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Number of even solutions to phi(k) = prime(n) - 1.
4

%I #18 Jan 21 2013 15:02:26

%S 1,2,3,2,1,4,5,2,1,1,1,5,6,2,1,1,1,5,1,1,11,1,1,4,13,2,1,1,5,4,1,1,1,

%T 1,1,1,4,2,1,1,1,5,1,17,1,1,1,1,1,1,4,1,21,1,9,1,1,1,5,5,1,1,1,1,10,1,

%U 1,13,1,3,9,1,1,1,1,1,1,7,9,4,1,7,1,23,1,1,9

%N Number of even solutions to phi(k) = prime(n) - 1.

%C a(n) >= A210501(n).

%D Alexander S. Karpenko, Lukasiewicz's Logics and Prime Numbers, Luniver Press, Beckington, 2006, pp. 52-56.

%H Arkadiusz Wesolowski, <a href="/A210500/b210500.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A058339(n) - A210501(n).

%e The set {k: phi(k) = 12} is {13, 21, 26, 28, 36, 42}. Thus, if phi(k) = prime(6) - 1, the equation has exactly four even solutions. Hence, a(6) = 4.

%t r = 87; lst1 = Table[EulerPhi[n], {n, (Prime[r] - 1)^2 + 2}]; lst2 = {}; Do[p = Prime[n]; AppendTo[lst2, Length@Select[Flatten@Position[Take[lst1, {p - 1, (p - 1)^2 + 2}], Prime[n] - 1], OddQ]], {n, r}]; lst2

%Y Cf. A000010, A032446, A058339, A066080, A210501, A210502.

%K nonn

%O 1,2

%A _Arkadiusz Wesolowski_, Jan 19 2013