cp's OEIS Frontend

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

%I A237523 #22 Feb 09 2014 09:31:42
%S A237523 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,
%T A237523 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,
%U A237523 9,7,5,8,5,9,8,6,4,5,6,11,5
%N A237523 a(n) = |{0 < k < n/2: phi(k*(n-k)) + 1 is a square}|, where phi(.) is Euler's totient function.
%C A237523 Conjecture: (i) a(n) > 0 for all n > 7, and a(n) = 1 only for n = 8, 9, 13, 15, 20, 132.
%C A237523 (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.
%C A237523 See also A237524 for a similar conjecture involving higher powers.
%H A237523 Zhi-Wei Sun, <a href="/A237523/b237523.txt">Table of n, a(n) for n = 1..10000</a>
%e A237523 a(8) = 1 since phi(3*5) + 1 = 8 + 1 = 3^2.
%e A237523 a(9) = 1 since phi(4*5) + 1 = 8 + 1 = 3^2.
%e A237523 a(13) = 1 since phi(3*10) + 1 = 8 + 1 = 3^2.
%e A237523 a(15) = 1 since phi(7*8) + 1 = 24 + 1 = 5^2.
%e A237523 a(20) = 1 since phi(6*14) + 1 = 24 + 1 = 5^2.
%e A237523 a(132) = 1 since phi(46*(132-46)) + 1 = 1848 + 1 = 43^2.
%t A237523 SQ[n_]:=IntegerQ[Sqrt[n]]
%t A237523 s[n_]:=SQ[EulerPhi[n]+1]
%t A237523 a[n_]:=Sum[If[s[k(n-k)],1,0],{k,1,(n-1)/2}]
%t A237523 Table[a[n],{n,1,80}]
%Y A237523 Cf. A000010, A000290, A234246, A236998, A237497, A237524.
%K A237523 nonn
%O A237523 1,10
%A A237523 _Zhi-Wei Sun_, Feb 08 2014