A237523 a(n) = |{0 < k < n/2: phi(k*(n-k)) + 1 is a square}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 1, 2, 1, 2, 5, 4, 4, 1, 3, 3, 3, 2, 4, 4, 4, 2, 4, 5, 6, 5, 3, 3, 3, 6, 5, 4, 4, 6, 6, 2, 6, 6, 6, 2, 6, 5, 5, 2, 4, 4, 7, 7, 4, 3, 5, 5, 9, 5, 5, 3, 5, 2, 3, 10, 10, 9, 7, 5, 8, 5, 9, 8, 6, 4, 5, 6, 11, 5

Offset: 1

Author

Zhi-Wei Sun, Feb 08 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 7, and a(n) = 1 only for n = 8, 9, 13, 15, 20, 132.

(ii) If n > 5 is not among 10, 15, 20, 60, 105, then phi(k*(n-k)) is a square for some 0 < k < n/2.

See also A237524 for a similar conjecture involving higher powers.

Examples

			a(8) = 1 since phi(3*5) + 1 = 8 + 1 = 3^2.
a(9) = 1 since phi(4*5) + 1 = 8 + 1 = 3^2.
a(13) = 1 since phi(3*10) + 1 = 8 + 1 = 3^2.
a(15) = 1 since phi(7*8) + 1 = 24 + 1 = 5^2.
a(20) = 1 since phi(6*14) + 1 = 24 + 1 = 5^2.
a(132) = 1 since phi(46*(132-46)) + 1 = 1848 + 1 = 43^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000010, A000290, A234246, A236998, A237497, A237524.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_]:=SQ[EulerPhi[n]+1]
a[n_]:=Sum[If[s[k(n-k)],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,80}]