cp's OEIS Frontend

A234451 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) + 3 is prime, where phi(.) is Euler's totient function.

%I A234451 #32 Jan 23 2014 11:15:42
%S A234451 0,0,0,0,0,0,0,0,0,1,1,2,2,2,2,3,4,4,4,3,5,4,5,4,6,4,4,5,5,5,6,6,6,5,
%T A234451 6,8,7,6,5,7,8,7,10,6,7,9,7,5,5,8,6,6,7,9,3,7,10,9,3,8,6,8,6,9,9,12,5,
%U A234451 8,8,10,9,10,9,8,8,8,10,9,12,10,13,11,9,10
%N A234451 Number of ways to write n = k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) + 3 is prime, where phi(.) is Euler's totient function.
%C A234451 Conjecture: (i) a(n) > 0 for all n > 9. Also, any integer n > 13 can be written as k + m with k > 0 and m > 0 such that 2^(phi(k)/2 + phi(m)/6) - 3 is prime.
%C A234451 (ii) Each integer n > 25 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/8) + 1 (or 3*2^(phi(k)/2 + phi(m)/12) + 1 when n > 38) is prime. Also, any integer n > 14 can be written as k + m with k > 0 and m > 0 such that 3*2^(phi(k)/2 + phi(m)/12) - 1 is prime.
%C A234451 This conjecture implies that there are infinitely many primes in any of the four forms 2^n + 3, 2^n - 3, 3*2^n + 1, 3*2^n - 1.
%C A234451 We have verified the conjecture for n up to 50000.
%H A234451 Zhi-Wei Sun, <a href="/A234451/b234451.txt">Table of n, a(n) for n = 1..10000</a>
%e A234451 a(10) = 1 since 10 = 3 + 7 with 2^(phi(3)/2 + phi(7)/6) + 3 = 7 prime.
%e A234451 a(11) = 1 since 11 = 4 + 7 with 2^(phi(4)/2 + phi(7)/6) + 3 = 7 prime.
%e A234451 a(12) = 2 since 12 = 3 + 9 = 5 + 7 with 2^(phi(3)/2 + phi(9)/6) + 3 = 7 and 2^(phi(5)/2 + phi(7)/6) + 3 = 11 both prime.
%e A234451 a(769) = 1 since 769 = 31 + 738 with 2^(phi(31)/2 + phi(738)/6) + 3 = 2^(55) + 3 prime.
%e A234451 a(787) = 1 since 787 = 112 + 675 with 2^(phi(112)/2 + phi(675)/6) + 3 = 2^(84) + 3 prime.
%e A234451 a(867) = 1 since 867 = 90 + 777 with 2^(phi(90)/2 + phi(777)/6) + 3 = 2^(84) + 3 prime.
%e A234451 a(869) = 1 since 869 = 51 + 818 with 2^(phi(51)/2 + phi(818)/6) + 3 = 2^(84) + 3 prime.
%e A234451 a(913) = 1 since 913 = 409 + 504 with 2^(phi(409)/2 + phi(504)/6) + 3 = 2^(228) + 3 prime.
%e A234451 a(1085) = 1 since 1085 = 515 + 570 with 2^(phi(515)/2 + phi(570)/6) + 3 = 2^(228) + 3 prime.
%t A234451 f[n_,k_]:=2^(EulerPhi[k]/2+EulerPhi[n-k]/6)+3
%t A234451 a[n_]:=Sum[If[PrimeQ[f[n,k]],1,0],{k,1,n-1}]
%t A234451 Table[a[n],{n,1,100}]
%Y A234451 Cf. A000010, A000040, A000079, A050415, A057733, A234309, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361, A234388, A234399, A234470, A236358.
%K A234451 nonn
%O A234451 1,12
%A A234451 _Zhi-Wei Sun_, Dec 26 2013