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A127524 Number of unordered rooted trees where each subtree from given node has the same number of nodes.

%I A127524 #11 Feb 25 2017 12:16:57
%S A127524 1,1,2,3,5,6,11,12,20,25,42,43,81,82,150,192,287,288,563,564,982,1277,
%T A127524 2182,2183,3658,3785,7108,8659,13101,13102,27827,27828,47768,61025,
%U A127524 102355,105689,170882,170883,329651,421547,606283,606284,1193038,1193039,2158117
%N A127524 Number of unordered rooted trees where each subtree from given node has the same number of nodes.
%H A127524 Alois P. Heinz, <a href="/A127524/b127524.txt">Table of n, a(n) for n = 1..1000</a>
%F A127524 a(1) = 1; a(n+1) = Sum_{d|n} C(a(n/d) + d-1, d).
%e A127524 The tree shown below left counts, because the subtree shown on the left has 3 nodes and so does the one on the right and a similar condition holds for the subtrees. The tree shown on the right is not counted, because the subtree shown on the left has 3 nodes, while the one on the right has 4.
%e A127524 O..........O...O...O
%e A127524 |..........|....\./.
%e A127524 O...O...O..O.....O..
%e A127524 .\...\./....\....|..
%e A127524 .O...O......O...O..
%e A127524 ..\./........\./...
%e A127524 ...O..........O....
%p A127524 with(numtheory):
%p A127524 a:= proc(n) option remember; `if`(n<2, n,
%p A127524       add(binomial(a((n-1)/d)+d-1, d), d=divisors(n-1)))
%p A127524     end:
%p A127524 seq(a(n), n=1..50);  # _Alois P. Heinz_, May 16 2013
%t A127524 a[1] = 1; a[n_] := a[n] = DivisorSum[n-1, Binomial[a[(n-1)/#]+#-1, #]&]; Table[a[n], {n, 1, 50}] (* _Jean-François Alcover_, Feb 25 2017 *)
%Y A127524 Cf. A000081, A127525.
%K A127524 nonn
%O A127524 1,3
%A A127524 _Franklin T. Adams-Watters_, Jan 17 2007