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a(n) = |{0 < m < 2*n: m is a square with 2*n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).
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%I #16 Dec 17 2013 08:46:19

%S 0,1,1,1,2,1,2,1,1,3,2,1,4,2,1,3,2,1,3,3,1,4,2,1,6,2,3,4,1,3,4,2,3,3,

%T 3,2,6,3,1,6,3,3,6,2,2,6,2,4,2,3,4,5,3,3,6,4,5,7,2,3,7,3,3,3,5,1,6,2,

%U 3,6,4,5,5,4,4,7,3,4,6,4,3,5,2,2,8,5,3,5,3,6,6,4,5,5,4,4,7,2,5,9

%N a(n) = |{0 < m < 2*n: m is a square with 2*n - 1 - phi(m) prime}|, where phi(.) is Euler's totient function (A000010).

%C Conjecture: (i) a(n) > 0 for all n > 1.

%C (ii) For any odd number 2*n - 1 > 4, there is a positive integer k < 2*n such that 2*n - 1 - phi(k) and 2*n - 1 + phi(k) are both prime.

%C By Goldbach's conjecture, 2*n > 2 could be written as p + q with p and q both prime, and hence 2*n - 1 = p + (q - 1) = p + phi(q).

%C By induction, phi(k^2) (k = 1,2,3,...) are pairwise distinct.

%H Zhi-Wei Sun, <a href="/A233867/b233867.txt">Table of n, a(n) for n = 1..10000</a>

%e a(29) = 1 since 2*29 - 1 = 37 + phi(5^2) with 37 prime.

%e a(39) = 1 since 2*39 - 1 = 71 + phi(3^2) with 71 prime.

%e a(66) = 1 since 2*66 - 1 = 89 + phi(7^2) with 89 prime.

%e a(128) = 1 since 2*128 - 1 = 223 + phi(8^2) with 223 prime.

%e a(182) = 1 since 2*182 - 1 = 331 + phi(8^2) with 331 prime.

%e a(413) = 1 since 2*413 - 1 = 823 + phi(2^2) with 823 prime.

%e a(171) = 3 since 2*171 - 1 = 233 + phi(18^2) = 257 + phi(14^2) = 293 + phi(12^2) with 233, 257, 293 all prime.

%t a[n_]:=Sum[If[PrimeQ[2n-1-EulerPhi[k^2]],1,0],{k,1,Sqrt[2n-1]}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000010, A000040, A002372, A002375, A233542, A233544, A233547, A233654, A233793, A233864.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Dec 17 2013