%I #23 Feb 04 2024 08:58:12
%S 0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,3,0,0,0,1,0,3,0,0,0,0,
%T 0,0,0,0,0,2,0,2,0,0,0,0,0,5,0,0,0,0,0,2,0,1,0,0,0,17,0,0,0,0,0,2,0,0,
%U 0,1,0,0,0,0,0,0,0,2,0,3,0,0,0,14,0,0,0,1,0,13,0,0,0,0,0,11,0,0,0,0,0,2,0,1
%N a(n) is the number of ways of partitioning the divisors of n into two disjoint sets with equal sum.
%C a(n)=0 for deficient numbers n (A005100), but the converse is not true, as 18 is abundant (A005101) and a(18)=0, see A083211;
%C a(n)=1 for perfect numbers n (A000396), see A083209 for all numbers with a(n)=1;
%C records: A083213(k)=a(A083212(k)).
%C In order that a(n)>0, the sum of divisors of n must be even by definition: a(n) = half the number of partitions of A000203(n)/2 into divisors of n, see formula. [_Reinhard Zumkeller_, Jul 10 2010]
%H T. D. Noe, <a href="/A083206/b083206.txt">Table of n, a(n) for n=1..10000</a> (terms 1..800 from Reinhard Zumkeller).
%H Reinhard Zumkeller, <a href="/A083206/a083206.txt">Illustration of initial terms</a>
%F a(n) = if sigma(n) mod 2 = 1 then 0 else f(n,sigma(n)/2,2), where sigma=A000203 and f(n,m,k) = if k<=m then f(n,m,k+1)+f(n,m-k,k+1)*0^(n mod k) else 0^m, cf. A033630, also using f. [_Reinhard Zumkeller_, Jul 10 2010]
%F a(n) is half the coefficient of x^0 in Product_{d|n} (x^d + 1/x^d). - _Ilya Gutkovskiy_, Feb 04 2024
%e a(24)=3: 1+2+3+4+8+12=6+24, 1+3+6+8+12=2+4+24, 4+6+8+12=1+2+3+24.
%t a[n_] := (s = DivisorSigma[1, n]; If[Mod[s, 2] == 1, 0, f[n, s/2, 2]]); f[n_, m_, k_] := f[n, m, k] = If[k <= m, f[n, m, k+1] + f[n, m-k, k+1]*Boole[Mod[n, k] == 0], Boole[m == 0]]; Array[a, 105] (* _Jean-François Alcover_, Jul 29 2015, after _Reinhard Zumkeller_ *)
%Y Cf. A083207, A083208, A083209, A083210, A083211, A000005, A000203, A082729, A033630, A065205.
%K nonn
%O 1,24
%A _Reinhard Zumkeller_, Apr 22 2003