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Number of ways to write n = i_1 + i_2 + i_3 + i_4 + i_5 (0 < i_1 <= i_2 <= i_3 <= i_4 <= i_5) with i_1, i_2, ..., i_5 not all equal such that the product i_1*i_2*i_3*i_4*i_5 is a fifth power.
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%I #9 Oct 12 2017 10:05:33

%S 0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1,0,1,1,1,1,1,0,1,1,1,1,2,2,2,3,

%T 3,3,3,5,4,3,3,3,5,4,5,7,3,5,3,4,3,3,4,6,4,4,4,4,2,4,3,5,5,3,5,4,8,7,

%U 7,9,10,9,12,7,6,9,10,9,9,8,8,7,10,7,10,10,10,10,5,8,13,10,9,8,12,15,10,12,9,8

%N Number of ways to write n = i_1 + i_2 + i_3 + i_4 + i_5 (0 < i_1 <= i_2 <= i_3 <= i_4 <= i_5) with i_1, i_2, ..., i_5 not all equal such that the product i_1*i_2*i_3*i_4*i_5 is a fifth power.

%C Conjecture: For each k = 3, 4, ... there is a positive integer M(k) such that any integer n > M(k) can be written as i_1 + i_2 + ... + i_k with i_1, i_2, ..., i_k positive and not all equal such that the product i_1*i_2*...*i_k is a k-th power. In particular, we may take M(3) = 486, M(4) = 23, M(5) = 26, M(6) = 36 and M(7) = 31.

%C This is motivated by the conjectures in A233386 and A237049.

%H Zhi-Wei Sun, <a href="/A237050/b237050.txt">Table of n, a(n) for n = 1..300</a>

%H Tianxin Cai and Deyi Chen, <a href="https://doi.org/10.1090/S0025-5718-2013-02685-3">A new variant of the Hilbert-Waring problem</a>, Math. Comp. 82 (2013), 2333-2341.

%e a(25) = 1 since 25 = 1 + 4 + 4 + 8 + 8 with 1*4*4*8*8 = 4^5.

%t QQ[n_]:=IntegerQ[n^(1/5)]

%t a[n_]:=Sum[If[QQ[i*j*h*k*(n-i-j-h-k)],1,0],{i,1,(n-1)/5},{j,i,(n-i)/4},{h,j,(n-i-j)/3},{k,h,(n-i-j-h)/2}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000584, A233386, A236998, A237016, A237049.

%K nonn

%O 1,31

%A _Zhi-Wei Sun_, Feb 02 2014