Displaying 1-10 of 13 results found.
Column k=2 of the triangle A286718; Sheffer ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)).
+20
0
1, 12, 159, 2485, 45474, 959070, 22963996, 616224492, 18331744896, 599061555136, 21339235262784, 823098817664448, 34183157124707200, 1520908498941532800, 72182781516370886400, 3640264913563748243200, 194408478299496756556800, 10961007293837647131724800
COMMENTS
a(n) is, for n >= 1, the total volume of the binomial(n+2, n) rectangular polytopes (hyper-cuboids) built from n orthogonal vectors with lengths of the sides from the set {1 + 3*j | j=0..n+1}. See the formula a(n) = sigma[3,1]^{(n+2)}_n and an example below.
FORMULA
E.g.f.: (d^2/dx^2)((1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^2/2!) = (2*(log(1-3*x))^2 - 15*log(1-3*x) + 9)/(3^2*(1-3*x)^(7/3)).
a(n) = sigma[3,1]^{(n+2)}_n, n >= 0, with the elementary symmetric function sigma[3,1]^{n+2}_n of degree n of the n+2 numbers 1, 4, 7, ..., (1 + 3*(n+1)).
EXAMPLE
a(2) = 159 because sigma[3,1]^{(4)}_2 = 1*(4 + 7 + 10) + 4*(7 + 10) + 7*10 = 159. There are six rectangles (2D rectangular polytopes) built from two orthogonal vectors of different lengths from the set of {1,4,7,10} of total area 159.
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 A286718 (|S1hat[3,1]| generalized Stirling 1), for n >= 0.
+20
0
1, 1, 2, 4, 19, 4, 28, 222, 147, 8, 280, 3194, 4128, 887, 16, 3640, 55024, 113566, 52538, 4835, 32, 58240, 1107336, 3268788, 2562676, 555684, 25167, 64, 1106560, 25526192, 100544412, 117517960, 45415640, 5301150, 128203, 128, 24344320, 663605680, 3325767376, 5352311764, 3189383200, 695714590, 47537320, 646519, 256, 608608000, 19213911360, 118361719296, 248493947496, 208996478388, 72479948400, 9696965250, 410038434, 3245139, 512
COMMENTS
The ordinary generating function (o.g.f.) of the (n+1)-th diagonal sequence of the Sheffer triangle A286718 = ((1 - 3*x)^(-1/3), -log(1 - 3*x)/3), called |S1hat[3,1]|, is GD(3,1;n,x) = P(n, x)/(1 - x)^(2*n+1), with the row polynomials P(n, x) = Sum_{k=0..n} T(n, k)*x^k, n >= 0. For the two parameter Sheffer case |S1hat[d,a]| = ((1 - d*x)^{-a/d}, -log(1 - d*x)/d) (with gcd(d,a) = 1, d >=0, a >= 0, and for d = 1 one takes a = 0) the e.g.f. ED(t, x) of the o.g.f.s {GD(d,a;n,x)}_{n>=0} of the diagonal sequences with elements D(d,a;n,m) = |S1hat[d,a]|(n+m, m) (n=0 for the main diagonal) is of interest. It can be computed via Lagrange's theorem. For the special Sheffer case (1, f(x)) this has been done by P. Bala (see the link). This method can be generalized for Sheffer (g(x), f(x)), as shown in the W. Lang link.
FORMULA
T(n, k) = [x^k] P(n, x) with the numerator polynomials of the o.g.f. GD(n, x) = P(n, x)/(1-x)^(2*n+1) of the (n+1)-th diagonal sequence of the triangle A286718. 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: 4 19 4
3: 28 222 147 8
4: 280 3194 4128 887 16
5: 3640 55024 113566 52538 4835 32
6: 58240 1107336 3268788 2562676 555684 25167 6
7: 1106560 25526192 100544412 117517960 45415640 5301150 128203 128
...
n = 8: 24344320 663605680 3325767376 5352311764 3189383200 695714590 47537320 646519 256,
n = 9: 608608000 19213911360 118361719296 248493947496 208996478388 72479948400 9696965250 410038434 3245139 512.
n = 3: The o.g.f. of the 4th diagonal sequence of A286718, [28, 418, 2485, ...] = A024213(n+1), n >= 0, is P(3, x) = (28 + 222*x + 147*x^2 + 8*x^3)/(1 - 3*x)^7.
Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.
+10
121
1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
COMMENTS
Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938. Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014 Also the Bell transform of the factorial numbers ( A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015 This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017 T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017 The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End) For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019
REFERENCES
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150
FORMULA
T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007 Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008 Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008 a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008 Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008 Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008 E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/ A027642. The e.g.f. for the z-sequence is 0. (End) The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1))_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/ A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017 T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)* A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023 n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
EXAMPLE
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 6, 11, 6, 1;
0, 24, 50, 35, 10, 1;
0, 120, 274, 225, 85, 15, 1;
0, 720, 1764, 1624, 735, 175, 21, 1;
0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
...
---------------------------------------------------
Production matrix is
0, 1
0, 1, 1
0, 1, 2, 1
0, 1, 3, 3, 1
0, 1, 4, 6, 4, 1
0, 1, 5, 10, 10, 5, 1
0, 1, 6, 15, 20, 15, 6, 1
0, 1, 7, 21, 35, 35, 21, 7, 1
...
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - Wolfdieter Lang, May 28 2017 O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - Wolfdieter Lang, Jul 20 2017 Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - Wolfdieter Lang, Aug 11 2017
MAPLE
a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x, n)), x, k), k=0..n) end: # Peter Luschny, Nov 28 2010
MATHEMATICA
p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *) Flatten[Table[Abs[StirlingS1[n, i]], {n, 0, 10}, {i, 0, n}]] (* Harvey P. Dale, Feb 04 2014 *)
PROG
(Maxima) create_list(abs(stirling1(n, k)), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */ (Haskell)
a132393 n k = a132393_tabl !! n !! k
a132393_row n = a132393_tabl !! n
a132393_tabl = map (map abs) a048994_tabl
Triple factorial numbers (3*n-2)!!! with leading 1 added.
(Formerly M3627)
+10
120
1, 1, 4, 28, 280, 3640, 58240, 1106560, 24344320, 608608000, 17041024000, 528271744000, 17961239296000, 664565853952000, 26582634158080000, 1143053268797440000, 52580450364682240000, 2576442067869429760000, 133974987529210347520000, 7368624314106569113600000
COMMENTS
a(n) is the number of increasing quaternary trees on n vertices. (See A001147 for ternary and A000142 for binary trees.) - David Callan, Mar 30 2007 a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 1. - Peter Luschny, Jun 23 2011 a(n) is the number of generalized permutations of length n related to the degenerate Eulerian numbers (see arXiv:2007.13205), cf. A336633. - Orli Herscovici, Jul 28 2020
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = Product_{k=0..n-1} (3*k + 1).
a(n) = (3*n - 2)!!!.
E.g.f.: (1-3*x)^(-1/3).
a(n) ~ sqrt(2*Pi)/Gamma(1/3)*n^(-1/6)*(3*n/e)^n*(1 - (1/36)/n - ...). - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = 3^n*Pochhammer(1/3, n).
a(n) = n!*(1+Sum_{m=1..n} (m/n)*Sum_{k=1..n-m} binomial(k, n-m-k)*(-1/3)^(n-m-k)*binomial(k+n-1, n-1)), n>1. - Vladimir Kruchinin, Aug 09 2010 a(n) = upper left term in M^n, M = a variant of Pascal (1,3) triangle (Cf. A095660); as an infinite square production matrix: 1, 3, 0, 0, 0,...
1, 4, 3, 0, 0,...
1, 5, 7, 3, 0,...
...
a(n+1) = sum of top row terms of M^n. (End)
a(n) = (-2)^n*Sum_{k=0..n} (3/2)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012 G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+1)/( 1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 21 2013 G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013 E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+1) + (k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013 Let D(x) = 1/sqrt(1 - 2*x) be the e.g.f. for the sequence of double factorial numbers A001147. Then the e.g.f. A(x) for the triple factorial numbers satisfies D( Integral_{t=0..x} A(t) dt ) = A(x). Cf. A007696 and A008548. - Peter Bala, Jan 02 2015 O.g.f.: hypergeom([1, 1/3], [], 3*x). - Peter Luschny, Oct 08 2015 a(n) = sigma[3,1]^{(n)}_n, n >= 0, with the elementary symmetric function of degree n in the n numbers 1, 4, 7, ..., 1+3*(n-1), with sigma[3,1]^{(n)}_0 := 1. See the first formula. - Wolfdieter Lang, May 29 2017 a(n) = (-1)^n / A008544(n), 0 = a(n)*(+3*a(n+1) -a(n+2)) +a(n+1)*a(n+1) for all n in Z. - Michael Somos, Sep 30 2018 D-finite with recurrence: a(n) +(-3*n+2)*a(n-1)=0, n>=1. - R. J. Mathar, Feb 14 2020 Sum_{n>=1} 1/a(n) = (e/9)^(1/3) * (Gamma(1/3) - Gamma(1/3, 1/3)). - Amiram Eldar, Jun 29 2020
EXAMPLE
G.f. = 1 + x + 4*x^2 + 28*x^3 + 280*x^4 + 3640*x^5 + 58240*x^6 + ...
a(3) = 28 and a(4) = 280; with top row of M^3 = (28, 117, 108, 27), sum = 280.
MAPLE
# second Maple program:
b:= proc(n) option remember; `if`(n<1, 1, n*b(n-3)) end:
a:= n-> b(3*n-2):
MATHEMATICA
a[ n_] := If[ n < 0, 1 / Product[ k, {k, - 2, 3 n - 1, -3}],
Product[ k, {k, 1, 3 n - 2, 3}]]; (* Michael Somos, Oct 14 2011 *) FoldList[Times, 1, Range[1, 100, 3]] (* Harvey P. Dale, Jul 05 2013 *) Range[0, 19]! CoefficientList[Series[((1 - 3 x)^(-1/3)), {x, 0, 19}], x] (* Vincenzo Librandi, Oct 08 2015 *)
PROG
(Maxima) a(n):=if n=1 then 1 else (n)!*(sum(m/n*sum(binomial(k, n-m-k)*(-1/3)^(n-m-k)* binomial (k+n-1, n-1), k, 1, n-m), m, 1, n)+1); /* Vladimir Kruchinin, Aug 09 2010 */ (PARI) {a(n) = if( n<0, (-1)^n / prod(k=0, -1-n, 3*k + 2), prod(k=0, n-1, 3*k + 1))}; /* Michael Somos, Oct 14 2011 */ (PARI) my(x='x+O('x^33)); Vec(serlaplace((1-3*x)^(-1/3))) /* Joerg Arndt, Apr 24 2011 */ (Sage)
def A007559(n) : return mul(j for j in range(1, 3*n, 3)) (Haskell)
a007559 n = a007559_list !! n
a007559_list = scanl (*) 1 a016777_list
(Magma)
b:= func< n | (n lt 2) select n else (3*n-2)*Self(n-1) >;
(GAP) List([0..20], n-> Product([0..n-1], k-> 3*k+1 )); # G. C. Greubel, Aug 20 2019
CROSSREFS
a(n)= A035469(n, 1), n >= 1, (first column of triangle A035469(n, m)).
Triangle of coefficients in expansion of (x+1)*(x+3)*...*(x + 2n - 1) in rising powers of x.
+10
27
1, 1, 1, 3, 4, 1, 15, 23, 9, 1, 105, 176, 86, 16, 1, 945, 1689, 950, 230, 25, 1, 10395, 19524, 12139, 3480, 505, 36, 1, 135135, 264207, 177331, 57379, 10045, 973, 49, 1, 2027025, 4098240, 2924172, 1038016, 208054, 24640, 1708, 64, 1, 34459425, 71697105, 53809164, 20570444, 4574934, 626934, 53676, 2796, 81, 1
COMMENTS
Exponential Riordan array (1/sqrt(1-2*x), log(1/sqrt(1-2*x))). - Paul Barry, May 09 2011 The o.g.f.s D(d, x) of the column sequences, for d, d >= 0,(d=0 for the main diagonal) are P(d, x)/(1 - x)^(2*d+1), with the row polynomial P(d, x) = Sum_{m=0..d} A288875(d, m)*x^m. See A288875 for details. - Wolfdieter Lang, Jul 21 2017
FORMULA
Triangle T(n, k), read by rows, given by [1, 2, 3, 4, 5, 6, 7, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 20 2005 T(n, k) = Sum_{i=k..n} (-2)^(n-i) * binomial(i, k) * s(n, i) where s(n, k) are signed Stirling numbers of the first kind. - Francis Woodhouse (fwoodhouse(AT)gmail.com), Nov 18 2005
G.f. of row polynomials in y: 1/(1-(x+x*y)/(1-2*x/(1-(3*x+x*y)/(1-4*x/(1-(5*x+x*y)/(1-6*x*y/(1-... (continued fraction). - Paul Barry, Feb 07 2009 T(n, m) = (2*n-1)*T(n-1,m) + T(n-1,m-1) with T(n, 0) = (2*n-1)!! and T(n, n) = 1. - Johannes W. Meijer, Jun 08 2009 E.g.f. of row polynomials in y: (1/sqrt(1-2*x))*exp(-y*log(sqrt(1-2*x))) = exp(-(1+y)*log(sqrt(1-2*x))) = 1/sqrt(1-2*x)^(1+y).
E.g.f. of column m sequence: (1/sqrt(1-2*x))* (-log(sqrt(1-2*x)))^m/m!. For the special Sheffer, also known as exponential Riordan array, see a comment above. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 2^(n-1-p)*(1 + 2*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/ A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 09 2017
EXAMPLE
G.f. for n = 4: (x + 1)*(x + 3)*(x + 5)*(x + 7) = 105 + 176*x + 86*x^2 + 16*x^3 + x^4.
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9
0: 1
1: 1 1
2: 3 4 1
3: 15 23 9 1
4: 105 176 86 16 1
5: 945 1689 950 230 25 1
6: 10395 19524 12139 3480 505 36 1
7: 135135 264207 177331 57379 10045 973 49 1
8: 2027025 4098240 2924172 1038016 208054 24640 1708 64 1
9: 34459425 71697105 53809164 20570444 4574934 626934 53676 2796 81 1
...
row n = 10: 654729075 1396704420 1094071221 444647600 107494190 16486680 1646778 106800 4335 100 1.
O.g.f.s of diagonals d >= 0: D(2, x) = (3 + 8*x + x^2)/(1 - x)^5 generating [3, 23, 86, ...] = A024196(n+1), from the row d=2 entries of A288875 [3, 8, 1]. - Wolfdieter Lang, Jul 21 2017 Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(2*(1+4*(5/12))*T(2,2)/2! + 1*(1 + 4*(1/2))*T(3,2)/3!) = (4!/2)*(8/3*1 + 3*9/3!) = 86. - Wolfdieter Lang, Aug 11 2017
MAPLE
nmax:=8; for n from 0 to nmax do a(n, 0) := doublefactorial(2*n-1) od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (2*n-1)*a(n-1, m) + a(n-1, m-1) od; od: seq(seq(a(n, m), m=0..n), n=0..nmax); # Johannes W. Meijer, Jun 08 2009, revised Nov 25 2012
MATHEMATICA
T[n_, k_] := Sum[(-2)^(n-i) Binomial[i, k] StirlingS1[n, i], {i, k, n}] (* Woodhouse *)
Join[{1}, Flatten[Table[CoefficientList[Expand[Times@@Table[x+i, {i, 1, 2n+1, 2}]], x], {n, 0, 10}]]] (* Harvey P. Dale, Jan 29 2013 *)
CROSSREFS
A161198 is a scaled triangle version and A109692 is a transposed triangle version.
Sheffer triangle (exp(x), exp(3*x) - 1). Named S2[3,1].
+10
22
1, 1, 3, 1, 15, 9, 1, 63, 108, 27, 1, 255, 945, 594, 81, 1, 1023, 7380, 8775, 2835, 243, 1, 4095, 54729, 109890, 63180, 12393, 729, 1, 16383, 395388, 1263087, 1151010, 387828, 51030, 2187, 1, 65535, 2816865, 13817034, 18752391, 9658278, 2133054, 201204, 6561, 1, 262143, 19914660, 146620935, 285232185, 210789621, 69502860, 10825650, 767637, 19683
COMMENTS
For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references). The e.g.f. for the sequence of column m is (Sheffer property) exp(x)*(exp(3*x) - 1)^m/m!.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0]. The a-sequence for this Sheffer triangle has e.g.f. 3*x/log(1+x) and is 3* A006232(n)/ A006233(n) (Cauchy numbers of the first kind). The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)) and is A284857(n) / A284858(n). The triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 3*m)^n}_{m>=0}, as G(n, x) = Sum_{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 3*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225117 with row reversion. The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,m)*x^m of the Sheffer triangle S2[d,a] satisfy, as special polynomials of the Boas-Buck class, the identity (see the reference, and 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*a*R(d,a;n-1,x) - Sum_{k=0..n-1} binomial(n, k+1)*(-d)^(k+1)*Bernoulli(k+1)*E_x*R(d,a;n-1-k,x), with E_x = x*d/dx (Euler operator).
This entails a recurrence for the sequence of column m, for n > m:
T(d,a;n,m) = (1/(n - m))*[(n/2)*(2*a + d*m)*T(d,a;n-1,m) + m*Sum_{p=m..n-2} binomial(n,p)(-d)^(n-p)*Bernoulli(n-p)*T(d,a;p,m)], with input T(d,a;n,n) = d^n. For the present [d,a] = [3,1] case see the formula and example sections below. - Wolfdieter Lang, Aug 09 2017 (End) The inverse of this triangular Sheffer matrix S2[3,1] is S1[3,1] with rational elements S1[3,1](n, k) = (-1)^(n-k)* A286718(n, k)/3^k. - Wolfdieter Lang, Nov 15 2018 Named after the American mathematician Isador Mitchell Sheffer (1901-1992). - Amiram Eldar, Jun 19 2021
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
A nontrivial recurrence for the column m=0 entries T(n, 0) = 1 from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+1) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 3*k)^n/m!, 0 <= m <= n.
E.g.f. of triangle: exp(z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column m is exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
Standard three-term recurrence: T(n, m) = 0 if n < m, T(n,-1) = 0, T(0, 0) = 1, T(n, m) = 3*T(n-1, m-1) + (1+3*m)*T(n-1, m) for n >= 1. From the T(n, m) formula. Compare with the recurrence of S2[3,2] given in A225466. The o.g.f. for sequence of column m is 3^m*x^m/Product_{j=0..m} (1 - (1+3*j)*x). (End)
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^k*Stirling2(k, m), 0 <= m <= n. - Wolfdieter Lang, Apr 13 2017 Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*[(n/2)*(2 + 3*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n,p)(-3)^(n-p)*Bernoulli(n-p)*T(p, m)], for n > m >= 0, with input T(m, m) = 3^m. See a comment above. - Wolfdieter Lang, Aug 09 2017
EXAMPLE
The triangle T(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9
0: 1
1: 1 3
2: 1 15 9
3: 1 63 108 27
4: 1 255 945 594 81
5: 1 1023 7380 8775 2835 243
6: 1 4095 54729 109890 63180 12393 729
7: 1 16383 395388 1263087 1151010 387828 51030 2187
8: 1 65535 2816865 13817034 18752391 9658278 2133054 201204 6561
9: 1 262143 19914660 146620935 285232185 210789621 69502860 10825650 767637 19683
...
------------------------------------------------------------------------------------
Nontrivial recurrence for m=0 column from z-sequence:T(4,0) = 4*(1*1 + 63*(-1/6) + 108*(11/54) + 27*(-49/108)) = 1.
Recurrence for m=2 column from a-sequence: T(4, 2) = (4/2)*(1*63*3 + 2*108*(3/2) + 3*27*(-3/6)) = 945.
Recurrence for row polynomial R(3, x) (Meixner type): ((3*x + 1) + 3*x*d_x)*(1 + 15*x + 9*x^2) = 1 + 63*x + 108*x^2 + 27*x^3.
E.g.f. and o.g.f. of n = 1 powers {(1 + 3*m)^1}_{m>=0} A016777: E(1, x) = exp(x) * (T(1, 0) + T(1, 1)*x) = exp(x)*(1+3*x). O.g.f.: G(1, x) = T(1, 0)*0!/(1-x) + T(1, 1)*1!*x/(1-x)^2 = (1+2*x)/(1-x)^2. Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) =(1/2)*[2*(2 + 3*2)*T(3, 2) + 2*6*(-3)^2*bernoulli(2)*T(2, 2))] = (1/2)*(16*108 + 12*9*(1/6)*9) = 945. - Wolfdieter Lang, Aug 09 2017
MATHEMATICA
Table[Sum[Binomial[m, k] (-1)^(k - m) (1 + 3 k)^n/m!, {k, 0, m}], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 08 2017 *)
PROG
(PARI) T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
for(n=0, 9, for(m=0, n, print1(T(n, m), ", "); ); print(); ) \\ Indranil Ghosh, Apr 08 2017
CROSSREFS
Cf. A000012, A000244, A006232/ A006233, A016777, A024036, A111577, A225117, A225466, A284857, A284858, A284859, A284860, A284861, A286718.
2nd elementary symmetric function of first n+1 positive integers congruent to 1 mod 3.
+10
10
4, 39, 159, 445, 1005, 1974, 3514, 5814, 9090, 13585, 19569, 27339, 37219, 49560, 64740, 83164, 105264, 131499, 162355, 198345, 240009, 287914, 342654, 404850, 475150, 554229, 642789, 741559, 851295, 972780, 1106824, 1254264, 1415964, 1592815, 1785735
FORMULA
a(n) = n*(n+1)*(9*n^2+9*n-2)/8.
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Clark Kimberling, Aug 18 2012 E.g.f.: exp(x)*x*(32+124*x+72*x^2+9*x^3)/8 = exp(x)*x*(2 + x)*(16 + 54*x + 9*x^2)/8.
a(n) = A286718(n+1, n-1), n >= 1. (End)
MATHEMATICA
Table[n(n+1)(9n^2+9n-2)/8, {n, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {4, 39, 159, 445, 1005}, 40] (* Harvey P. Dale, Oct 16 2023 *)
Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.
+10
10
1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1
COMMENTS
The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468. Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... ( A007494)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015
FORMULA
For a recurrence see the Maple program.
This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(3,0,2)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.
E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+2)*R(n-1,x+3), with R(0, x) = 1.
R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End) Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/ A002209(k+1), beginning {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6]
[0] 1,
[1] 2, 1,
[2] 10, 7, 1,
[3] 80, 66, 15, 1,
[4] 880, 806, 231, 26, 1,
[5] 12320, 12164, 4040, 595, 40, 1,
[6] 209440, 219108, 80844, 14155, 1275, 57, 1.
...
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)
MAPLE
SF_C := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or k < 0 then return(0) fi;
SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);
MATHEMATICA
SFC[0, 0, _] = 1; SFC[n_, k_, _] /; (k > n || k < 0) = 0; SFC[n_, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)
Triangle read by rows, s_4(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.
+10
7
1, 3, 1, 21, 10, 1, 231, 131, 21, 1, 3465, 2196, 446, 36, 1, 65835, 45189, 10670, 1130, 55, 1, 1514205, 1105182, 290599, 36660, 2395, 78, 1, 40883535, 31354119, 8951355, 1280419, 101325, 4501, 105, 1, 1267389585, 1012861224, 308846124, 48644344, 4421494, 240856, 7756, 136, 1
COMMENTS
The Stirling-Frobenius cycle numbers are defined in A225470. Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, 19, 20, ... ( A014601)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015
FORMULA
For a recurrence see the Sage program.
This is the Sheffer triangle (1/(1 - 4*x)^{-3/4}, -(1/4)*log(1-4*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(4,0,3)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 4*z)^{-(3+x)/4}.
E.g.f. of column k: (1-4*x)^(-3/4)*(-(1/4)*log(1-4*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+3)*R(n-1,x+4), with R(0, x) = 1.
R(n, x) = risefac(4,3;x,n) := Product_{j=0..(n-1)} (x + (3 + 4*j)). (See the P. Bala link, eq. (16) for the signed s_{4,0,3} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*3^(n-k-j)*4^j, with S1p(n, m) = A132393(n, m). T(n, k) = sigma[4,3]^{(n)}_{n-k}, with the elementary symmetric functions sigma[4,3]^{(n)}_m of degree m in the n numbers 3, 7, 11, ..., 3+4*(n-1), with sigma[4,3]^{(n)}_0 := 1. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(3 + 8*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/ A002209(k+1), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 3, 1,
[2] 21, 10, 1,
[3] 231, 131, 21, 1,
[4] 3465, 2196, 446, 36, 1,
[5] 65835, 45189, 10670, 1130, 55, 1,
[6] 1514205, 1105182, 290599, 36660, 2395, 78, 1.
...
Recurrence: T(4, 2) = T(3, 1) + (4*4 - 1)*T(3, 2) = 131 +15*21 = 446.
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(4*(3+8*(5/12)) *T(2, 2)/2! + 1*(3 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(4*(19/3)/2 + 7*21/3!) = 446.
(End)
MATHEMATICA
T[0, 0] = 1; T[n_, k_] := Sum[Binomial[n - j, k]*Abs[StirlingS1[n, n - j]]* 3^(n - k - j)*4^j, {j, 0, n - k}];
PROG
(Sage)
@CachedFunction
def SF_C(n, k, m):
if k > n or k < 0 : return 0
if n == 0 and k == 0: return 1
return SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m)
for n in (0..8): [SF_C(n, k, 4) for k in (0..n)]
a(n) = 3rd elementary symmetric function of first n+2 positive integers congruent to 1 mod 3.
+10
6
28, 418, 2485, 9605, 28700, 72128, 159978, 322770, 604560, 1066450, 1790503, 2884063, 4484480, 6764240, 9936500, 14261028, 20050548, 27677490, 37581145, 50275225, 66355828, 86509808, 111523550, 142292150, 179829000, 225275778
FORMULA
a(n) = n*(n+1)*(n+2)*(3n+5)*(9n^2+21*n-2)/48.
G.f. -x*(28+222*x+147*x^2+8*x^3) / (x-1)^7 . - R. J. Mathar, Oct 08 2011 E.g.f.: x*exp(x)*(1344 + 8688*x + 10520*x^2 + 4122*x^3 + 594*x^4 + 27*x^5)/48.
a(n) = A286718(n+2, n-1), n >= 1. (End)
PROG
(Magma) [n*(n+1)*(n+2)*(3*n+5)*(9*n^2+21*n-2)/48: n in [1..30]]; // Vincenzo Librandi, Oct 10 2011
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