A237050 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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

• nonn

Comments

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.

This is motivated by the conjectures in A233386 and A237049.

Examples

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..300
• Tianxin Cai and Deyi Chen, A new variant of the Hilbert-Waring problem, Math. Comp. 82 (2013), 2333-2341.

Crossrefs

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

Programs

Mathematica

QQ[n_]:=IntegerQ[n^(1/5)]
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}]
Table[a[n],{n,1,100}]