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%I #29 Mar 08 2018 06:44:48
%S -240,26760,-4096240,708938760,-130880766192,25168873498760,
%T -4978357936128240,1005225129317834760,-206195878414962688240,
%U 42824436296045618358408,-8983966738037593190400240,1900416270294787067711818760,-404814256845771786255876096240,86744167089111545378556727322760
%N Exponents a(1), a(2), ... such that theta series of E_8 lattice, 1 + 240 q + 2160 q^2 + ... (A004009) is equal to (1-q)^a(1) (1-q^2)^a(2) (1-q^3)^a(3) ...
%C Negative of inverse Euler transform of [240, 2160, ...].
%H Seiichi Manyama, <a href="/A110163/b110163.txt">Table of n, a(n) for n = 1..424</a>
%F a(n) = A013953(n^2) for n>=1. - _Seiichi Manyama_, Jun 17 2017
%F a(n) = 8 + (1/(3*n)) * Sum_{d|n} A008683(n/d) * A288261(d). - _Seiichi Manyama_, Jun 17 2017
%F a(n) = (1/n) * Sum_{d|n} A008683(n/d) * A289636(d). - _Seiichi Manyama_, Jul 09 2017
%F a(n) ~ (-1)^n * exp(Pi*sqrt(3)*n) / n. - _Vaclav Kotesovec_, Mar 08 2018
%e From _Seiichi Manyama_, Jun 17 2017: (Start)
%e a(1) = 8 + 1/3 * A008683(1/1) * A288261(1) = 8 + 1/3 * (-744) = -240,
%e a(2) = 8 + 1/6 * (A008683(2/1) * A288261(1) + A008683(2/2) * A288261(2)) = 8 + 1/6 * (744 + 159768) = 26760. (End)
%t terms = 14; Clear[a, sol];
%t a4009[n_] := If[n == 0, 1, 240 DivisorSigma[3, n]];
%t sol[0] = {}; sol[kmax_] := sol[kmax] = Join[sol[kmax-1], SolveAlways[ Sum[ a4009[k] q^k, {k, 0, kmax}] == Normal[Product[(1-q^k)^a[k], {k, 1, kmax}] + O[q]^(kmax+1)] /. sol[kmax-1], q][[1]] ];
%t A110163 = Array[a, terms] /. sol[terms] (* _Jean-François Alcover_, Jul 03 2017 *)
%Y Cf. A288968 (k=2), this sequence (k=4), A288851 (k=6), A288471 (k=8), A289024 (k=10), A288990/A288989 (k=12), A289029 (k=14).
%Y Cf. A004009, A008683, A013953, A288261, A289636.
%K sign
%O 1,1
%A _N. J. A. Sloane_, Sep 16 2005
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