1)^k*T(n,k)/(k + 1) = 2^n*Bernoulli(n,1/2), the n-th cosecant number.
The corresponding Type A result is considered in A028246 as Worpitzky's algorithm. Also for n >= 0, Sum_{k = 0..2*n} (-1)^k*T(2*n,k)/((k + 1)*(k + 2)) = 1/2*2^(2*n)*Bernoulli(2*n,1/2) and for n >= 1, Sum_{k = 0..2*n-1} (-1)^k*T(2*n - 1,k)/((k + 1)*(k + 2)) = -1/2 * 2^(2*n)* Bernoulli(2*n,1/2).
The row polynomials R(n,x) satisfy the recurrence equation R(n+1,x) = D(R(n,x)) with R(0,x) = 1, where D is the operator 1 + 2*x + 2*x(1 + x)*d/dx.
R(n,x) = 1/(1 + x)* Sum_{k = 0..inf} (2*k + 1)^n*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf. A019538. The shifted row polynomial x*R(n,x) = (1 + x)^n*P(n,x/(1 + x)) where P(n,x) denotes the n-th row polynomial of A060187. The row polynomials R(n,x) have only real zeros.
Symmetry: R(n,x) = (-1)^n*R(n,-1 - x). Consequently the zeros of R(n,x) lie in the open interval (-1, 0). (End)
Recurrence for row polynomials: R(n,x) = 1 + x*Sum_{k = 0..n-1} binomial(n,k)2^(n-k)*R(k,x) with R(0,x) = 1.
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(k,z))) )^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). A(k,z) = A(-(k + 1),-z). Cf. A019538. n-th row polynomial R(n,x) = (1 + 2*x) o (1 + 2*x) o ... o (1 + 2*x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E13 in the Bala link.
R(n,x) = Sum_{k = 0..n} binomial(n,k)*2^k*F(k,x) where F(k,x) is the Fubini polynomial of order k, the k-th row polynomial of A019538. (End)
EXAMPLE
The triangle begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1
1..|..1.....2
2..|..1.....8.....8
3..|..1....26....72....48
4..|..1....80...464...768...384
5..|..1...242..2640..8160..9600..3840
...
Row 2: the permutohedron of type B_2 is an octagon with 8 vertices and 8 edges. Its dual, also an octagon, has f-vector (1,8,8) - row 3 of this triangle.
Row 3: for an appropriate choice of generic point in R_3, the permutohedron of type B_3 is realized as the great rhombicuboctahedron, also known as the truncated cuboctahedron, with 48 vertices, 72 edges and 26 faces (12 squares, 8 regular hexagons and 6 regular octagons). See the Wikipedia entry and also [Fomin and Reading p.22]. Its dual polyhedron is a simplicial polyhedron, the disdyakis dodecahedron, with 26 vertices, 72 edges and 48 triangular faces and so its f-vector is (1,26,72,48) - row 4 of this triangle.
Examples of falling factorials identities for odd numbered rows: Let (x)_n = x*(x - 1)*...*(x - n + 1) with (x)_0 = 1 denote the falling factorial power.
Row 1: 2*(x)_1 + (0 - 2*x)_1 = 0.
Row 3: 48*(x)_3 + 72*(x)_2 * (2 - 2*x)_1 + 26*(x)_1 * (2 - 2*x)_2 + (2 - 2*x)_3 = 0
Row 5: 3840*(x)_5 + 9600*(x)_4 * (4 - 2*x)_(1) + 8160*(x)_3 * (4 - 2*x)_2 + 2640*(x)_2 * (4 - 2*x)_3 + 242*(x)_1 * (4 - 2*x)_4 + (4 - 2*x)_5 = 0. (End)
MAPLE
with(combinat):
T:= (n, k) -> add((-1)^(k-i)*binomial(k, i)*(2*i+1)^n, i = 0..k):
for n from 0 to 9 do
seq(T(n, k), k = 0..n);
end do;
MATHEMATICA
T[n_, k_] := Sum[(-1)^(k - i)*Binomial[k, i]*(2*i + 1)^n, {i, 0, k}];
Inverse of a generalized Stirling number triangle of first kind.
+10
15
1, -1, 1, 1, -1, 1, -1, 1, 0, 1, 1, -1, 1, 2, 1, -1, 1, 0, 5, 5, 1, 1, -1, 1, 10, 20, 9, 1, -1, 1, 0, 21, 70, 56, 14, 1, 1, -1, 1, 42, 231, 294, 126, 20, 1, -1, 1, 0, 85, 735, 1407, 924, 246, 27, 1, 1, -1, 1, 170, 2290, 6363, 6027, 2400, 435, 35, 1
COMMENTS
Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296. T(n,k)*(-1)^(n-k) gives the inverse Sheffer matrix of A094645. In the umbral notation (cf. Roman, p. 17, quoted in A094645) this is Sheffer for (1-t,-log(1-t)). - Wolfdieter Lang, Jun 20 2011 For a graph G and a positive integer k, the graphical Stirling number S(G;k) is the number of partitions of the vertex set of G into k nonempty independent sets (an independent set in G is a subset of the vertices, no two elements of which are adjacent). Omitting the first two rows and columns of this triangle produces the triangle of graphical Stirling numbers of cycles on n vertices (interpreting a 2-cycle as a single edge). In comparison, the classical Stirling numbers of the second kind, A008277, are the graphical Stirling numbers of path graphs on n vertices. See Galvin and Than. See also A227341. This is the triangle of connection constants for expressing the polynomial sequence (x-1)^n in terms of the falling factorial polynomials, that is, (x-1)^n = Sum_{k = 0..n} T(n,k)*x_(k), where x_(k) = x*(x-1)*...*(x-k+1) denotes the falling factorial.
The row polynomials are a particular case of the Actuarial polynomials - see Roman 4.3.4. (End)
REFERENCES
S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.
FORMULA
Term k in row n is given by {(-1)^(k+n) * (Sum_{j=0..k} (-1)^j * binomial(k,j) * (1-j)^n) / k! }; i.e., a finite difference. - Tom Copeland, Jun 05 2006 O.g.f. for row n = n-th finite difference of the Touchard (Bell) polynomials, T(x,j) and so the e.g.f. for these finite differences and therefore the sequence = exp{x*[exp(t)-1]-t} = exp{t*[T(x,.)-1]} umbrally. - Tom Copeland, Jun 05 2006 The e.g.f. A(x,t) = exp(x*(exp(t)-1)-t) satisfies the partial differential equation x*dA/dx - dA/dt = (1-x)*A.
Recurrence relation: T(n+1,k) = T(n,k-1) + (k-1)*T(n,k).
Let f(x) = ((x-1)/x)*exp(x). For n >= 1, the n-th row polynomial R(n,x) = x*exp(-x)*(x*d/dx)^(n-1)(f(x)) and satisfies the recurrence equation R(n+1,x) = (x-1)*R(n,x) + x*R'(n,x). - Peter Bala, Oct 28 2011 T(n,k) = Sum_{i = 0..n-1} (-1)^i*Stirling2(n-1-i,k-1), for n >= 1, k >= 1.
The k-th column o.g.f. is (1/(1+x))*x^k/((1-x)*(1-2*x)*...*(1-(k-1)*x)) (the empty product occurring in the denominator when k = 0 and k = 1 is taken equal to 1).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k >= 0} (k-1)^n*x^k/k!.
Sum_{k = 0..n} binomial(n,k)*R(k,x) = n-th Bell polynomial (n-th row polynomial of A048993). (End) T(n,k) = Sum_{i = 0..n} (-1)^(n-i)*binomial(n,i)*Stirling2(i,k).
T(n,k) = Sum_{i = 0..n} (-2)^(n-i)*binomial(n,i)*Stirling2(i+1,k+1).
Conjecture 1: Sum_{k=0..n} (-1)^(n-k) * T(n, k) * (k+m)! / m! = (m+2)^n for m >= 0.
Conjecture 2: (-1)^n - B(n) = Sum_{k=1..n} (-1)^k * T(n, k) * (k-1)! / (k+1) where B(n) = B(n, 0) is n-th Bernoulli number. (End)
EXAMPLE
The triangle starts with
n=0: 1;
n=1: -1, 1;
n=2: 1, -1, 1;
n=3: -1, 1, 0, 1;
n=4: 1, -1, 1, 2, 1;
n=5: -1, 1, 0, 5, 5, 1;
MAPLE
B:= Matrix(12, 12, shape=triangular[lower], (n, k) -> combinat:-stirling1(n-1, k-1)+(n-1)*combinat:-stirling1(n-2, k-1)):
A:= B^(-1):
T := (n, k) -> add((-1)^(n - i)*binomial(n, i)*Stirling2(i, k), i=0..n):
MATHEMATICA
Table[Sum[(-1)^(n - i)*Binomial[n, i] StirlingS2[i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 14 2019 *)
Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.
+10
15
1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
COMMENTS
In triangles of analogs to Stirling numbers of the second kind, the multipliers of T(n-1,k) in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318, the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers are in the set (1,2,3...). For this sequence here, the multipliers are from A016777. Riordan array [exp(x), (exp(3x)-1)/3]. - Paul Barry, Nov 26 2008 Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498. This triangle is the particular case a = 3, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
FORMULA
T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)). - Paul Barry, Nov 26 2008 Let f(x) = exp(1/3*exp(3*x)+x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012 T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013 T(n,k) = sum {i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*sum {k >= 0} (3*k + 1)^(k + n - 1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)
EXAMPLE
T(5,3) = T(4,2)+7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n=1 as:
1;
1,1;
1,5,1;
1,21,12,1;
1,85,105,22,1;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - Peter Bala, Jan 27 2015
MAPLE
A111577 := proc(n, k) option remember; if k = 1 or k = n then 1; else procname(n-1, k-1)+(3*k-2)*procname(n-1, k) ; fi; end:
MATHEMATICA
T[_, 1] = 1; T[n_, n_] = 1;
T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.
+10
13
1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
COMMENTS
This is a generalization of the unsigned Stirling1 triangle A132393. In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/ A002209(k+1). This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018
REFERENCES
Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
FORMULA
Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}_{n-k}(a_0,a_1,...,a_{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393. Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/ A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 ...
O: 1
1: 1 1
2: 4 5 1
3: 28 39 12 1
4: 280 418 159 22 1
5: 3640 5714 2485 445 35 1
6: 58240 95064 45474 9605 1005 51 1
7: 1106560 1864456 959070 227969 28700 1974 70 1
8: 24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
MATHEMATICA
T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[_, -1] = 0; T[0, 0] = 1; T[n_, k_] /; n<k = 0;
CROSSREFS
S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively. |S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
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