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 A116646 Number of doubletons in all partitions of n. By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses). 6
 0, 0, 1, 0, 2, 2, 4, 5, 10, 11, 20, 25, 38, 49, 73, 91, 131, 167, 228, 291, 392, 493, 653, 822, 1065, 1336, 1714, 2131, 2706, 3354, 4209, 5193, 6471, 7934, 9817, 11990, 14725, 17909, 21875, 26477, 32172, 38797, 46893, 56339, 67804, 81147, 97260, 116017 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS a(n) = (the number of 2's in all partitions of n) - (the number of 3's in all partitions of n). - Gregory L. Simay, Jul 28 2020 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..1000 FORMULA G.f.: x^2 / ((1+x)*(1-x^3)*(Product_{j>=1} 1-x^j)). a(n) = Sum_{k>=0} k * A116644(n,k). a(n) ~ exp(Pi*sqrt(2*n/3)) / (3*2^(5/2)*Pi*sqrt(n)). - Vaclav Kotesovec, Mar 07 2016 EXAMPLE a(6) = 4 because in the partitions of 6, namely [6],[5,1],[4,2],[4,(1,1)],[(3,3)],[3,2,1],[3,1,1,1],[2,2,2],[(2,2),(1,1)],[2,1,1,1,1] and [1,1,1,1,1,1], we have a total of 4 doubletons (shown between parentheses). MAPLE f:= x^2/(1+x)/(1-x^3)/product(1-x^j, j=1..70): fser:= series(f, x=0, 70): seq(coeff(fser, x, n), n=0..55); MATHEMATICA nmax = 50; CoefficientList[Series[x^2/((1+x)*(1-x^3)) * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *) Table[Sum[PartitionsP[n-6*m-2] - PartitionsP[n-6*m-3] + PartitionsP[n-6*m-4], {m, 0, Floor[n/6]}], {n, 0, 50}] (* Vaclav Kotesovec, Mar 07 2016 *) CROSSREFS Cf. A116644. Column k=2 of A197126. Sequence in context: A116651 A135586 A168542 * A306318 A091188 A147678 Adjacent sequences: A116643 A116644 A116645 * A116647 A116648 A116649 KEYWORD nonn AUTHOR Emeric Deutsch, Feb 20 2006 STATUS approved

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Last modified September 26 15:35 EDT 2023. Contains 365660 sequences. (Running on oeis4.)