A225467 - OEIS

%I #45 Mar 14 2024 15:35:48

%S 1,3,4,9,40,16,27,316,336,64,81,2320,4960,2304,256,243,16564,63840,

%T 54400,14080,1024,729,116920,768496,1071360,485120,79872,4096,2187,

%U 821356,8921136,19144384,13502720,3777536,430080,16384,6561,5758240,101417920,322850304

%N 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.

%C The definition of the Stirling-Frobenius subset numbers of order m is in A225468.

%C 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

%H Vincenzo Librandi, <a href="/A225467/b225467.txt">Rows n = 0..50, flattened</a>

%H Paweł Hitczenko, <a href="https://arxiv.org/abs/2403.03422">A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality</a>, arXiv:2403.03422 [math.CO], 2024. See p. 9.

%H Peter Luschny, <a href="http://www.luschny.de/math/euler/GeneralizedEulerianPolynomials.html">Eulerian polynomials.</a>

%H Peter Luschny, <a href="http://www.luschny.de/math/euler/StirlingFrobeniusNumbers.html">The Stirling-Frobenius numbers.</a>

%F 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.

%F For a recurrence see the Maple program.

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

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

%F From _Wolfdieter Lang_, Apr 13 2017: (Start)

%F T(n, k) = Sum_{m=0..k} binomial(k,m)*(-1)^(m-k)*((3+4*m)^n)/k!, 0 <= k <= n.

%F 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.

%F E.g.f. exp(3*z)*exp(x*(exp(4*z) - 1)) (Sheffer property).

%F E.g.f. column k: exp(3*x)*((exp(4*x) - 1)^k)/k!, k >= 0.

%F O.g.f. column k: (4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.

%F (End)

%F 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

%e [n\k][  0,      1,       2,        3,        4,       5,      6,     7]

%e [0]     1,

%e [1]     3,      4,

%e [2]     9,     40,      16,

%e [3]    27,    316,     336,       64,

%e [4]    81,   2320,    4960,     2304,      256,

%e [5]   243,  16564,   63840,    54400,    14080,    1024,

%e [6]   729, 116920,  768496,  1071360,   485120,   79872,   4096,

%e [7]  2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384.

%e ...

%e From _Wolfdieter Lang_, Aug 11 2017: (Start)

%e Recurrence (see the Maple program): T(4, 2) = 4*T(3, 1) + (4*2+3)*T(3, 2) = 4*316 + 11*336 = 4960.

%e 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)

%p SF_SS := proc(n, k, m) option remember;

%p if n = 0 and k = 0 then return(1) fi;

%p if k > n or  k < 0 then return(0) fi;

%p m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:

%p seq(print(seq(SF_SS(n, k, 4), k=0..n)), n=0..5);

%t 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 *)

%o (Sage)

%o @CachedFunction

%o def EulerianNumber(n, k, m) :

%o     if n == 0: return 1 if k == 0 else 0

%o     return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m)+(m*k+1)*EulerianNumber(n-1,k,m)

%o def SF_SS(n, k, m):

%o     return add(EulerianNumber(n,j,m)*binomial(j,n-k) for j in (0..n))/factorial(k)

%o def A225467(n): return SF_SS(n, k, 4)

%o (PARI) T(n, k) = sum(m=0, k, binomial(k, m)*(-1)^(m - k)*((3 + 4*m)^n)/k!);

%o for(n = 0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ _Indranil Ghosh_, Apr 13 2017

%o (Python)

%o from sympy import binomial, factorial

%o def T(n, k): return sum(binomial(k, m)*(-1)**(m - k)*((3 + 4*m)**n)//factorial(k) for m in range(k + 1))

%o for n in range(11): print([T(n, k) for k in range(n + 1)]) # _Indranil Ghosh_, Apr 13 2017

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

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

%K nonn,easy,tabl

%O 0,2

%A _Peter Luschny_, May 08 2013