A319944 - OEIS

%I #8 Oct 13 2020 08:25:55

%S 1,225,714000,10430111250,455589570897000,46993311212615010000,

%T 9839324906977709480400000,3761494651833327732316790250000,

%U 2427487105139453587868600367048750000,2489491831933123075592260875312720412500000,3867129529486594159007141093572270035942600000000

%N O.g.f. A(x) satisfies: [x^n] exp( n^5*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="/A319944/b319944.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) ~ sqrt(1-c) * 5^(5*n - 1) * n^(4*n - 3/2) / (sqrt(2*Pi) * c^n * (5-c)^(4*n - 1) * exp(4*n)), where c = -LambertW(-5*exp(-5)) = 0.03488576825572369630124086867... - _Vaclav Kotesovec_, Oct 13 2020

%e G.f.: A(x) = x + 225*x^2 + 714000*x^3 + 10430111250*x^4 + 455589570897000*x^5 + 46993311212615010000*x^6 + 9839324906977709480400000*x^7 + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp( n^5*x - n*A(x) ) begins:

%e n=1: [1, 0, -450, -4284000, -250322062500, ...];

%e n=2: [1, 30, 0, -8622000, -501675120000, ...];

%e n=3: [1, 240, 56250, 0, -760449262500, ...];

%e n=4: [1, 1020, 1038600, 1038564000, 0, ...];

%e n=5: [1, 3120, 9732150, 30328848000, 93108209197500, 0, ...];

%e n=6: [1, 7770, 60370200, 469008792000, 3641608218960000, 27906215370093360000, 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 + 451*x^2/2! + 4285351*x^3/3! + 250340416201*x^4/4! + 54672019444872001*x^5/5! + 33835513974650405264251*x^6/6! + ...

%e The 5th root of A(x)/x appears to be an integer sequence:

%e (A(x)/x)^(1/5) = 1 + 45*x + 138750*x^2 + 2060865000*x^3 + 90706765441275*x^4 + 9381160956625666875*x^5 + 1966116273013953349582500*x^6 + 751938952953001936098785681250*x^7 + 485360862323214790797483583171389375*x^8 + 497810555195750107907248882311441377821875*x^9 + ...

%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^5*x +x*O(x^#A)) / Ser(A)^m )[m+1]/m ); polcoeff( log(Ser(A)), n)}

%o for(n=1, 15, print1(a(n), ", "))

%K nonn

%O 1,2

%A _Paul D. Hanna_, Oct 02 2018