A053256 Coefficients of the '5th-order' mock theta function f_0(q).
1, 1, -1, 1, 0, 0, -1, 1, 0, 1, -2, 1, -1, 2, -2, 2, -1, 1, -3, 2, -1, 3, -3, 2, -2, 3, -4, 3, -3, 4, -5, 5, -3, 5, -7, 5, -5, 6, -7, 7, -6, 7, -9, 9, -7, 9, -11, 9, -9, 11, -13, 12, -11, 13, -15, 15, -13, 16, -19, 17, -17, 19, -21, 21, -20, 22, -26, 25, -23, 27, -30, 29, -28, 32, -35, 34, -34, 36, -41, 40, -38, 44, -48, 46
Offset: 0
Examples
G.f. = 1 + x - x^2 + x^3 - x^6 + x^7 + x^9 - 2*x^10 + x^11 - x^12 + 2*x^13 - ...
References
- Srinivasa Ramanujan, Collected Papers, Chelsea, New York, 1962, pp. 354-355.
- Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, pp. 19, 21, 22, 23.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000
- George E. Andrews, The fifth and seventh order mock theta functions, Trans. Amer. Math. Soc., 293 (1986) 113-134.
- George E. Andrews and Frank G. Garvan, Ramanujan's "lost" notebook VI: The mock theta conjectures, Advances in Mathematics, 73 (1989) 242-255.
- Dean Hickerson, A proof of the mock theta conjectures, Inventiones Mathematicae, 94 (1988) 639-660.
- George N. Watson, The mock theta functions (2), Proc. London Math. Soc., series 2, 42 (1937) 274-304.
Crossrefs
Programs
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Maple
N:= 100: # for a(0)..a(N) g:= add(q^(k^2)/mul(1+q^i,i=1..k),k=0..floor(sqrt(N))): S:= series(g,q,N+1): seq(coeff(S,q,k),k=0..N)]; # Robert Israel, Mar 27 2018
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Mathematica
Series[Sum[q^n^2/Product[1+q^k, {k, 1, n}], {n, 0, 10}], {q, 0, 100}] a[ n_] := SeriesCoefficient[ Sum[ x^k^2 / QPochhammer[ -x, x, k] // FunctionExpand, {k, 0, Sqrt@ n}], {x, 0, n}]; (* Michael Somos, Feb 13 2017 *) -
PARI
{a(n) = my(t); if( n<0, 0, t = 1 + O(x^n); polcoeff( sum( k=1, sqrtint(n), t *= x^(2*k-1) / (1 + x^k + O(x^(n - (k-1)^2 + 1))), 1), n))}; /* Michael Somos, Mar 12 2006 */
Formula
G.f.: 1 + Sum_{k>0} q^k^2 / ((1 + q) * (1 + q^2) * ... * (1 + q^k)).
Consider partitions of n into parts differing by at least 2. For n > 0: a(n) is the number of them with largest part odd minus number with largest part even.
a(n) ~ -(-1)^n * exp(Pi*sqrt(n/15)) / (2*5^(1/4)*sqrt(phi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 15 2019
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