A317344 - OEIS

%I #11 Aug 06 2018 08:54:55

%S 1,1,6,74,1400,35676,1140328,43740848,1954336608,99561874080,

%T 5691393582336,360561583177440,25064797000415744,1896477768217789120,

%U 155128714525468598400,13639839907494680219648,1282811359778733608826368,128498290985443181787800064,13657938489514600713859515392,1535272989503239280608301470720,181975961346350933380240113192960

%N O.g.f. A(x) satisfies: [x^n] exp( n^2*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="/A317344/b317344.txt">Table of n, a(n) for n = 1..300</a>

%F a(n) ~ c * d^n * n! / n^2, where d = -4 / (LambertW(-2*exp(-2)) * (2 + LambertW(-2*exp(-2)))) = 6.17655460948348035823168... and c = (2 + LambertW(-2*exp(-2))) * sqrt(-LambertW(-2*exp(-2)) * (1 + LambertW(-2*exp(-2)))) / (4*sqrt(2)*Pi) = 0.0440433939... - _Vaclav Kotesovec_, Aug 06 2018

%e O.g.f.: 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 + ...

%e such that [x^n] exp( n^2*x - n*A(x) ) = 0  for n >= 1.

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp( n^2*x - n*A(x) ) 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 RELATED SERIES.

%e exp(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! + ... + A317343(n)*x^n/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 ); polcoeff( log(Ser(A)),n)}

%o for(n=1,30,print1(a(n),", "))

%Y Cf. A317343, A317346, A317347.

%K nonn

%O 1,3

%A _Paul D. Hanna_, Jul 26 2018