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A237524 Number of ordered ways to write n = i + j + k with 0 < i <= j <= k such that phi(i*j*k) is a cube, where phi(.) is Euler's totient function.

%I A237524 #9 Feb 09 2014 09:32:10
%S A237524 0,0,1,1,0,0,0,1,4,3,2,2,1,1,1,1,2,5,2,3,2,3,6,5,4,4,4,5,4,5,4,6,6,5,
%T A237524 5,9,6,10,8,7,7,5,5,4,11,10,8,10,5,8,8,10,10,8,11,16,11,13,14,16,18,
%U A237524 19,18,16,24,19,21,18,15,21,9,15,14,13,15,18,19,20,15,19
%N A237524 Number of ordered ways to write n = i + j + k with 0 < i <= j <= k such that phi(i*j*k) is a cube, where phi(.) is Euler's totient function.
%C A237524 Conjecture: For each k = 3, 4, ..., any integer n > 2*k + 1 can be written as a sum of k positive integers n_1, n_2, ..., n_k such that phi(n_1*n_2*...*n_k) is a k-th power.
%C A237524 Note that 2*k + 2 = (k-1)*2 + 4 with phi(2^(k-1)*4) = 2^k.
%C A237524 See also A237523 for a similar conjecture with k = 2.
%H A237524 Zhi-Wei Sun, <a href="/A237524/b237524.txt">Table of n, a(n) for n = 1..1500</a>
%e A237524 a(4) = 1 since 4 = 1 + 1 + 2 with phi(1*1*2) = 1^3.
%e A237524 a(13) = 1 since 13 = 1 + 2 + 10 with phi(1*2*10) = 2^3.
%e A237524 a(16) = 1 since 16 = 4 + 4 + 8 with phi(4*4*8) = phi(2^7) = 4^3.
%t A237524 CQ[n_]:=IntegerQ[n^(1/3)]
%t A237524 q[n_]:=CQ[EulerPhi[n]]
%t A237524 a[n_]:=Sum[If[q[i*j(n-i-j)],1,0],{i,1,n/3},{j,i,(n-i)/2}]
%t A237524 Table[a[n],{n,1,80}]
%Y A237524 Cf. A000010, A000578, A233386, A237050, A237123, A237523.
%K A237524 nonn
%O A237524 1,9
%A A237524 _Zhi-Wei Sun_, Feb 09 2014