A169975 Expansion of Product_{i>=0} (1 + x^(4*i+1)).
1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 3, 2, 0, 1, 3, 3, 1, 1, 4, 4, 1, 1, 4, 5, 2, 1, 5, 7, 3, 1, 5, 8, 5, 2, 6, 10, 6, 2, 6, 12, 9, 3, 7, 14, 11, 4, 7, 16, 15, 6, 8, 19, 18, 8, 9, 21, 23, 11, 10, 24, 27, 14, 11, 27, 34, 19, 13, 30, 39, 24, 15, 33, 47, 31, 18, 37, 54, 38
Offset: 0
Keywords
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..10000
- Günter Meinardus, Über Partitionen mit Differenzenbedingungen, Mathematische Zeitschrift (1954/55), Volume: 61, page 289-302
Crossrefs
Programs
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Mathematica
nmax = 200; CoefficientList[Series[Product[(1 + x^(4*k+1)), {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 26 2015 *) nmax = 200; poly = ConstantArray[0, nmax + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[If[Mod[k, 4] == 1, Do[poly[[j + 1]] += poly[[j - k + 1]], {j, nmax, k, -1}]; ], {k, 2, nmax}]; poly (* Vaclav Kotesovec, Jan 18 2017 *)
Formula
G.f.: Sum_{n>=0} (x^(2*n^2 - n) / Product_{k=1..n} (1 - x^(4*k))). - Joerg Arndt, Mar 10 2011
G.f.: G(0)/x where G(k) = 1 - 1/(1 - 1/(1 - 1/(1+(x)^(4*k+1))/G(k+1) )); (recursively defined continued fraction, see A006950). - Sergei N. Gladkovskii, Jan 28 2013
a(n) ~ exp(Pi*sqrt(n)/(2*sqrt(3))) / (2^(7/4) * 3^(1/4) * n^(3/4)) * (1 - (3*sqrt(3)/(4*Pi) + Pi/(192*sqrt(3))) / sqrt(n)). - Vaclav Kotesovec, Aug 26 2015, extended Jan 18 2017
Comments