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%I #24 Jan 24 2014 01:14:20
%S 0,0,0,0,0,0,0,1,1,0,2,1,1,3,2,3,2,3,2,3,4,5,5,4,5,6,7,4,5,6,7,6,5,7,
%T 8,5,7,9,8,8,6,8,7,10,7,10,10,9,9,8,9,10,5,10,10,9,10,10,9,10,9,7,12,
%U 14,10,9,5,11,7,13,8,13,6,9,11,11,14,15,9,13
%N Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k) + phi(m)/4) - 5 is prime, where phi(.) is Euler's totient function.
%C Conjecture: a(n) > 0 for all n > 10.
%C We have verified this for n up to 50000. The conjecture implies that there are infinitely many primes of the form 2^n - 5.
%H Zhi-Wei Sun, <a href="/A234504/b234504.txt">Table of n, a(n) for n = 1..10000</a>
%e a(15) = 2 since 2^(phi(2) + phi(13)/4) - 5 = 2^4 - 5 = 11 and 2^(phi(3) + phi(12)/4) - 5 = 2^3 - 5 = 3 are both prime.
%t f[n_,k_]:=2^(EulerPhi[k]+EulerPhi[n-k]/4)-5
%t a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A000079, A050522, A156560, A234309, A234451, A234470, A234475, A234503, A236358.
%K nonn
%O 1,11
%A _Zhi-Wei Sun_, Dec 26 2013
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