%I #13 Oct 13 2020 08:09:27
%S 1,49,22542,34776266,124857847020,863035137487572,
%T 10208133235178252640,190511518719216943969008,
%U 5284939084238999180631562560,208156037245304153601560603185040,11224507767787823723649649410800624768,804502870984274832989329177960786158548256
%N O.g.f. A(x) satisfies: [x^n] exp( n^4*x - n*A(x) ) = 0 for n >= 1.
%C It is remarkable that this sequence should consist entirely of integers.
%H Paul D. Hanna, <a href="/A319941/b319941.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) ~ sqrt(1-c) * 2^(8*n - 5/2) * n^(3*n - 3/2) / (sqrt(Pi) * exp(3*n) * c^n * (4-c)^(3*n - 1)), where c = -LambertW(-4*exp(-4)) = 0.079309605127113656439108647... - _Vaclav Kotesovec_, Oct 13 2020
%e G.f.: A(x) = x + 49*x^2 + 22542*x^3 + 34776266*x^4 + 124857847020*x^5 + 863035137487572*x^6 + 10208133235178252640*x^7 + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k/k! in exp( n^4*x - n*A(x) ) begins:
%e n=1: [1, 0, -98, -135252, -834601572, -14982809095440, ...];
%e n=2: [1, 14, 0, -275992, -1684485824, -30082728311616, ...];
%e n=3: [1, 78, 5790, 0, -2603944836, -45947242627272, ...];
%e n=4: [1, 252, 63112, 15165648, 0, -63525640595328, ...];
%e n=5: [1, 620, 383910, 236740340, 140783667580, 0, ...];
%e n=6: [1, 1290, 1663512, 2143601928, 2754163718208, 3423991878509760, 0, ...]; ...
%e in which the coefficient of x^n in row n forms a diagonal of zeros.
%e RELATED SERIES.
%e exp(A(x)) = 1 + x + 99*x^2/2! + 135547*x^3/3! + 835200793*x^4/4! + 14987248838841*x^5/5! + 621476619810599851*x^6/6! + ...
%o (PARI) {a(n) = my(A=[1], m); for(i=1, n+1, m=#A; A=concat(A, 0); A[m+1] = Vec( exp(m^4*x +x*O(x^#A)) / Ser(A)^m )[m+1]/m ); polcoeff( log(Ser(A)), n)}
%o for(n=1, 15, print1(a(n), ", "))
%Y Cf. A319942, A319943, A319944, A317344, A317346, A317347.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Oct 02 2018
|