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Number of 4's in all the partitions of n into distinct parts.
3

%I #25 May 16 2020 20:59:44

%S 0,0,0,1,1,1,2,1,2,3,3,5,6,7,9,10,12,15,18,22,26,31,36,42,50,58,68,80,

%T 92,107,124,142,164,189,216,248,284,323,369,420,476,541,613,693,784,

%U 885,997,1123,1264,1419,1593,1787,2000,2239,2504,2795,3120,3479

%N Number of 4's in all the partitions of n into distinct parts.

%H Alois P. Heinz, <a href="/A015739/b015739.txt">Table of n, a(n) for n = 1..1000</a>

%F G.f.: x^4*prod(j>=1, 1+x^j)/(1+x^4). - _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]

%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 4.

%p g:=x^4*product(1+x^j,j=1..60)/(1+x^4): gser:=series(g,x=0,57): seq(coeff(gser,x,n),n=1..54);

%p # _Emeric Deutsch_, Apr 17 2006

%p b:= proc(n, i) option remember; local g;

%p if n=0 then [1, 0]

%p elif i<1 then [0, 0]

%p else g:= `if`(i>n, [0$2], b(n-i, i-1));

%p b(n, i-1) +g +[0, `if`(i=4, g[1], 0)]

%p fi

%p end:

%p a:= n-> b(n, n)[2]:

%p seq (a(n), n=1..100);

%p # _Alois P. Heinz_, Oct 27 2012

%t $RecursionLimit = 1000; b[n_, i_] := b[n, i] = Module[{g}, If[n==0, {1, 0}, If[i<1 , {0, 0}, g = If[i>n, {0, 0}, b[n-i, i-1]]; b[n, i-1] + g + {0, If[i == 4, g[[1]], 0]}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Apr 01 2015, after _Alois P. Heinz_ *)

%t Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 4], {n, 58}] (* _Robert Price_, May 16 2020 *)

%K nonn

%O 1,7

%A _Clark Kimberling_