cp's OEIS Frontend

A143862 Number of compositions of n such that every part is divisible by number of parts.

%I A143862 #23 Mar 03 2025 13:31:20
%S A143862 1,1,1,1,2,1,3,1,4,2,5,1,9,1,7,7,9,1,19,1,14,16,11,1,43,2,13,29,34,1,
%T A143862 56,1,51,46,17,16,130,1,19,67,139,1,105,1,142,162,23,1,315,2,151,121,
%U A143862 246,1,219,211,321,154,29,1,1021,1,31,219,488,496,495,1,594,232,834,1,1439,1
%N A143862 Number of compositions of n such that every part is divisible by number of parts.
%H A143862 Alois P. Heinz, <a href="/A143862/b143862.txt">Table of n, a(n) for n = 0..10000</a>
%F A143862 G.f.: Sum_{k>=0} x^(k^2) / (1 - x^k)^k.
%F A143862 G.f.: 1 + Sum_{n>=1} (1 + x^n)^(n-1) * x^n. - _Paul D. Hanna_, Jul 09 2019
%F A143862 a(n) = Sum_{d|n} binomial(n/d-1, d-1) for n>0 with a(0) = 1. - _Paul D. Hanna_, Apr 25 2018
%F A143862 G.f.: 1 + Sum_{n>=1} (x^n/(1-x^n))^n (conjecture). - _Joerg Arndt_, Jan 04 2024
%F A143862 For prime p, a(p) = 1, a(2*p) = p and a(p^2) = 2. - _Peter Bala_, Mar 02 2025
%o A143862 (PARI) {a(n) = if(n==0,1, sumdiv(n,d, binomial(n/d-1,d-1) ))}
%o A143862 for(n=0,50, print1(a(n),", ")) \\ _Paul D. Hanna_, Apr 25 2018
%Y A143862 Cf. A143773, A261605.
%K A143862 easy,nonn
%O A143862 0,5
%A A143862 _Vladeta Jovovic_, Sep 03 2008
%E A143862 More terms from _Franklin T. Adams-Watters_, Apr 09 2009