cp's OEIS Frontend

A238878 a(n) = |{0 < k <= n: prime(prime(k)) - prime(k) + 1 and prime(prime(k*n)) - prime(k*n) + 1 are both prime}|.

%I A238878 #11 Mar 06 2014 09:03:48
%S A238878 1,2,3,1,1,4,3,2,5,5,3,4,2,2,3,3,5,3,1,3,4,4,2,5,2,2,7,3,2,4,4,7,4,4,
%T A238878 4,4,4,3,4,4,4,2,4,3,7,4,9,6,3,4,5,4,2,4,4,4,3,4,5,6,10,4,4,8,9,6,5,6,
%U A238878 5,7,8,9,5,2,5,7,1,7,4,5
%N A238878 a(n) = |{0 < k <= n: prime(prime(k)) - prime(k) + 1 and prime(prime(k*n)) - prime(k*n) + 1 are both prime}|.
%C A238878 Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 19, 77.
%C A238878 (ii) For any integer n > 0, there is a number k among 1, ..., n such that 2*k + 1 and prime(prime(k^2*n)) - prime(k^2*n) + 1 are both prime.
%H A238878 Zhi-Wei Sun, <a href="/A238878/b238878.txt">Table of n, a(n) for n = 1..5000</a>
%H A238878 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e A238878 a(5) = 1 since prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 and prime(prime(4*5)) - prime(4*5) + 1 = prime(71) - 71 + 1 = 353 - 70 = 283 are both prime.
%e A238878 a(77) = 1 since prime(prime(3)) - prime(3) + 1 = prime(5) - 5 + 1 = 11 - 4 = 7 and prime(prime(3*77)) - prime(3*77) + 1 = prime(1453) - 1453 + 1 = 12143 - 1452 = 10691 are both prime.
%t A238878 PQ[n_]:=PrimeQ[Prime[n]-n+1]
%t A238878 p[k_,n_]:=PQ[Prime[k]]&&PQ[Prime[k*n]]
%t A238878 a[n_]:=Sum[If[p[k,n],1,0],{k,1,n}]
%t A238878 Table[a[n],{n,1,80}]
%Y A238878 Cf. A000040, A234695, A237715, A238573, A238576, A238756, A238766, A238776, A238814, A238881.
%K A238878 nonn
%O A238878 1,2
%A A238878 _Zhi-Wei Sun_, Mar 06 2014