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A264593
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Let G[1](q) denote the g.f. for A003114 and G[2](q) the g.f. for A003106 (the two Rogers-Ramanujan identities). For i>=3, let G[i](q) = (G[i-1](q)-G[i-2](q))/q^(i-2). Sequence gives coefficients of G[6](q).
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9
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1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 8, 8, 10, 11, 13, 14, 17, 18, 21, 23, 26, 28, 33, 35, 40, 44, 50, 54, 62, 67, 76, 83, 93, 101, 114, 123, 138, 150, 167, 181, 202, 219, 243, 264, 292, 317, 351, 380, 419, 455, 500, 542, 596, 645, 707, 766, 838, 907, 992, 1072
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OFFSET
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0,15
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COMMENTS
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It is conjectured that G[i](q) = 1 + O(q^i) for all i.
For n >=1 a(n) gives the number of partitions of n without parts 1, 2, 3, 4, and 5, and the parts differ by at least 2. For the proof see a comment given in A264592. - Wolfdieter Lang, Nov 10 2016
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LINKS
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FORMULA
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G.f.: G[6](q) = GII_2(q) = Sum_{m>=0} q^(m*(m+5)) / Product_{j =1..m} (1 - q^j).
See Andrews and Baxter [A-B], eq. (5.1) for i=6.
G.f.: Sum_{m=0} ((-1)^m*(1 - q^(m+1))*(1 - q^(m+2))*(1 - q^(m+3))*(1 - q^(m+4))*(1 - q^(2*m+5))*q^((5*m+19)*m/2)) / Product_{j>=1} (1 - q^j). See [A-B] eq. (3.8) for i=6. (End)
a(n) ~ exp(2*Pi*sqrt(n/15)) / (2 * 3^(1/4) * sqrt(5) * phi^(9/2) * n^(3/4)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Dec 24 2016
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EXAMPLE
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a(18) = 4 because the four partitions of 18 without parts 1, 2, 3, 4 and 5, and the parts differ by at least 2 are [18], [12, 6], [11, 7], [10, 8]. - Wolfdieter Lang, Nov 10 2016
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MATHEMATICA
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nmax = 100; CoefficientList[Series[Sum[x^(k*(k+5))/Product[1-x^j, {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 24 2016 *)
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CROSSREFS
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For the generalized Rogers-Ramanujan series G[0], G[1], G[2], G[3], G[4], G[5], G[6], G[7], G[8] see A003113, A003114, A003106, A006141, A264591, A264592, A264593, A264594, A264595.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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