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%I #36 Sep 21 2018 22:41:10
%S 1,2,1,2,3,1,2,4,9,1,2,4,25,129,1,2,4,35,4225,32769,1,2,4,36,47905,
%T 268468225,2147483649,1,2,4,36,222241,733276217345,
%U 1152921506754330625,9223372036854775809,1
%N Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) is the number of partitions of 2^2^n into powers of 2 less than or equal to 2^k.
%C A(18,18) = 2797884726...4715787265 has 1420371 decimal digits and was computed by the algorithm given below.
%H Alois P. Heinz, <a href="/A172288/b172288.txt">Table of n, a(n) for n = 0..77</a>
%F A(n,k) = [x^2^(2^n-1)] 1/(1-x) * 1/Product_{j=0..k-1} (1-x^(2^j)).
%e A(2,1) = 9, because there are 9 partitions of 2^2^2=16 into powers of 2 less than or equal to 2^1=2: [2,2,2,2,2,2,2,2], [2,2,2,2,2,2,2,1,1], [2,2,2,2,2,2,1,1,1,1], [2,2,2,2,2,1,1,1,1,1,1], [2,2,2,2,1,1,1,1,1,1,1,1], [2,2,2,1,1,1,1,1,1,1,1,1,1], [2,2,1,1,1,1,1,1,1,1,1,1,1,1], [2,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1].
%e Square array A(n,k) begins:
%e 1, 2, 2, 2, 2, ...
%e 1, 3, 4, 4, 4, ...
%e 1, 9, 25, 35, 36, ...
%e 1, 129, 4225, 47905, 222241, ...
%e 1, 32769, 268468225, 733276217345, 751333186150401, ...
%p b:= proc(n,j) option remember; local nn, r;
%p if n<0 then 0
%p elif j=0 then 1
%p elif j=1 then n+1
%p elif n<j then b(n,j):= b(n-1, j) +b(2*n, j-1)
%p else nn:= 1 +floor(n);
%p r:= n-nn;
%p (nn-j) *binomial(nn, j) *add(binomial(j, h)
%p /(nn-j+h) *b(j-h+r, j) *(-1)^h, h=0..j-1)
%p fi
%p end:
%p A:= (n,k)-> b(2^(2^n-k), k):
%p seq(seq(A(n, d-n), n=0..d), d=0..8);
%t 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_, k_] := b[2^(2^n-k), k]; Table[Table[a[n, d-n] // FullSimplify, {n, 0, d}], {d, 0, 8}] // Flatten (* _Jean-François Alcover_, Dec 11 2013, translated from Maple *)
%Y Cf. A002577, A000123, A181322, A145515.
%Y Main diagonal gives: A182135.
%K nonn,tabl
%O 0,2
%A _Alois P. Heinz_, Jan 26 2011
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