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Number of rooted identity trees with 2n+1 nodes.
3

%I #17 May 02 2020 16:40:12

%S 1,1,3,12,52,247,1226,6299,33209,178618,976296,5407384,30283120,

%T 171196956,975662480,5599508648,32334837886,187737500013,

%U 1095295264857,6417886638389,37752602033079,222861754454841,1319834477009635,7839314017612273,46688045740233741

%N Number of rooted identity trees with 2n+1 nodes.

%H Alois P. Heinz, <a href="/A299113/b299113.txt">Table of n, a(n) for n = 0..1253</a>

%F a(n) = A004111(2n+1).

%e a(2) = 3:

%e o o o

%e | | / \

%e o o o o

%e | / \ |

%e o o o o

%e | | |

%e o o o

%e |

%e o

%p with(numtheory):

%p b:= proc(n) option remember; `if`(n<2, n, add(b(n-k)*add(

%p b(d)*d*(-1)^(k/d+1), d=divisors(k)), k=1..n-1)/(n-1))

%p end:

%p a:= n-> b(2*n+1):

%p seq(a(n), n=0..30);

%t b[n_] := b[n] = If[n < 2, n, Sum[b[n - k]*Sum[b[d]*d*(-1)^(k/d + 1), {d, Divisors[k]}], {k, 1, n - 1}]/(n - 1)];

%t a[n_] := b[2*n + 1];

%t Array[a, 30, 0] (* _Jean-François Alcover_, May 30 2019, from Maple *)

%o (Python)

%o from sympy import divisors

%o from sympy.core.cache import cacheit

%o @cacheit

%o def b(n): return n if n<2 else sum([b(n-k)*sum([b(d)*d*(-1)**(k//d+1) for d in divisors(k)]) for k in range(1, n)])//(n-1)

%o def a(n): return b(2*n+1)

%o print([a(n) for n in range(31)]) # _Indranil Ghosh_, Mar 02 2018

%Y Bisection of A004111 (odd part).

%Y Cf. A100427, A299098.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Feb 02 2018