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%I #12 Feb 25 2026 12:52:21
%S 1,1,16,480,20094,1051232,65024482,4603101446,365116402526,
%T 31976454542555,3058942014578602,317021252927702154,
%U 35364462798259624358,4223939351547783922564,537808539569014368308062,72721425272305729382146028,10408706387258804711360666654,1572419486478837318683324924440
%N O.g.f. A(x) satisfies: [x^n] exp( n*(n+1) * x*A(x)^7 ) / A(x)^(n+1) = 0 for n > 0.
%H Paul D. Hanna, <a href="/A393628/b393628.txt">Table of n, a(n) for n = 0..400</a>
%F a(n) ~ c * 7^(2*n) * n^(n + 17/10) / (exp(n) * 6^n), where c = 0.132594844139368414... - _Vaclav Kotesovec_, Feb 25 2026
%e O.g.f.: A(x) = 1 + x + 16*x^2 + 480*x^3 + 20094*x^4 + 1051232*x^5 + 65024482*x^6 + 4603101446*x^7 + 365116402526*x^8 + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k/k! in exp(n*(n+1)*x*A(x)^7) / A(x)^(n+1) for n >= 0 begins
%e n = 0: [1, -1, -30, -2694, -454200, -119739480, ...];
%e n = 1: [1, 0, -34, -3976, -737400, -204152768, ...];
%e n = 2: [1, 3, 0, -3624, -861480, -260085384, ...];
%e n = 3: [1, 8, 108, 0, -724248, -279054528, ...];
%e n = 4: [1, 15, 350, 11030, 0, -225030320, ...];
%e n = 5: [1, 24, 810, 37896, 2118240, 0, ...];
%e ...
%e in which the main diagonal is all zeros after the initial term.
%e The above table is derived from the following series
%e n = 1: exp(1*2*x*A(x)^7)/A(x)^2 = 1 + 0*x - 34*x^2/2! - 3976*x^3/3! - 737400*x^4/4! - 204152768*x^5/5! + ...
%e n = 2: exp(2*3*x*A(x)^7)/A(x)^3 = 1 + 3*x + 0*x^2/2! - 3624*x^3/3! - 861480*x^4/4! - 260085384*x^5/5! + ...
%e n = 3: exp(3*4*x*A(x)^7)/A(x)^4 = 1 + 8*x + 108*x^2/2! + 0*x^3/3! - 724248*x^4/4! - 279054528*x^5/5! + ...
%e n = 4: exp(4*5*x*A(x)^7)/A(x)^5 = 1 + 15*x + 350*x^2/2! + 11030*x^3/3! + 0*x^4/4! - 225030320*x^5/5! + ...
%e n = 5: exp(5*6*x*A(x)^7)/A(x)^6 = 1 + 24*x + 810*x^2/2! + 37896*x^3/3! + 2118240*x^4/4! + 0*x^5/5! + ...
%e ...
%e illustrating that [x^n] exp(n*(n+1)*x*A(x)^7) / A(x)^(n+1) = 0 for n > 0.
%o (PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp((m-1)*m*x*Ser(A)^7 +x*O(x^m)) / Ser(A)^m )[m]/m ); A[n+1]}
%o for(n=0, 25, print1( a(n), ", "))
%Y Cf. A393620, A337458, A393621, A393622, A393623, A393624, A393625, A393627.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 25 2026