|
%I #26 May 16 2020 21:01:53
%S 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,
%T 75,88,102,118,137,157,181,208,238,272,311,355,404,460,522,592,671,
%U 758,856,966,1088,1224,1377,1546,1734,1944
%N Number of 6's in all the partitions of n into distinct parts.
%C a(n+6) = A015753(n). - _Alois P. Heinz_, Aug 24 2011
%H Vaclav Kotesovec, <a href="/A015741/b015741.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: x^6 * product(j>=1, 1+x^j )/(1+x^6). - _Emeric Deutsch_, Apr 17 2006
%F 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]
%F a(n) ~ exp(Pi*sqrt(n/3)) / (8*3^(1/4)*n^(3/4)). - _Vaclav Kotesovec_, Oct 30 2015
%e 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.
%p 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
%t 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 *)
%t Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 6], {n, 54}] (* _Robert Price_, May 16 2020 *)
%Y Cf. A015753.
%K nonn
%O 1,9
%A _Clark Kimberling_
|
