A182135 Number of partitions of 2^2^n into powers of 2 less than or equal to 2^n.

1, 3, 25, 47905, 751333186150401, 371679100488302192208527928207947545444353

Offset

0,2

Comments

Lengths (in decimal digits) of the terms a(0), a(1), ... are: 1, 1, 2, 5, 15, 42, 107, 258, 602, 1369, 3060, 6755, 14765, 32022, 69007, 147915, 315599, 670702, 1420371, ... .

Links

Alois P. Heinz, Table of n, a(n) for n = 0..8

Formula

a(n) = [x^2^(2^n-1)] 1/(1-x) * 1/Product_{j=0..n-1} (1-x^(2^j)).

Example

a(1) = 3 because the number of partitions of 2^2^1 = 4 into powers of 2 less than or equal to 2^1 = 2 is 3: [2,2], [2,1,1], [1,1,1,1].

Maple

b:= proc(n, j) option remember; local nn, r;
      if n<0 then 0
    elif j=0 then 1
    elif j=1 then n+1
    elif n<j then b(n, j):= b(n-1, j) +b(2*n, j-1)
             else nn:= 1 +floor(n);
                  r:= n-nn;
                  (nn-j) *binomial(nn, j) *add(binomial(j, h)
                  /(nn-j+h) *b(j-h+r, j) *(-1)^h, h=0..j-1)
      fi
    end:
a:= n-> b(2^(2^n-n), n):
seq(a(n), n=0..10);

Mathematica

b[n_, j_] := b[n, j] = Module[{nn, r}, Which[n<0, 0, j==0, 1, j==1, n+1, n < j, b[n, j] = b[n-1, j] + b[2*n, j-1], True, nn = 1+Floor[n]; r = n-nn; (nn-j)*Binomial[nn, j]*Sum[Binomial[j, h]/(nn-j+h)*b[j-h+r, j]*(-1)^h, {h, 0, j-1}]]]; a[n_] := b[2^(2^n-n), n]; Table[a[n], {n, 0, 5}] (* Jean-François Alcover, Feb 05 2017, translated from Maple *)

Crossrefs

Main diagonal of A172288.

Sequence in context: A246536 A183248 A144788 * A307654 A307653 A326610

Adjacent sequences: A182132 A182133 A182134 * A182136 A182137 A182138

Keyword

nonn

Author

Alois P. Heinz, May 26 2012

Status

approved