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a(n) = |{0 < k < n: k*prime(n-k) + 1 is prime}|
5

%I #16 Aug 05 2019 09:26:33

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

%T 7,6,5,8,6,4,5,7,8,4,6,7,6,10,7,4,8,11,9,11,6,5,5,8,8,10,5,7,10,10,11,

%U 7,6,6,12,6,10,11,6,11,7,11,8,12,7,7,9,13,9,10,14,7,13,8,10,11,9,14,10,14,17,14,13,8,12,12

%N a(n) = |{0 < k < n: k*prime(n-k) + 1 is prime}|

%C Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 2, k*prime(n-k) - 1 (or k^2*prime(n-k) - 1) is prime for some 0 < k < n.

%C (ii) Let n > 3 be an integer. Then k + prime(n-k) is prime for some 0 < k < n. Also, if n is not equal to 13, then k^2 + prime(n-k)^2 is prime for some 0 < k < n.

%H Zhi-Wei Sun, <a href="/A233296/b233296.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1312.1166">On a^n+ bn modulo m</a>, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014-2016.

%e a(17) = 1 since 17 = 14 + 3 with 14*prime(3) + 1 = 14*5 + 1 = 71 prime.

%e a(19) = 1 since 19 = 18 + 1 with 18*prime(1) + 1 = 18*2 + 1 = 37 prime.

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

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

%Y Cf. A000040, A232465, A232502, A233150, A233183, A233204, A233206.

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Dec 07 2013