%I #9 May 18 2026 10:41:32
%S 1,1,8065,1723380877,2972352315820441,24013530904194819396970,
%T 661513372249008158305805082187,
%U 50423205212193924706520234988903231406,9172099526697986231143005697761988750382396953,3565599162605040989126220209472530594315226235614225958
%N O.g.f. A(x) satisfies: [x^n] exp( n^7 * x ) / A(x) = 0 for n>0.
%H Paul D. Hanna, <a href="/A393847/b393847.txt">Table of n, a(n) for n = 0..100</a>
%F a(n) ~ sqrt(1-c) * 7^(7*n) * n^(6*n - 1/2) / (sqrt(2*Pi) * c^n * (7-c)^(6*n) * exp(6*n)), where c = -LambertW(-7*exp(-7)). - _Vaclav Kotesovec_, May 18 2026
%e O.g.f.: A(x) = 1 + x + 8065*x^2 + 1723380877*x^3 + 2972352315820441*x^4 + 24013530904194819396970*x^5 + 661513372249008158305805082187*x^6 + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k/k! in exp(n^7*x)/A(x) begins
%e n = 1: [1, 0, -16129, -10340236874, ...];
%e n = 2: [1, 127, 0, -10344333640, ...];
%e n = 3: [1, 2186, 4762467, 0, ...];
%e n = 4: [1, 16383, 268386560, 4386108292792, 0, ...];
%e n = 5: [1, 78124, 6103343247, 476804727467762, 37175837330786900381, 0, ...];
%e ...
%e in which a diagonal, the coefficient of x^n in row n, is all zeros.
%o (PARI) {a(n,k=7) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp(x*(m-1)^k +x*O(x^m)) / Ser(A) )[m] ); A[n+1]}
%o for(n=0, 15, print1( a(n,7), ", "))
%Y Sequences below with the given parameter p have an o.g.f. F(x) satisfying [x^n] exp( n^p*x ) / F(x) = 0 for n >= 1.
%Y Cf. A304322 (p=2), A304323 (p=3), A304324 (p=4), A304325 (p=5).
%Y Cf. A393846 (p=6), A393848 (p=8), A393849 (p=9).
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 02 2026