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%I #5 Feb 02 2014 22:36:35
%S 0,0,0,0,0,0,0,0,1,0,0,0,1,0,2,0,1,2,1,0,0,2,2,1,1,1,3,4,2,3,3,1,2,3,
%T 2,3,2,2,3,2,1,0,5,3,4,3,1,2,4,1,2,3,5,7,5,6,3,4,6,7,6,7,3,8,2,7,6,4,
%U 3,8,7,6,6,2,7,5,7,2,8,4,8,6,5,7,7,9,10,5,9,7,11,3,6,7,8,8,7,5,6,5
%N Number of ordered ways to write n = i + j + k (0 < i <= j <= k) with i,j,k not all equal such that sigma(i)*sigma(j)*sigma(k) is a cube, where sigma(m) denotes the sum of all positive divisors of m.
%C Conjecture: For every k = 2, 3, ... there is a positive integer N(k) such that any integer n > N(k) can be written as n_1 + n_2 + ... + n_k with n_1, n_2, ..., n_k positive and distinct such that the product sigma(n_1)*sigma(n_2)*...*sigma(n_k) is a k-th power. In particular, we may take N(2) = 309, N(3) = 42, N(4) = 25, N(5) = 24, N(6) = 27 and N(7) = 32.
%C This is similar to the conjecture in A233386.
%H Zhi-Wei Sun, <a href="/A237049/b237049.txt">Table of n, a(n) for n = 1..1000</a>
%e a(9) = 1 since 9 = 1 + 1 + 7 with sigma(1)*sigma(1)*sigma(7) = 1*1*8 = 2^3.
%e a(41) = 1 since 41 = 2 + 6 + 33 with sigma(2)*sigma(6)*sigma(33) = 3*12*48 = 12^3.
%e a(50) = 1 since 50 = 2 + 17 + 31 with sigma(2)*sigma(17)*sigma(31) = 3*18*32 = 12^3.
%t sigma[n_]:=DivisorSigma[1,n]
%t CQ[n_]:=IntegerQ[n^(1/3)]
%t p[i_,j_,k_]:=CQ[sigma[i]*sigma[j]*sigma[k]]
%t a[n_]:=Sum[If[p[i,j,n-i-j],1,0],{i,1,(n-1)/3},{j,i,(n-i)/2}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000203, A000578, A233386, A236998, A237016.
%K nonn
%O 1,15
%A _Zhi-Wei Sun_, Feb 02 2014