A237123 Number of ways to write n = i + j + k with 0 < i < j < k such that phi(i), phi(j) and phi(k) are all cubes, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 0, 0, 1, 2, 1, 0, 1, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1

Offset: 1

Author

Zhi-Wei Sun, Feb 03 2014

Keywords

• nonn

Comments

Conjecture: For each k = 2, 3, ... there is a positive integer s(k) such that any integer n >= s(k) can be written as i_1 + i_2 + ... + i_k with 0 < i_1 < i_2 < ... < i_k such that all those phi(i_1), phi(i_2), ..., phi(i_k) are k-th powers. In particular, we may take s(2) = 70640, s(3) = 935 and s(4) = 3273.

Examples

			a(18) = 1 since 18 = 1 + 2 + 15 with phi(1) = 1^3, phi(2) = 1^3 and phi(15) = 2^3.
a(101) = 1 since 101 = 1 + 15 + 85 with phi(1) = 1^3, phi(15) = 2^3 and phi(85) = 4^3.
a(1613) = 1 since 1613 = 192 + 333 + 1088 with phi(192) = 4^3, phi(333) = 6^3 and phi(1088) = 8^3.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..7000

Crossrefs

Cf. A000010, A000578, A039771, A233386, A236998.

Programs

Mathematica

CQ[n_]:=IntegerQ[EulerPhi[n]^(1/3)]
a[n_]:=Sum[If[CQ[i]&&CQ[j]&&CQ[n-i-j],1,0],{i,1,n/3-1},{j,i+1,(n-1-i)/2}]
Table[a[n],{n,1,70}]