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A234963 Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function. 2

%I #17 Jan 06 2014 12:19:54

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

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

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

%N Number of ways to write n = k + m with k > 0 and m > 2 such that C(2*sigma(k) + phi(m), sigma(k) + phi(m)/2) - 1 is prime, where sigma(k) is the sum of all positive divisors of k and phi(.) is Euler's totient function.

%C Conjecture: a(n) > 0 for all n >= 180.

%C Clearly, this implies that there are infinitely many primes of the form C(2*n,n) - 1. We have verified the conjecture for n up to 10000.

%C Note that every n = 400, ..., 9123 can be written as k + m with k > 0 and m > 0 such that f(k, m) = sigma(k) + phi(m) is even and C(f(k, m) + 2, f(k, m)/2 + 1) + 1 is prime, but this fails for n = 9124.

%H Zhi-Wei Sun, <a href="/A234963/b234963.txt">Table of n, a(n) for n = 1..2000</a>

%e a(5) = 1 since 5 = 1 + 4 with C(2*sigma(1) + phi(4), sigma(1) + phi(4)/2) - 1 = C(4, 2) - 1 = 5 prime.

%e a(28) = 1 since 28 = 2 + 26 with C(2*sigma(2) + phi(26), sigma(2) + phi(26)/2) - 1 = C(18, 9) - 1 = 48619 prime.

%t sigma[n_] := DivisorSigma[1, n];

%t f[n_,k_] := Binomial[2*sigma[k] + EulerPhi[n-k], sigma[k] + EulerPhi[n-k]/2] - 1;

%t a[n_] := Sum[If[PrimeQ[f[n,k]], 1, 0], {k, 1, n-3}];

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

%Y Cf. A000010, A000040, A000203, A000984, A066726, A066699, A232270, A233544, A234451, A234470, A234503, A234514, A234567, A234615.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Jan 01 2014

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