%I #106 Feb 16 2025 08:33:25
%S 1,1,5,33,285,3081,40005,606033,10491885,204343641,4422082005,
%T 105265315233,2733583519485,76902684021801,2329889536156005,
%U 75629701786875633,2618654297178083085,96336948993312237561,3752590641305604502005,154294551397830418471233,6677999524135208461382685
%N a(n) = (3/4) * Sum_{k>=0} (3*k)^n/4^k.
%H Alois P. Heinz, <a href="/A255927/b255927.txt">Table of n, a(n) for n = 0..400</a>
%H P. Blasiak, K. A. Penson and A. I. Solomon, <a href="http://arxiv.org/abs/quant-ph/0303030">Dobinski-type relations and the Log-normal distribution</a>, arXiv:quant-ph/0303030, 2003.
%H P. Blasiak, K. A. Penson and A. I. Solomon, <a href="http://dx.doi.org/10.1088/0305-4470/36/18/101">Dobinski-type relations and the Log-normal distribution</a>, J. Phys. A: Math. Gen. 36, (2003), L273.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/LerchTranscendent.html">Lerch Transcendent</a>
%F a(n) = Sum_{k>=0} Stirling2(n,k)*k!*3^(n-k).
%F E.g.f.: 3/(4-exp(3*x)).
%F Special values of the generalized hypergeometric function n_F_(n-1):
%F a(n) = (3^(n+1)/16) * hypergeom([2,2,..2],[1,1,..1],1/4), where the sequence in the first square bracket ("upper" parameters) has n elements all equal to 2 whereas the sequence in the second square bracket ("lower" parameters) has n-1 elements all equal to 1.
%F Example: a(5) = 729 * hypergeom([2,2,2,2,2],[1,1,1,1],1/4)/16 = 3081.
%F a(n) is the n-th moment of the discrete weight function W(x) = (3/4)*sum(k>=0, Dirac(x-3*k)/4^k), n>=1.
%F a(n) ~ n! * 3^(n+1) / ((log(2))^(n+1) * 2^(n+3)). - _Vaclav Kotesovec_, Jul 09 2018
%F G.f.: Sum_{j>=0} j!*x^j / Product_{k=1..j} (1 - 3*k*x). - _Ilya Gutkovskiy_, Apr 04 2019
%F a(n) = A_{4}(n) where A_{n}(x) are the Eulerian polynomials as defined in A326323. - _Peter Luschny_, Jun 27 2019
%e a(5) = 729*hypergeom([2,2,2,2,2],[1,1,1,1],1/4)/16 = 3081.
%p S:= series(3/(4-exp(3*x)), x, 51):
%p seq(coeff(S,x,n)*n!, n=0..50); # _Robert Israel_, Sep 03 2015
%p seq(add(combinat:-eulerian1(n,k)*4^k, k=0..n), n=0..20); # _Peter Luschny_, Jun 27 2019
%t a[n_] := 3^(n+1)/4 HurwitzLerchPhi[1/4, -n, 0];
%t Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Sep 18 2018 *)
%t Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k] 4^k, {k, 0, n}], {n, 0, 20}] (* _Jean-François Alcover_, Jul 13 2019, after _Peter Luschny_ *)
%o (PARI) a(n) = sum(k=0, n, stirling(n,k,2)*k!*3^(n-k)); \\ _Michel Marcus_, Sep 03 2015
%Y Cf. A000670, A094418, A004123 A032033, A094417, A122704, A326323.
%K nonn,changed
%O 0,3
%A _Karol A. Penson_, Sep 03 2015
%E a(0)=1 prepended by _Alois P. Heinz_, Sep 18 2018