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A233549 Number of ways to write n = p + q (q > 0) with p and (phi(p)*phi(q))^4 + 1 prime, where phi(.) is Euler's totient function (A000010).

%I A233549 #12 Dec 13 2013 03:16:51
%S A233549 0,0,1,2,2,3,3,2,3,2,1,3,1,4,3,3,4,4,6,1,1,1,4,1,2,2,4,4,1,6,7,3,4,3,
%T A233549 4,3,3,5,2,3,5,3,1,3,5,3,3,5,6,4,4,5,4,3,4,6,4,4,3,4,5,4,2,2,4,3,6,1,
%U A233549 4,2,8,9,2,5,5,4,2,3,4,3,6,1,7,5,8,5,4,4,4,10,10,6,4,8,4,3,4,6,6,2
%N A233549 Number of ways to write n = p + q (q > 0) with p and (phi(p)*phi(q))^4 + 1 prime, where phi(.) is Euler's totient function (A000010).
%C A233549 Conjecture: (i) a(n) > 0 for all n > 2.
%C A233549 (ii) If n > 2 is not equal to 26, then there is a prime p < n with (phi(p)*phi(n-p))^2 + 1 prime.
%C A233549 (iii) If n > 3 is different from 9 and 16, then there is a prime p < n with ((p+1)*phi(n-p))^2 + 1 prime.
%C A233549 Part (i) of the conjecture implies that there are infinitely many primes of the form x^4 + 1. We have verified it for n up to 10^7.
%H A233549 Zhi-Wei Sun, <a href="/A233549/b233549.txt">Table of n, a(n) for n = 1..10000</a>
%e A233549 a(11) = 1 since 11 = 2 + 9 with 2 and (phi(2)*phi(9))^4 + 1 = 6^4 + 1 = 1297 both prime.
%e A233549 a(13) = 1 since 13 = 5 + 8 with 5 and (phi(5)*phi(8))^4 + 1 = 16^4 + 1 = 65537 both prime.
%e A233549 a(258) = 1 since 258 = 167 + 91 with 167 and (phi(167)*phi(91))^4 + 1 = (166*72)^4 + 1 = 20406209352892417 both prime.
%t A233549 a[n_]:=Sum[If[PrimeQ[((Prime[k]-1)*EulerPhi[n-Prime[k]])^4+1],1,0],{k,1,PrimePi[n-1]}]
%t A233549 Table[a[n],{n,1,100}]
%Y A233549 Cf. A000010, A000040, A000068, A037896, A233542, A233544, A233547.
%K A233549 nonn
%O A233549 1,4
%A A233549 _Zhi-Wei Sun_, Dec 12 2013