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A317345 E.g.f. A(x) satisfies: [x^n] exp(n^3*x) / A(x)^(n^2) = 0 for n >= 1. 4

%I #13 Aug 07 2018 03:46:13

%S 1,1,5,445,196105,221673401,501981700621,1983064113021685,

%T 12488526496641458705,117611695946767352571505,

%U 1578802193598207376026165781,29098684071572000208903027320621,714476480265312671332820625804579865,22796869288656035590303941174243615386665,925701505348044648968634173494720540556875805

%N E.g.f. A(x) satisfies: [x^n] exp(n^3*x) / A(x)^(n^2) = 0 for n >= 1.

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

%C Periodic modulo 10: a(5*n+k) = [1,1,5,5,5](k) (mod 10), for n>=0 and k = 0..4 (conjecture).

%H Paul D. Hanna, <a href="/A317345/b317345.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) ~ sqrt(1-c) * 3^(3*n - 7/3) * n^(3*n - 2) / (exp(3*n) * c^(n - 1/3) * (3-c)^(2*n - 2)), where c = -LambertW(-3*exp(-3)) = 0.1785606278779211065968086697... = -A226750. - _Vaclav Kotesovec_, Aug 07 2018

%e E.g.f.: A(x) = 1 + x + 5*x^2/2! + 445*x^3/3! + 196105*x^4/4! + 221673401*x^5/5! + 501981700621*x^6/6! + 1983064113021685*x^7/7! + 12488526496641458705*x^8/8! + ...

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients of x^k/k! in exp(n^3*x) / A(x)^(n^2) begins:

%e n=1: [1, 0, -4, -432, -194256, -220662720, -500627544000, ...];

%e n=2: [1, 4, 0, -1856, -805376, -898258176, -2023715201024, ...];

%e n=3: [1, 18, 288, 0, -1989792, -2154563712, -4727980751616, ...];

%e n=4: [1, 48, 2240, 94464, 0, -4244861952, -9137589559296, ...];

%e n=5: [1, 100, 9900, 959200, 84852400, 0, -15901448888000, ...];

%e n=6: [1, 180, 32256, 5738688, 1003636224, 161358324480, 0, ...];

%e n=7: [1, 294, 86240, 25218144, 7335234144, 2103824749824, 557359956846336, 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 + 2*x^2 + 72*x^3 + 8096*x^4 + 1839000*x^5 + 695334816*x^6 + 392764566208*x^7 + 309340607492096*x^8 + ... + A317346(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^3*x +x*O(x^#A)) / Ser(A)^(m^2) )[m+1]/m^2 ); n!*A[n+1]}

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

%Y Cf. A317343, A317346.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 26 2018

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Last modified August 27 03:02 EDT 2024. Contains 375462 sequences. (Running on oeis4.)