A255927 - OEIS

%I #105 Jul 13 2019 09:02:11

%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="http://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

%O 0,3

%A _Karol A. Penson_, Sep 03 2015

%E a(0)=1 prepended by _Alois P. Heinz_, Sep 18 2018