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A238756 Number of ordered ways to write n = k + m (k > 0 and m > 0) such that 2*k + 1, prime(prime(k)) - prime(k) + 1 and prime(prime(m)) - prime(m) + 1 are all prime. 3

%I #12 Mar 05 2014 03:17:46

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

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

%U 9,7,10,9,7,7,6,9,10,4,5,5

%N Number of ordered ways to write n = k + m (k > 0 and m > 0) such that 2*k + 1, prime(prime(k)) - prime(k) + 1 and prime(prime(m)) - prime(m) + 1 are all prime.

%C Conjecture: a(n) > 0 for all n > 1.

%C We have verified this for n up to 10^7.

%C The conjecture suggests that there are infinitely many primes p with 2*pi(p) + 1 and prime(p) - p + 1 both prime.

%H Zhi-Wei Sun, <a href="/A238756/b238756.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/A1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.

%e a(6) = 2 since 6 = 2 + 4 with 2*2 + 1 = 5, prime(prime(2)) - prime(2) + 1 = prime(3) - 3 + 1 = 3 and prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 all prime, and 6 = 3 + 3 with 2*3 + 1 = 7 and prime(prime(3)) - prime(3) + 1 = prime(5) - 5 + 1 = 11 - 4 = 7 both prime.

%t p[k_]:=PrimeQ[Prime[Prime[k]]-Prime[k]+1]

%t a[n_]:=Sum[If[PrimeQ[2k+1]&&p[k]&&p[n-k],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,80}]

%Y Cf. A000040, A218829, A234694, A234695, A235189, A236832, A237413, A238134.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Mar 05 2014

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Last modified April 25 09:08 EDT 2024. Contains 371964 sequences. (Running on oeis4.)