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%I #51 Aug 23 2023 08:41:35
%S 1,-19,91,-179,126,-1,344,-1459,2521,-2394,1332,-737,2198,-6536,11466,
%T -11699,4914,485,6860,-22554,31304,-25308,12168,-6625,15751,-41762,
%U 68131,-61576,24390,-126,29792,-93619,121212,-93366,43344,-15803,50654,-130340
%N Expansion of (E_4(q) - 28*E_4(q^2) + 63*E_4(q^3) - 36*E(q^6)) / 240.
%C Define f(q) = (eta(q^2)*eta(q^3))^7/(eta(q)*eta(q^6))^5, g(q) = Sum_{n>=1} a(n)/n^3 * q^n and t(q) = (eta(q)*eta(q^6)/(eta(q^2)*eta(q^3))^12.
%C And define the sequence {b(n)} = {0, 6, 351/4, 62531/36, ...} as the solutions of the recursion (n+1)^3*b(n+1) = (34*n^3 + 51*n^2 + 27*n +5)*b(n) - n^3*b(n-1), n >= 1 with b(0) = 0, b(1) = 6.
%C The following equation holds: 6*f(q)*g(q) = Sum_{n>=0} b(n)*t(q)^n.
%D D. Zagier, "Elliptic modular forms and their applications." The 1-2-3 of modular forms. Springer Berlin Heidelberg, 2008. 1-103.
%H Seiichi Manyama, <a href="/A288485/b288485.txt">Table of n, a(n) for n = 1..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_theorem">Apéry's theorem</a>.
%e 6*f(q)*g(q)
%e = 6*(1 + 5*q + 13*q^2 + 23*q^3 + 29*q^4 + 30*q^5 + 31*q^6 + 40*q^7 + ... )
%e *(q - 19/8*q^2 + 91/27*q^3 - 179/64*q^4 + 126/125*q^5 - 1/216*q^6 + 344/343q^7 - ... )
%e = 6*q + 63/4*q^2 + 971/36*q^3 + 10679/288*q^4 + 1126103/36000*q^5 + 105401/2400*q^6 + 536870027/12348000*q^7 + ...
%e = 6 * (q - 12*q^2 + 66*q^3 - 220*q^4 + 495*q^5 - 804*q^6 + 1068*q^7 - ... )
%e + 351/4 * (q^2 - 24*q^3 + 276*q^4 - 2024*q^5 + 10626*q^6 - 42528*q^7 + ... )
%e + 62531/36 * (q^3 - 36*q^4 + 630*q^5 - 7140*q^6 + 58905*q^7 - ... )
%e + 11424695/288 * (q^4 - 48*q^5 + 1128*q^6 - 17296*q^7 + ... )
%e + 35441662103/36000 * (q^5 - 60*q^6 + 1770*q^7 - ... )
%e + ...
%t a[n_Integer] := Module[{b, ary}, b = Join[{0}, Table[DivisorSigma[3, i], {i, 1, n}]]; ary = b; Do[If[Mod[i, 2] == 0, ary[[i + 1]] -= 28*b[[i/2 + 1]]]; If[Mod[i, 3] == 0, ary[[i + 1]] += 63*b[[i/3 + 1]]]; If[Mod[i, 6] == 0, ary[[i + 1]] -= 36*b[[i/6 + 1]]];, {i, 1, n}]; Rest[ary]]; a[38] (* _Robert P. P. McKone_, Aug 23 2023 *)
%o (Ruby)
%o def A001158(n)
%o s = 0
%o (1..n).each{|i| s += i * i * i if n % i == 0}
%o s
%o end
%o def A288485(n)
%o a = [0] + (1..n).map{|i| A001158(i)}
%o ary = a.clone
%o (1..n).each{|i|
%o ary[i] -= 28 * a[i / 2] if i % 2 == 0
%o ary[i] += 63 * a[i / 3] if i % 3 == 0
%o ary[i] -= 36 * a[i / 6] if i % 6 == 0
%o }
%o ary[1..-1]
%o end
%o p A288485(100)
%Y {b(n)} = {A059415(n)/A059416(n)} = {0, 6, 351/4, 62531/36, ...}.
%Y Cf. A001158 (sigma_3(n)), A004009 (E_4), A006353 (f(q)), A226235 (t(q)).
%K sign
%O 1,2
%A _Seiichi Manyama_, Jun 09 2017
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