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A317343 E.g.f. A(x) satisfies: [x^n] exp(n^2*x) / A(x)^n = 0 for n >= 1. 4

%I #7 Aug 06 2018 06:46:05

%S 1,1,3,43,1945,178041,26792971,5940440563,1812303908913,

%T 725818277844145,368664266359757971,231291789356545214331,

%U 175516846850044015048393,158383499766971409675254953,167568661592262324239839114395,205391322843896611716680298647491,288693442696565330249751284373878881,461220054642000786943064376254671287393

%N E.g.f. A(x) satisfies: [x^n] exp(n^2*x) / A(x)^n = 0 for n >= 1.

%C It is remarkable that the logarithm of the e.g.f. A(x) should be an integer series.

%H Paul D. Hanna, <a href="/A317343/b317343.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) ~ sqrt(1-c) * 2^(2*n - 3/2) * n^(2*n-1) / (exp(2*n) * c^(n - 1/2) * (2-c)^(n-1)), where c = -LambertW(-2*exp(-2)) = -A226775 = 0.4063757399599599... - _Vaclav Kotesovec_, Aug 06 2018

%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 43*x^3/3! + 1945*x^4/4! + 178041*x^5/5! + 26792971*x^6/6! + 5940440563*x^7/7! + 1812303908913*x^8/8! + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp(n^2*x) / A(x)^n begins:

%e n=1: [1, 0, -2, -36, -1764, -167280, -25620600, -5737974480, ...];

%e n=2: [1, 2, 0, -88, -4160, -371328, -55329536, -12201990400, ...];

%e n=3: [1, 6, 30, 0, -7812, -698184, -97733304, -20791334880, ...];

%e n=4: [1, 12, 136, 1296, 0, -1171968, -168658176, -33909447168, ...];

%e n=5: [1, 20, 390, 7220, 113020, 0, -265712600, -55963975600, ...];

%e n=6: [1, 30, 888, 25704, 709056, 16600320, 0, -84622337280, ...];

%e n=7: [1, 42, 1750, 72072, 2909340, 112245672, 3684715944, 0, ...];

%e n=8: [1, 56, 3120, 172640, 9455488, 508540416, 26199517696, 1150524892160, 0, ...]; ...

%e in which the coefficient of x^n in row n forms a diagonal of zeros.

%e LOGARITHMIC PROPERTY.

%e The logarithm of the e.g.f. is the integer series:

%e log(A(x)) = x + x^2 + 6*x^3 + 74*x^4 + 1400*x^5 + 35676*x^6 + 1140328*x^7 + 43740848*x^8 + 1954336608*x^9 + 99561874080*x^10 + ... + A317344(n)*x^n + ...

%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^2*x +x*O(x^#A)) / Ser(A)^m )[m+1]/m ); n!*A[n+1]}

%o for(n=0,20,print1(a(n),", "))

%Y Cf. A317344, A317345.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 26 2018

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