A233386 Number of ways to write n = i + j + k with 0 < i <= j <= k and i, j, k not all equal such that phi(i)*phi(j)*phi(k) is a cube.

0, 0, 0, 1, 1, 0, 0, 0, 1, 3, 2, 3, 4, 3, 4, 4, 4, 5, 3, 3, 1, 2, 6, 5, 5, 6, 7, 8, 8, 7, 8, 13, 7, 9, 4, 6, 8, 10, 7, 11, 14, 12, 8, 9, 10, 14, 12, 9, 9, 8, 8, 11, 8, 9, 19, 14, 12, 9, 11, 19, 12, 19, 10, 15, 13, 22, 18, 27, 22, 31, 20, 22, 18, 25, 25, 24, 18, 22, 19, 21, 24, 22, 30, 31, 35, 25, 28, 32, 23, 27, 28, 29, 23, 24, 30, 30, 29, 30, 33, 31

Offset: 1

Author

Zhi-Wei Sun, Feb 02 2014

Keywords

• nonn

Comments

Conjecture: For each k = 3, 4, ..., any integer n >= 3*k can be written as n_1 + n_2 + ... + n_k with n_1, n_2, ..., n_k positive and not all equal such that the product phi(n_1)*phi(n_2)*...*phi(n_k) is a k-th power.

We have verified this conjecture with k = 3 for n up to 10^5 and with k = 4, 5, 6 for n up to 30000.

See also A236998 for a similar conjecture with k = 2.

Examples

			a(9) = 1 since 9 = 1 + 3 + 5 with phi(1)*phi(3)*phi(5) = 1*2*4 = 2^3.
a(21) = 1 since 21 = 5 + 8 + 8 with phi(5)*phi(8)*phi(8) = 4*4*4 = 4^3.
		

Links

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

Crossrefs

Cf. A000010, A000578, A236998.

Programs

Mathematica

CQ[n_]:=IntegerQ[n^(1/3)]
p[i_,j_,k_]:=CQ[EulerPhi[i]*EulerPhi[j]*EulerPhi[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}]