A225467 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.

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

Offset: 0

Author

Peter Luschny, May 08 2013

Keywords

• nonn
• easy
• tabl

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

Examples

			[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.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
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 k = 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)
		

Links

• Vincenzo Librandi, Rows n = 0..50, flattened
• Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
• Peter Luschny, Eulerian polynomials.
• Peter Luschny, The Stirling-Frobenius numbers.

Crossrefs

Cf. A048993 (m=1), A154537 (m=2), A225466 (m=3). A225469 (scaled).

Cf. Columns: A000244, 4*A016138, 16*A018054. A225118.

Programs

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 *)

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

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)

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, 0) ~ A000244; T(n, 1) ~ A190541.

T(n, n) ~ A000302; T(n, n-1) ~ A002700.

From Wolfdieter Lang, Apr 13 2017: (Start)

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 k: 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