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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).
5

%I #12 Dec 13 2013 03:16:51

%S 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 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 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 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 Conjecture: (i) a(n) > 0 for all n > 2.

%C (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 (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 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 Zhi-Wei Sun, <a href="/A233549/b233549.txt">Table of n, a(n) for n = 1..10000</a>

%e a(11) = 1 since 11 = 2 + 9 with 2 and (phi(2)*phi(9))^4 + 1 = 6^4 + 1 = 1297 both prime.

%e a(13) = 1 since 13 = 5 + 8 with 5 and (phi(5)*phi(8))^4 + 1 = 16^4 + 1 = 65537 both prime.

%e a(258) = 1 since 258 = 167 + 91 with 167 and (phi(167)*phi(91))^4 + 1 = (166*72)^4 + 1 = 20406209352892417 both prime.

%t a[n_]:=Sum[If[PrimeQ[((Prime[k]-1)*EulerPhi[n-Prime[k]])^4+1],1,0],{k,1,PrimePi[n-1]}]

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

%Y Cf. A000010, A000040, A000068, A037896, A233542, A233544, A233547.

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Dec 12 2013