A237049 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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

• nonn

Comments

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.

This is similar to the conjecture in A233386.

Examples

			 a(9) = 1 since 9 = 1 + 1 + 7 with sigma(1)*sigma(1)*sigma(7) = 1*1*8 = 2^3.
a(41) = 1 since 41 = 2 + 6 + 33 with sigma(2)*sigma(6)*sigma(33) = 3*12*48 = 12^3.
a(50) = 1 since 50 = 2 + 17 + 31 with sigma(2)*sigma(17)*sigma(31) = 3*18*32 = 12^3.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000203, A000578, A233386, A236998, A237016.

Programs

Mathematica

sigma[n_]:=DivisorSigma[1,n]
CQ[n_]:=IntegerQ[n^(1/3)]
p[i_,j_,k_]:=CQ[sigma[i]*sigma[j]*sigma[k]]
a[n_]:=Sum[If[p[i,j,n-i-j],1,0],{i,1,(n-1)/3},{j,i,(n-i)/2}]
Table[a[n],{n,1,100}]