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A236256 a(n) = |{0 < k < n: m = phi(k) + phi(n-k)/4 is an integer with C(2*m, m) - prime(m) prime}|, where C(2*m, m) = (2*m)!/(m!)^2. 9

%I #13 Jan 24 2014 01:13:12

%S 0,0,0,0,0,1,1,0,1,2,1,1,3,1,2,2,3,3,2,5,2,2,2,4,3,3,3,2,2,3,4,5,1,5,

%T 7,5,2,4,6,7,4,3,3,4,5,6,3,3,3,5,3,4,1,5,3,0,4,2,1,3,2,4,2,5,1,4,3,5,

%U 1,5,1,2,0,2,3,1,3,4,1,2,3,3,3,2,3,2,2

%N a(n) = |{0 < k < n: m = phi(k) + phi(n-k)/4 is an integer with C(2*m, m) - prime(m) prime}|, where C(2*m, m) = (2*m)!/(m!)^2.

%C Conjecture: a(n) > 0 for all n > 410.

%C This implies that there are infinitely many positive integers m with C(2*m, m) - prime(m) prime. We have verified the conjecture for n up to 51000.

%C See A236248 for a list of known numbers m with C(2*m, m) - prime(m) prime.

%C See also A236249 for those primes of the form C(2*m, m) - prime(m).

%H Zhi-Wei Sun, <a href="/A236256/b236256.txt">Table of n, a(n) for n = 1..10000</a>

%e a(12) = 1 since phi(2) + phi(10)/4 = 1 + 1 = 2 with C(2*2, 2) - prime(2) = 6 - 3 = 3 prime.

%e a(33) = 1 since phi(1) + phi(32)/4 = 1 + 4 = 5 with C(2*5, 5) - prime(5) = 252 - 11 = 241 prime.

%e a(697) = 1 since phi(452) + phi(697-452)/4 = 224 + 42 = 266 with C(2*266, 266) - prime(266) = C(532, 266) - 1699 prime.

%t p[n_]:=IntegerQ[n]&&PrimeQ[Binomial[2n,n]-Prime[n]]

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/4

%t a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]

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

%Y Cf. A000010, A000040, A000984, A236241, A236242, A236245, A236248, A236249.

%K nonn

%O 1,10

%A _Zhi-Wei Sun_, Jan 21 2014

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