A263897 Number of anagram compositions of 2n that can be formed from the compositions of n.
1, 1, 2, 6, 16, 44, 134, 414, 1290, 4102, 13306, 43718, 145176, 487384, 1651154, 5636702, 19381392, 67074420, 233483430, 817106622, 2873589060, 10151255976, 36009769278, 128231072994, 458268615966, 1643227382022, 5910606009330, 21322486928518, 77132729929864
Offset: 0
Keywords
Examples
For n=3, the compositions are [3], [1,2], [2,1], [1,1,1]. The anagram compositions of 2*3=6 are [3][3], [1,2][1,2], [1,2][2,1], [2,1][1,2], [2,1][2,1], [1,1,1][1,1,1], so there are a(3)=6 anagram compositions in all. For n=4, the individual partitions are [4], [3,1], [2,2], [2,1,1] and [1,1,1,1]. The corresponding number of compositions are 1, 2, 1, 3, and 1. The corresponding squares of these numbers are 1, 4, 1, 9, and 1. The sum of these squares is a(4)=16.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Miklos Bona, Rebecca Smith, An Involution on Involutions and a Generalization of Layered Permutations, arXiv preprint arXiv:1605.06158 [math.CO], 2016. See Theorem 4.3.
Crossrefs
First differences of A097085.
Programs
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Maple
b:= proc(n, i, p) option remember; `if`(n=0, p!^2, `if`(i<1, 0, add(b(n-i*j, i-1, p+j)/j!^2, j=0..n/i))) end: a:= n-> b(n$2, 0): seq(a(n), n=0..35); # Alois P. Heinz, Oct 29 2015
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Mathematica
b[n_, i_, p_] := b[n, i, p] = If[n == 0, p!^2, If[i<1, 0, Sum[b[n-i*j, i-1, p+j]/j!^2, {j, 0, n/i}]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Nov 05 2015, after Alois P. Heinz *)
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Python
from sympy.core.cache import cacheit from sympy import factorial as f @cacheit def b(n, i, p): return f(p)**2 if n==0 else 0 if i<1 else sum(b(n - i*j, i - 1, p + j)//f(j)**2 for j in range(n//i + 1)) def a(n): return b(n, n, 0) print([a(n) for n in range(36)]) # Indranil Ghosh, Aug 18 2017, after Maple code
Extensions
a(9)-a(28) from Alois P. Heinz, Oct 29 2015
Comments