A015739 Number of 4's in all the partitions of n into distinct parts.

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, 92, 107, 124, 142, 164, 189, 216, 248, 284, 323, 369, 420, 476, 541, 613, 693, 784, 885, 997, 1123, 1264, 1419, 1593, 1787, 2000, 2239, 2504, 2795, 3120, 3479

Offset

1,7

Links

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Formula

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

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 4.

Maple

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);
# Emeric Deutsch, Apr 17 2006
b:= proc(n, i) option remember; local g;
      if n=0 then [1, 0]
    elif i<1 then [0, 0]
    else g:= `if`(i>n, [0$2], b(n-i, i-1));
         b(n, i-1) +g +[0, `if`(i=4, g[1], 0)]
      fi
    end:
a:= n-> b(n, n)[2]:
seq (a(n), n=1..100);
# Alois P. Heinz, Oct 27 2012

Mathematica

$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 *)
Table[Count[Flatten@Select[IntegerPartitions[n], DeleteDuplicates[#] == # &], 4], {n, 58}] (* Robert Price, May 16 2020 *)

Crossrefs

Sequence in context: A180235 A239493 A331849 * A015746 A109266 A022876

Adjacent sequences: A015736 A015737 A015738 * A015740 A015741 A015742

Keyword

nonn

Author

Clark Kimberling

Status

approved