%I #61 Mar 14 2024 15:43:32
%S 1,2,3,4,21,9,8,117,135,27,16,609,1431,702,81,32,3093,13275,12015,
%T 3240,243,64,15561,115479,171990,81405,13851,729,128,77997,970515,
%U 2238327,1655640,479682,56133,2187,256,390369,7998111,27533142,29893941,13121514,2561706
%N 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.
%C The definition of the Stirling-Frobenius subset numbers of order m is in A225468.
%C From _Wolfdieter Lang_, Apr 09 2017: (Start)
%C 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].
%C 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).
%C The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(2/3)) and gives 2*A284862/A284863.
%C The first column k sequences divided by 3^k are A000079, A016127, A016297, A025999. For the e.g.f.s and o.g.f.s see below.
%C The row sums give A284864. The alternating row sums give A284865.
%C 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.
%C 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)
%H Vincenzo Librandi, <a href="/A225466/b225466.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>
%H Shi-Mei Ma, Toufik Mansour, and Matthias Schork, <a href="http://arxiv.org/abs/1308.0169">Normal ordering problem and the extensions of the Stirling grammar</a>, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv:1308.0169 [math.CO], 2013, p. 12.
%F 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.
%F For a recurrence see the Maple program.
%F T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A000244.
%F From _Wolfdieter Lang_, Apr 09 2017: (Start)
%F T(n, k) = Sum_{j=0..k} binomial(k,j)*(-1)^(j-k)*(2 + 3*j)^n/k!, 0 <= k <= n.
%F E.g.f. of triangle: exp(2*z)*exp(x*(exp(3*z)-1)) (Sheffer type).
%F E.g.f. for sequence of column k is exp(2*x)*((exp(3*x) - 1)^k)/k! (Sheffer property).
%F O.g.f. for sequence of column k is 3^k*x^k/Product_{j=0..k} (1 - (2+3*j)*x).
%F 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.
%F 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).
%F 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.
%F (End)
%F 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
%e [n\k][ 0, 1, 2, 3, 4, 5, 6, 7]
%e [0] 1,
%e [1] 2, 3,
%e [2] 4, 21, 9,
%e [3] 8, 117, 135, 27,
%e [4] 16, 609, 1431, 702, 81,
%e [5] 32, 3093, 13275, 12015, 3240, 243,
%e [6] 64, 15561, 115479, 171990, 81405, 13851, 729,
%e [7] 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187.
%e ...
%e From _Wolfdieter Lang_, Aug 11 2017: (Start)
%e Recurrence (see the Maple program): T(4, 2) = 3*T(3, 1) + (3*2+2)*T(3, 2) = 3*117 + 8*135 = 1431.
%e 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)
%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, 3), 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, 3], {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 A225466(n): return SF_SS(n, k, 3)
%o (PARI) T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);
%o for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ _Indranil Ghosh_, Apr 10 2017
%o (Python)
%o from sympy import binomial, factorial
%o def T(n, k): return sum(binomial(k, j)*(-1)**(j - k)*(2 + 3*j)**n//factorial(k) for j in range(k + 1))
%o for n in range(11): print([T(n, k) for k in range(n + 1)]) # _Indranil Ghosh_, Apr 10 2017
%Y Cf. A048993 (m=1), A154537 (m=2), A225467 (m=4), A225468.
%Y Cf. A000079, A000244, A005057, A016127, A016297, A025999, A006232/A006233, A225117, A225472, A225468, A282629, A284862/A284863, A284864, A284865.
%K nonn,easy,tabl
%O 0,2
%A _Peter Luschny_, May 08 2013