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A225467
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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.
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9
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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
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
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EXAMPLE
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[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)
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MAPLE
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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);
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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