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A236241 a(n) = |{0 < k < n: m = phi(k) + phi(n-k)/8 is an integer with C(2*m, m) + prime(m) prime}|, where C(2*m, m) = (2*m)!/(m!)^2, and phi(.) is Euler's totient function. 10

%I #19 Apr 06 2014 04:12:40

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

%T 3,2,4,4,6,3,5,8,9,6,6,4,5,5,4,5,6,6,4,4,4,10,9,7,4,4,5,7,2,2,3,7,7,5,

%U 7,6,7,5,4,7,5,5,3,8,6,4,6,5,8,9,5,4,3

%N a(n) = |{0 < k < n: m = phi(k) + phi(n-k)/8 is an integer with C(2*m, m) + prime(m) prime}|, where C(2*m, m) = (2*m)!/(m!)^2, and phi(.) is Euler's totient function.

%C Conjecture: a(n) > 0 for every n = 20, 21, ... .

%C We have verified this for n up to 75000.

%C The conjecture implies that there are infinitely many primes of the form C(2*m, m) + prime(m).

%C See A236245 for primes of the form C(2*m, m) + prime(m). See also A236242 for a list of known numbers m with C(2*m, m) + prime(m) prime.

%H Zhi-Wei Sun, <a href="/A236241/b236241.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014

%e a(20) = 1 since phi(5) + phi(15)/8 = 4 + 1 = 5 with C(2*5,5) + prime(5) = 252 + 11 = 263 prime.

%e a(330) = 1 since phi(211) + phi(330-211)/8 = 210 + 96/8 = 222 with C(2*222,222) + prime(222) = C(444,222) + 1399 prime.

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

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/8

%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, A236242, A236245, A236248, A236249, A236256.

%K nonn

%O 1,22

%A _Zhi-Wei Sun_, Jan 20 2014

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