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

0,3

Links

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

Formula

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

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

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

a(A091892(n)) = 1.

Example

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

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 *)

Prog

(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

Crossrefs

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

Cf. A091889, A091890, A091892, A091893.

Sequence in context: A235872 A100762 A059147 * A258127 A181982 A070194

Adjacent sequences: A091888 A091889 A091890 * A091892 A091893 A091894

Keyword

nonn,look

Author

Reinhard Zumkeller, Feb 10 2004

Extensions

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

Status

approved