| Displaying 1-10 of 10 results found.
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Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.
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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.
Triangle read by rows, 3^k*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
11
1, 2, 3, 4, 21, 9, 8, 117, 135, 27, 16, 609, 1431, 702, 81, 32, 3093, 13275, 12015, 3240, 243, 64, 15561, 115479, 171990, 81405, 13851, 729, 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187, 256, 390369, 7998111, 27533142, 29893941, 13121514, 2561706
COMMENTS
The definition of the Stirling-Frobenius subset numbers of order m is in A225468. This is the Sheffer triangle (exp(2*x), exp(3*x) - 1), denoted by S2[3,2]. See also A282629 for S2[3,1]. The stirling2 triangle A048993 is in this notation denoted by 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). For a- and z-sequences for Sheffer triangles see the W. Lang link under A006232, also with references). The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(2/3)) and gives 2* A284862/ A284863. This triangle appears in the o.g.f. G(n, x) of the sequence {(2 + 3*m)^n}_{m>=0}, as G(n, x) = Sum_{k=0..n} T(n, k)*k!*x^k/(1-x)^(k+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0} (2 + 3*m)^n t^m/m! = exp(t)*Sum_{k=0..n} T(n, k)*t^k.
The corresponding Eulerian number triangle is A225117(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, m)*m!, 0 <= k <= n. (End)
FORMULA
T(n, k) = (1/k!)*Sum_{j=0..n} binomial(j, n-k)*A_3(n, j) where A_m(n, j) are the generalized Eulerian numbers A225117. For a recurrence see the Maple program.
T(n, k) = Sum_{j=0..k} binomial(k,j)*(-1)^(j-k)*(2 + 3*j)^n/k!, 0 <= k <= n.
E.g.f. of triangle: exp(2*z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column k is exp(2*x)*((exp(3*x) - 1)^k)/k! (Sheffer property).
O.g.f. for sequence of column k is 3^k*x^k/Product_{j=0..k} (1 - (2+3*j)*x).
A nontrivial recurrence for the column m=0 entries T(n, 0) = 2^n from the z-sequence given above: T(n,0) = n*Sum_{k=0..n-1} z(k)*T(n-1,k), n >= 1, T(0, 0) = 1.
The corresponding recurrence for columns k >= 1 from the a-sequence is T(n, k) = (n/k)* Sum_{j=0..n-k} binomial(k-1+j, k-1)*a(j)*T(n-1, k-1+j).
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+2) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
(End)
Boas-Buck recurrence for column sequence m: T(n, k) = (1/(n - m))*[(n/2)*(4 + 3*m)*T(n-1, k) + m* Sum_{p=m..n-2} binomial(n, p)(-3)^(n-p)*Bernoulli(n-p)*T(p, k)], for n > k >= 0, with input T(k, k) = 3^k. See a comment and references in A282629, An example is given below. - Wolfdieter Lang, Aug 11 2017
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6, 7]
[0] 1,
[1] 2, 3,
[2] 4, 21, 9,
[3] 8, 117, 135, 27,
[4] 16, 609, 1431, 702, 81,
[5] 32, 3093, 13275, 12015, 3240, 243,
[6] 64, 15561, 115479, 171990, 81405, 13851, 729,
[7] 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187.
...
Recurrence (see the Maple program): T(4, 2) = 3*T(3, 1) + (3*2+2)*T(3, 2) = 3*117 + 8*135 = 1431.
Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) = (1/2)*[2*(4 + 3*2)*T(3, 2) + 2*6*(-3)^2*Bernoulli(2)*T(2, 2))] = (1/2)*(20*135 + 12*9*(1/6)*9) = 1431. (End)
MAPLE
SF_SS := 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;
m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:
seq(print(seq(SF_SS(n, k, 3), k=0..n)), n=0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + (m*k+1)*EulerianNumber(n-1, k, m)
def SF_SS(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n-k) for j in (0..n))/ factorial(k)
def A225466(n): return SF_SS(n, k, 3) (PARI) T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 10 2017 (Python)
from sympy import binomial, factorial
def T(n, k): return sum(binomial(k, j)*(-1)**(j - k)*(2 + 3*j)**n//factorial(k) for j in range(k + 1))
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017
CROSSREFS
Cf. A000079, A000244, A005057, A016127, A016297, A025999, A006232/ A006233, A225117, A225472, A225468, A282629, A284862/ A284863, A284864, A284865.
Triangle read by rows, S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
10
1, 3, 1, 9, 10, 1, 27, 79, 21, 1, 81, 580, 310, 36, 1, 243, 4141, 3990, 850, 55, 1, 729, 29230, 48031, 16740, 1895, 78, 1, 2187, 205339, 557571, 299131, 52745, 3689, 105, 1, 6561, 1439560, 6338620, 5044536, 1301286, 137592, 6524, 136, 1
COMMENTS
The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers (see the links for details).
This is the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x -1))). See the P. Bala link where this is called exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017
FORMULA
T(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_4(n, j)) / (4^k*k!) where A_4(n,j) = A225118. For a recurrence see the Maple program.
E.g.f.: exp(3*z)*exp((x/4)*(exp(4*z -1))). Sheffer triangle (see a comment above).
E.g.f. column k: exp(3*x)*(exp(4*x) -1)^k/(4^k*k!), k >= 0 (Sheffer property).
O.g.f. column k: x^m/Product_{j=0..k} (1 - (3+4*j)*x), k >= 0.
(End)
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6]
[0] 1,
[1] 3, 1,
[2] 9, 10, 1,
[3] 27, 79, 21, 1,
[4] 81, 580, 310, 36, 1,
[5] 243, 4141, 3990, 850, 55, 1,
[6] 729, 29230, 48031, 16740, 1895, 78, 1.
MAPLE
SF_S := 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_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
seq(print(seq(SF_S(n, k, 4), k=0..n)), n = 0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + (m*k+1)*EulerianNumber(n-1, k, m)
def SF_S(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/(factorial(k)*m^k)
for n in (0..6): [SF_S(n, k, 4) for k in (0..n)]
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, T(n, k) = 4^k*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
9
1, 3, 4, 9, 40, 16, 27, 316, 336, 64, 81, 2320, 4960, 2304, 256, 243, 16564, 63840, 54400, 14080, 1024, 729, 116920, 768496, 1071360, 485120, 79872, 4096, 2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384, 6561, 5758240, 101417920, 322850304
COMMENTS
The definition of the Stirling-Frobenius subset numbers of order m is in A225468. This is the Sheffer triangle (exp(3*x), exp(4*x) - 1). See also the P. Bala link under A225469, the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x) - 1)), which is named there exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017
FORMULA
T(n, k) = (1/k!)*sum_{j=0..n} binomial(j, n-k)*A_4(n, j) where A_m(n, j) are the generalized Eulerian numbers A225118. For a recurrence see the Maple program.
T(n, k) = Sum_{m=0..k} binomial(k,m)*(-1)^(m-k)*((3+4*m)^n)/k!, 0 <= k <= n.
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^(n-k)*4^k*Stirling2(k, m), 0 <= m <= n. E.g.f. exp(3*z)*exp(x*(exp(4*z) - 1)) (Sheffer property).
E.g.f. column k: exp(3*x)*((exp(4*x) - 1)^k)/k!, k >= 0.
O.g.f. column k: (4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
Boas-Buck recurrence for column sequence m: T(n, k) = (1/(n - k))*[(n/2)*(6 + 4*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)(-4)^(n-p)*Bernoulli(n-p)*T(p, k)], for n > k >= 0, with input T(k, k) = 4^k. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6, 7]
[0] 1,
[1] 3, 4,
[2] 9, 40, 16,
[3] 27, 316, 336, 64,
[4] 81, 2320, 4960, 2304, 256,
[5] 243, 16564, 63840, 54400, 14080, 1024,
[6] 729, 116920, 768496, 1071360, 485120, 79872, 4096,
[7] 2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384.
...
Recurrence (see the Maple program): T(4, 2) = 4*T(3, 1) + (4*2+3)*T(3, 2) = 4*316 + 11*336 = 4960.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) =(1/2)*[2*(6 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2))] = (1/2)*(28*336 + 12*16*(1/6)*16) = 4960. (End)
MAPLE
SF_SS := 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;
m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:
seq(print(seq(SF_SS(n, k, 4), k=0..n)), n=0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+(m*k+1)*EulerianNumber(n-1, k, m)
def SF_SS(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n-k) for j in (0..n))/factorial(k)
def A225467(n): return SF_SS(n, k, 4) (PARI) T(n, k) = sum(m=0, k, binomial(k, m)*(-1)^(m - k)*((3 + 4*m)^n)/k!);
for(n = 0, 10, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 13 2017 (Python)
from sympy import binomial, factorial
def T(n, k): return sum(binomial(k, m)*(-1)**(m - k)*((3 + 4*m)**n)//factorial(k) for m in range(k + 1))
for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 13 2017
Triangle read by rows, k!*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
5
1, 2, 3, 4, 21, 18, 8, 117, 270, 162, 16, 609, 2862, 4212, 1944, 32, 3093, 26550, 72090, 77760, 29160, 64, 15561, 230958, 1031940, 1953720, 1662120, 524880, 128, 77997, 1941030, 13429962, 39735360, 57561840, 40415760, 11022480, 256, 390369, 15996222, 165198852
COMMENTS
The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
FORMULA
For a recurrence see the Maple program.
E.g.f. for sequence of column k: exp(2*x)*(exp(3*x) - 1)^k, k >= 0. From the Sheffer triangle S2[3,2] = A225466 with column k multiplied with k!. O.g.f. for sequence of column k is 3^k*k!*x^k/Product_{j=0..k} (1 - (2+3*j)*x), k >= 0.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k, j)*(2+3*j)^n, 0 <= k <= n.
Three term recurrence (see the Maple program): T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = 3*k*T(n-1, k-1) + (2 + 3*k)*T(n-1, k) for n >= 1, k=0..n.
For the column scaled triangle (with diagonal 1s) see A225468, and the Bala link with (a,b,c) = (3,0,2), where Sheffer triangles are called exponential Riordan triangles. (End)
The e.g.f. of the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k is exp(2*z)/(1 - x*(exp(3*z) - 1)). - Wolfdieter Lang, Jul 12 2017
EXAMPLE
[n\k][0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 2, 3,
[2] 4, 21, 18,
[3] 8, 117, 270, 162,
[4] 16, 609, 2862, 4212, 1944,
[5] 32, 3093, 26550, 72090, 77760, 29160,
[6] 64, 15561, 230958, 1031940, 1953720, 1662120, 524880.
MAPLE
SF_SO := 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;
m*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
seq(print(seq(SF_SO(n, k, 3), k=0..n)), n = 0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+ (m*k+1)*EulerianNumber(n-1, k, m)
def SF_SO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
for n in (0..6): [SF_SO(n, k, 3) for k in (0..n)]
Triangle read by rows, k!*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
5
1, 3, 4, 9, 40, 32, 27, 316, 672, 384, 81, 2320, 9920, 13824, 6144, 243, 16564, 127680, 326400, 337920, 122880, 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120, 2187, 821356, 17842272, 114866304, 324065280, 453304320, 309657600, 82575360, 6561
COMMENTS
The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
FORMULA
For a recurrence see the Maple program.
T(n, k) = Sum_{m=0..n} binomial(k,m)*(-1)^(k-m)*(3 + 4*m)^n.
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, k) = 0 if n < k and T(n, k) =
4*k*T(n-1, k-1) + (3 + 4*k)*T(n-1, k) for n >= 1, k = 0..n (see the Maple program).
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, k)*x^k: exp(3*z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column k: exp(3*x)*(exp(4*x) - 1)^k, k >= 0.
O.g.f. column k: k!*(4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
EXAMPLE
[n\k][0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 3, 4,
[2] 9, 40, 32,
[3] 27, 316, 672, 384,
[4] 81, 2320, 9920, 13824, 6144,
[5] 243, 16564, 127680, 326400, 337920, 122880,
[6] 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120.
MAPLE
SF_SO := 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;
m*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
seq(print(seq(SF_SO(n, k, 4), k=0..n)), n = 0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+ (m*k+1)*EulerianNumber(n-1, k, m)
def SF_SO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
for n in (0..6): [SF_SO(n, k, 4) for k in (0..n)]
Triangle read by rows. A generalization of unsigned Lah numbers, called L[3,2].
+10
2
1, 4, 1, 28, 14, 1, 280, 210, 30, 1, 3640, 3640, 780, 52, 1, 58240, 72800, 20800, 2080, 80, 1, 1106560, 1659840, 592800, 79040, 4560, 114, 1, 24344320, 42602560, 18258240, 3043040, 234080, 8778, 154, 1, 608608000, 1217216000, 608608000, 121721600, 11704000, 585200, 15400, 200, 1, 17041024000, 38342304000, 21909888000, 5112307200, 589881600, 36867600, 1293600, 25200, 252, 1, 528271744000, 1320679360000, 849008160000, 226402176000, 30477216000, 2285791200, 100254000, 2604000, 39060, 310, 1
COMMENTS
For the general L[d,a] triangles see A286724, also for references. This is the generalized signless Lah number triangle L[3,2], the Sheffer triangle ((1 - 3*t)^(-4/3), t/(1 - 3*t)). It is defined as transition matrix risefac[3,2](x, n) = Sum_{m=0..n} L[3,2](n, m)*fallfac[3,2](x, m), where risefac[3,2](x, n):= Product_{0..n-1} (x + (2 + 3*j)) for n >= 1 and risefac[3,2](x, 0) := 1, and fallfac[3,2](x, n):= Product_{0..n-1} (x - (2 + 3*j)) for n >= 1 and fallfac[3,2](x, 0) := 1.
In matrix notation: L[3,2] = S1phat[3,2]*S2hat[3,2] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A225470 and A225468, respectively. The a- and z-sequences for this Sheffer matrix have e.g.f.s 1 + 3*t and (1 + 3*t)*(1 - (1 + 3*t)^(-4/3))/t, respectively. That is, a = {1, 3, repeat(0)} and z(n) = A290603(n)/ A038500(n+1). See a W. Lang link under A006232 for these types of sequences with a reference, and also the present link, eq. (142). The inverse matrix T^(-1) = L^(-1)[3,2] is Sheffer ((1 + 3*t)^(-4/3), t/(1 + 3*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[3,2](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[3,2](x, m), n >= 0.
REFERENCES
Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.
FORMULA
T(n, m) = L[3,2](n,m) = Sum_{k=m..n} A225470(n, k) * A225468(k, m), 0 <= m <= n. E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 3*t)^(-4/3)*exp(x*t/(1 - 3*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 3*t)^(-4/3)*(t/(1 - 3*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 3*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), from the a-sequence {1, 3 repeat(0)} and the z-sequence given above.
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(3*n - 1)*T(n-1, m) - 3*(n-1)*(3*n - 2)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 3*D_x)) * R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-3)^k*{D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(4 + x)*1 + 6*(2 + x)*D_x + 3^2*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(4 + 3*m)*Sum_{p=0..n-1-m} 3^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
EXAMPLE
The triangle T(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 ...
0: 1
1: 4 1
2: 28 14 1
3: 280 210 30 1
4: 3640 3640 780 52 1
5: 58240 72800 20800 2080 80 1
6: 1106560 1659840 592800 79040 4560 114 1
7: 24344320 42602560 18258240 3043040 234080 8778 154 1
8: 608608000 1217216000 608608000 121721600 11704000 585200 15400 200 1
...
n = 9: 17041024000 38342304000 21909888000 5112307200 589881600 36867600 1293600 25200 252 1,
n = 10: 528271744000 1320679360000 849008160000 226402176000 30477216000 2285791200 100254000 2604000 39060 310 1.
...
Recurrence from a-sequence: T(4, 2) = (4/2)*T(3, 1) + 3*4*T(3, 2) = 2*210 + 12*30 = 780.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2) + z(3)*T(3, 3)) = 4*(4* 280 - 2*210 + (28/3)*30 - 70*1) = 3640.
Four term recurrence: T(4, 2) = T(3, 1) + 2*11*T(3, 2) - 3*3*10*T(2, 2) = 210 + 22*30 - 90*1 = 780.
Meixner type identity for n = 2: (D_x - 3*(D_x)^2)*(28 + 14*x + x^2) = (14 + 2*x) - 3*2 = 2*(4 + x).
Sheffer recurrence for R(3, x): [(4 + x) + 6*(2 + x)*D_x + 9*x*(D_x)^2] (28 + 14*x + x^2) = (4 + x)*(28 + 14*x + x^2) + 6*(2 + x)*(14 + 2*x) + 9*2*x= 280 + 210*x + 30*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*(4 + 3*2)/2)*(1*30/3! + 3*1/2!) = 780.
CROSSREFS
Cf. A007559(n+1) (column m=0), A225468, A225470, A271703 L[1,0], A286724 L[2,1], A290596, L[3,1], A290603.
Triangle read by rows, k!*2^k*S_2(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
1
1, 1, 1, 1, 4, 2, 1, 13, 18, 6, 1, 40, 116, 96, 24, 1, 121, 660, 1020, 600, 120, 1, 364, 3542, 9120, 9480, 4320, 720, 1, 1093, 18438, 74466, 121800, 94920, 35280, 5040, 1, 3280, 94376, 576576, 1394064, 1653120, 1028160, 322560, 40320, 1, 9841, 478440, 4319160
COMMENTS
The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
FORMULA
T(n, k) = sum_{j=0..n} A_2(n, j)*binomial(j, n-k), where A_2(n, j) are the generalized Eulerian numbers of order m=2.
For a recurrence see the Maple program.
EXAMPLE
[n\k][0, 1, 2, 3, 4, 5 ]
[0] 1,
[1] 1, 1,
[2] 1, 4, 2,
[3] 1, 13, 18, 6,
[4] 1, 40, 116, 96, 24,
[5] 1, 121, 660, 1020, 600, 120.
MAPLE
SF_SSO := 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;
k*SF_SSO(n-1, k-1, m) + (m*(k+1)-1)*SF_SSO(n-1, k, m) end:
seq(print(seq(SF_SSO(n, k, 2), k=0..n)), n = 0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n - k) + m - 1)*EulerianNumber[n - 1, k - 1, m] + (m*k + 1)*EulerianNumber[n - 1, k, m]]); SFSSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n - k], {j, 0, n}]/m^k; Table[ SFSSO[n, k, 2], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+(m*k+1)*EulerianNumber(n-1, k, m)
def SF_SSO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/m^k
for n in (0..6): [SF_SSO(n, k, 2) for k in (0..n)]
CROSSREFS
Alternating row sum ~ A000364 (Euler secant numbers).
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