%I #14 Oct 13 2020 11:58:39
%S 1,1,-5,-197,-65111,-62390159,-125012786669,-447082993406405,
%T -2583111044504384687,-22511408975342644804991,
%U -281350305428215911326408789,-4850582201056517165575319399909,-111834955668396093904661955538037255,-3361788412998032560821833199260880942287,-128987969989211586699135087535153035663946301,-6203990036027464835833031041177436339788197962789
%N E.g.f. A(x) satisfies: [x^(n-1)] A(x)^(n^3) = [x^n] A(x)^(n^3) for n>=1.
%C Compare e.g.f. to: [x^(n-1)] exp(x)^n = [x^n] exp(x)^n for n>=1.
%H Paul D. Hanna, <a href="/A296172/b296172.txt">Table of n, a(n) for n = 0..180</a>
%F The logarithm of the e.g.f. A(x) is an integer series:
%F _ log(A(x)) = Sum_{n>=1} A296173(n) * x^n.
%F E.g.f. A(x) satisfies:
%F _ 1/n! * d^n/dx^n A(x)^(n^3) = 1/(n-1)! * d^(n-1)/dx^(n-1) A(x)^(n^3) for n>=1, when evaluated at x = 0.
%F a(n) ~ -sqrt(1-c) * 3^(3*n - 3) * n^(3*n - 3) / (c^n * (3-c)^(2*n - 3) * exp(3*n)), where c = -LambertW(-3*exp(-3)) = -A226750. - _Vaclav Kotesovec_, Oct 13 2020
%e E.g.f.: A(x) = 1 + x - 5*x^2/2! - 197*x^3/3! - 65111*x^4/4! - 62390159*x^5/5! - 125012786669*x^6/6! - 447082993406405*x^7/7! - 2583111044504384687*x^8/8! - 22511408975342644804991*x^9/9! - 281350305428215911326408789*x^10/10! - 4850582201056517165575319399909*x^11/11! - 111834955668396093904661955538037255*x^12/12! +...
%e To illustrate [x^(n-1)] A(x)^(n^3) = [x^n] A(x)^(n^3), form a table of coefficients of x^k in A(x)^(n^3) that begins as
%e n=1: [(1), (1), -5/2, -197/6, -65111/24, -62390159/120, -125012786669/720, ...];
%e n=2: [1, (8), (8), -1040/3, -71152/3, -64676744/15, -63817770776/45, ...];
%e n=3: [1, 27, (567/2), (567/2), -787941/8, -648507951/40, -405807483249/80, ...];
%e n=4: [1, 64, 1856, (88448/3), (88448/3), -689015872/15, -611019817664/45, ...];
%e n=5: [1, 125, 14875/2, 1649375/6, (156207625/24), (156207625/24), ...];
%e n=6: [1, 216, 22680, 1533168, 73812816, (12455715384/5), (12455715384/5), ...];
%e n=7: [1, 343, 115591/2, 38174185/6, 12294445009/24, 3808296195823/120, (1051338418817239/720), (1051338418817239/720), ...];
%e ...
%e in which the diagonals indicated by parenthesis are equal.
%e Dividing the coefficients of x^(n-1)/(n-1)! in A(x)^(n^3) by n^3, we obtain the following sequence:
%e [1, 1, 21, 2764, 1249661, 1383968376, 3065126585473, 11913154589356672, 74286423963211939641, 696469981042645688972800, ...].
%e LOGARITHMIC PROPERTY.
%e Amazingly, the logarithm of the e.g.f. A(x) is an integer series:
%e log(A(x)) = x - 3*x^2 - 30*x^3 - 2686*x^4 - 517311*x^5 - 173118807*x^6 - 88535206152*x^7 - 63977172334344*x^8 - 61971659588102940*x^9 - 77470793599569049440*x^10 - 121439997599825393413344*x^11 - 233353875172602479932391040*x^12 - 539638027429765922735002220880*x^13 - 1479049138515818646669055218090480*x^14 - 4742815067612592169849894663392228480*x^15 +...
%o (PARI) {a(n) = my(A=[1]); for(i=1,n+1, A=concat(A,0); V=Vec(Ser(A)^((#A-1)^3)); A[#A] = (V[#A-1] - V[#A])/(#A-1)^3 ); n!*A[n+1]}
%o for(n=0,20,print1(a(n),", "))
%Y Cf. A296173, A296170, A296174, A296176.
%K sign
%O 0,3
%A _Paul D. Hanna_, Dec 07 2017
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