|
%I #31 Jun 21 2024 18:37:16
%S 1,2,18,274,5938,167122,5786418,237857874,11319677618,612109819602,
%T 37069480301618,2485356833141074,182753029186750898,
%U 14623552941626800082,1265002802597606144818,117633823750542653153874,11701922865351577653913778
%N Expansion of e.g.f. 1 / (7 - 6 * exp(x))^(1/3).
%F a(n) = Sum_{k=0..n} 2^k * (Product_{j=0..k-1} (3*j+1)) * Stirling2(n,k) = Sum_{k=0..n} (Product_{j=0..k-1} (6*j+2)) * Stirling2(n,k).
%F a(0) = 1; a(n) = Sum_{k=1..n} (6 - 4*k/n) * binomial(n,k) * a(n-k).
%F a(n) ~ sqrt(2*Pi) * n^(n - 1/6) / (Gamma(1/3) * 7^(1/3) * exp(n) * log(7/6)^(n + 1/3)). - _Vaclav Kotesovec_, Sep 09 2023
%F From _Peter Bala_, Sep 24 2023: (Start)
%F O.g.f. (conjectural): 1/(1 - 2*x/(1 - 7*x/(1 - 8*x/(1 - 14*x/(1 - 14*x/(1 - 21*x/(1 - ... - (6*n - 4)*x/(1 - 7*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type (S-fraction).
%F More generally, it appears that the o.g.f. of the sequence whose e.g.f. is 1/(r+1 - r*exp(s*x))^(m/s) corresponds to the S-fraction 1/(1 - r*m*x/(1 - s*(r+1)*x/(1 - r*(m+s)*x/(1 - 2*s(r+1)*x/(1 - r*(m+2*s)*x/(1 - 3*s(r+1)*x/( 1 - ... ))))))). This is the case r = 6, s = 1, m = 1/3. (End)
%F a(0) = 1; a(n) = 2*a(n-1) - 7*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - _Seiichi Manyama_, Nov 17 2023
%t a[n_] := Sum[Product[6*j + 2, {j, 0, k - 1}] * StirlingS2[n, k], {k, 0, n}]; Array[a, 17, 0] (* _Amiram Eldar_, Sep 11 2023 *)
%t With[{nn=20},CoefficientList[Series[1/(7-6Exp[x])^(1/3),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Jun 21 2024 *)
%o (PARI) a(n) = sum(k=0, n, prod(j=0, k-1, 6*j+2)*stirling(n, k, 2));
%Y Cf. A094419, A346985, A354252, A365556, A365557.
%K nonn
%O 0,2
%A _Seiichi Manyama_, Sep 09 2023
|
