A091891 Number of partitions of n into parts which are a sum of exactly as many distinct powers of 2 as n has 1's in its binary representation.

1, 1, 2, 1, 4, 1, 2, 1, 10, 3, 2, 1, 5, 1, 2, 1, 36, 6, 12, 1, 11, 3, 2, 1, 24, 3, 3, 1, 5, 1, 2, 1, 202, 67, 55, 9, 93, 4, 5, 1, 112, 8, 13, 1, 10, 3, 2, 1, 304, 22, 18, 1, 20, 3, 3, 1, 34, 3, 3, 1, 5, 1, 2, 1, 1828, 1267, 1456, 71, 1629, 77, 100, 2, 2342, 99, 123, 9, 132, 4, 3, 1

Offset: 0

Author

Reinhard Zumkeller, Feb 10 2004

Keywords

• nonn
• look

Examples

			a(9) = 3 because there are 3 partitions of 9 into parts of size 3, 5, 6, 9 which are the numbers that have two 1's in their binary representations. The 3 partitions are: 9, 6 + 3 and 3 + 3 + 3. - _Andrew Howroyd_, Apr 20 2021
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16384 (terms n = 1..1000 from Andrew Howroyd)

Crossrefs

Cf. A000079, A000120, A000041, A018819, A018900, A014311.

Cf. A091889, A091890, A091892, A091893.

Programs

Maple

H:= proc(n) option remember; add(i, i=Bits[Split](n)) end:
v:= proc(n, k) option remember; `if`(n<1, 0,
      `if`(H(n)=k, n, v(n-1, k)))
    end:
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, v(i-1, k), k)+b(n-i, v(min(n-i, i), k), k)))
    end:
a:= n-> b(n$2, H(n)):
seq(a(n), n=0..80);  # Alois P. Heinz, Dec 12 2021

Mathematica

etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}] b[n - j], {j, 1, n}]/n]; b];
EulerT[v_List] := With[{q = etr[v[[#]]&]}, q /@ Range[Length[v]]];
a[n_] := EulerT[Table[DigitCount[k, 2, 1] == DigitCount[n, 2, 1] // Boole, {k, 1, n}]][[n]];
Array[a, 100] (* Jean-François Alcover, Dec 12 2021, after Andrew Howroyd *)

PARI

EulerT(v)={Vec(exp(x*Ser(dirmul(v,vector(#v,n,1/n))))-1, -#v)}
a(n) = {EulerT(vector(n,k,hammingweight(k)==hammingweight(n)))[n]} \\ Andrew Howroyd, Apr 20 2021

Formula

a(A000079(n)) = A018819(n);

a(A018900(n)) = A091889(n);

a(A014311(n)) = A091890(n);

a(A091892(n)) = 1.

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 12 2021