A299113 Number of rooted identity trees with 2n+1 nodes.
1, 1, 3, 12, 52, 247, 1226, 6299, 33209, 178618, 976296, 5407384, 30283120, 171196956, 975662480, 5599508648, 32334837886, 187737500013, 1095295264857, 6417886638389, 37752602033079, 222861754454841, 1319834477009635, 7839314017612273, 46688045740233741
Offset: 0
Keywords
Examples
a(2) = 3: o o o | | / \ o o o o | / \ | o o o o | | | o o o | o
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1253
Programs
-
Maple
with(numtheory): b:= proc(n) option remember; `if`(n<2, n, add(b(n-k)*add( b(d)*d*(-1)^(k/d+1), d=divisors(k)), k=1..n-1)/(n-1)) end: a:= n-> b(2*n+1): seq(a(n), n=0..30);
-
Mathematica
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)]; a[n_] := b[2*n + 1]; Array[a, 30, 0] (* Jean-François Alcover, May 30 2019, from Maple *)
-
Python
from sympy import divisors from sympy.core.cache import cacheit @cacheit 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) def a(n): return b(2*n+1) print([a(n) for n in range(31)]) # Indranil Ghosh, Mar 02 2018
Formula
a(n) = A004111(2n+1).