A249941 E.g.f.: Sum_{n>=0} exp(n^3*x) / 2^(n+1).
1, 13, 4683, 7087261, 28091567595, 230283190977853, 3385534663256845323, 81124824998504073881821, 2958279121074145472650648875, 155897763918621623249276226253693, 11403568794011880483742464196184901963, 1120959742203056268267494209293006882589981
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + 13*x + 4683*x^2/2! + 7087261*x^3/3! + 28091567595*x^4/4! +... where the e.g.f. equals the infinite series: A(x) = 1/2 + exp(x)/2^2 + exp(8*x)/2^3 + exp(27*x)/2^4 + exp(64*x)/2^5 + exp(125*x)/2^6 + exp(216*x)/2^7 + exp(343*x)/2^8 +...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..141
Programs
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Mathematica
Table[Sum[k! * StirlingS2[3*n, k],{k,0,3*n}],{n,0,20}] (* Vaclav Kotesovec, May 04 2015 *) Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; a[n_] := Fubini[3n, 1]; a[0] = 1; Table[a[n], {n, 0, 11}] (* Jean-François Alcover, Mar 30 2016 *) -
PARI
/* E.g.f.: Sum_{n>=0} exp(n^3*x)/2^(n+1) */ \p100 \\ set precision {a(n) = round( n!*polcoeff(sum(m=0,600,exp(m^3*x +x*O(x^n))/2^(m+1)*1.),n) )} for(n=0,20,print1(a(n),", ")) -
PARI
/* Formula for a(n): */ {a(n) = sum(k=0, 3*n, k! * stirling(3*n, k, 2))} for(n=0, 20, print1(a(n), ", ")) -
PARI
/* Formula for a(n): */ {a(n) = if(n==0,1, sum(k=1,(3*n+1)\2, (2*k-1)! * stirling(3*n+1, 2*k, 2)))} for(n=0,20,print1(a(n),", "))
Formula
a(n) = Sum_{k=0..3*n} k! * Stirling2(3*n, k) for n>=0.
a(n) = Sum_{k=1..[(3*n+1)/2]} (2*k-1)! * Stirling2(3*n+1, 2*k) for n>0 with a(0)=1.
a(n) ~ (3*n)! / (2 * (log(2))^(3*n+1)). - Vaclav Kotesovec, May 04 2015
a(n) = Sum_{k>=0} k^(3*n) / 2^(k + 1). - Ilya Gutkovskiy, Dec 19 2019
Comments