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 A015741 Number of 6's in all the partitions of n into distinct parts. 2
 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 8, 9, 12, 14, 17, 21, 24, 29, 34, 40, 47, 55, 65, 75, 88, 102, 118, 137, 157, 181, 208, 238, 272, 311, 355, 404, 460, 522, 592, 671, 758, 856, 966, 1088, 1224, 1377, 1546, 1734, 1944 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,9 COMMENTS a(n+6) = A015753(n). - Alois P. Heinz, Aug 24 2011 LINKS Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 FORMULA G.f.: x^6 * product(j>=1, 1+x^j )/(1+x^6). - Emeric Deutsch, Apr 17 2006 Corresponding g.f. for "number of k's" is x^k/(1+x^k)*prod(n>=1, 1+x^n ). [Joerg Arndt, Feb 20 2014] a(n) ~ exp(Pi*sqrt(n/3)) / (8*3^(1/4)*n^(3/4)). - Vaclav Kotesovec, Oct 30 2015 EXAMPLE a(9) = 2 because in the 8 (=A000009(9)) partitions of 9 into distinct parts, namely [9], [8,1], [7,2], [6,3], [6,2,1], [5,4], [5,3,1] and [4,3,2] we have altogether two parts equal to 6. MAPLE g:=x^6*product(1+x^j, j=1..60)/(1+x^6): gser:=series(g, x=0, 57): seq(coeff(gser, x, n), n=1..54); # Emeric Deutsch, Apr 17 2006 MATHEMATICA nmax = 100; Rest[CoefficientList[Series[x^6/(1+x^6) * Product[1+x^k, {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 30 2015 *) Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 6], {n, 54}] (* Robert Price, May 16 2020 *) CROSSREFS Cf. A015753. Sequence in context: A261770 A096792 A335746 * A015753 A005686 A328675 Adjacent sequences:  A015738 A015739 A015740 * A015742 A015743 A015744 KEYWORD nonn AUTHOR STATUS approved

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Last modified April 20 16:51 EDT 2021. Contains 343135 sequences. (Running on oeis4.)