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%I #35 Jun 12 2019 03:10:02
%S 1,2,2,3,4,4,6,7,8,10,12,14,16,20,22,26,31,34,40,46,52,60,68,76,87,98,
%T 110,124,140,156,174,196,216,242,270,298,332,368,406,449,496,546,602,
%U 664,728,800,880,962,1056,1156,1262,1381,1508,1644,1794,1956,2128
%N Coefficients of the '10th-order' mock theta function phi(q).
%C The alternating sum of the same series, namely phi(q) = Sum_{n>=0} (-1)^n q^(n(n+1)/2)/((1-q)(1-q^3)...(1-q^(2n+1))) = 1 + x^3 - x^7 - x^16 + x^24 + x^39 - x^51 - ..., where the exponents are given by 5n^2 +- 2n. See the Amer. Math. Monthly reference.
%D Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 9.
%H Vaclav Kotesovec, <a href="/A053281/b053281.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..1000 from Seiichi Manyama)
%H Youn-Seo Choi, <a href="http://dx.doi.org/10.1007/s002220050318">Tenth order mock theta functions in Ramanujan's lost notebook</a>, Inventiones Mathematicae, 136 (1999) p. 497-569.
%H David Newman, <a href="http://www.jstor.org/stable/2589364">A Recurrence inside a Generating Function: Solution to problem 10681</a>, American Mathematical Monthly, vol. 107 (2000), p. 569.
%F G.f.: phi(q) = Sum_{n >= 0} q^(n(n+1)/2)/((1-q)(1-q^3)...(1-q^(2n+1))).
%F a(n) ~ sqrt(phi) * exp(Pi*sqrt(n/5)) / (2*5^(1/4)*sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Jun 12 2019
%t Series[Sum[q^(n(n+1)/2)/Product[1-q^(2k+1), {k, 0, n}], {n, 0, 13}], {q, 0, 100}]
%t nmax = 100; CoefficientList[Series[Sum[x^(k*(k+1)/2) / Product[1-x^(2*j+1), {j, 0, k}], {k, 0, Floor[Sqrt[2*nmax]]}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Jun 11 2019 *)
%Y Other '10th-order' mock theta functions are at A053282, A053283, A053284.
%K nonn,easy
%O 0,2
%A _Dean Hickerson_, Dec 19 1999
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