cp's OEIS Frontend

A285813 Let p_1 be primes such that 2*n-p_i is prime. a(n) is the smallest i such that 2*n+(odd part of p_i-1) is prime or a(n)=0 if there is no such i.

%I A285813 #27 May 14 2017 22:51:39
%S A285813 0,1,1,0,1,1,2,1,1,2,1,3,2,1,4,3,6,1,1,1,1,2,1,3,2,1,2,2,1,3,2,10,1,1,
%T A285813 1,1,4,5,1,1,1,2,2,1,5,3,6,3,1,1,1,2,1,1,2,1,9,6,0,6,2,5,2,1,1,4,2,1,
%U A285813 7,4,4,7,1,2,8,3,7,1,2,4,1,1,1,2,2,1
%N A285813 Let p_1<p_2<... be primes such that 2*n-p_i is prime. a(n) is the smallest i such that 2*n+(odd part of p_i-1) is prime or a(n)=0 if there is no such i.
%C A285813 The sequence of positions of zeros either grows very fast or is finite. We are inclined to the latter option (cf. our comments in A284919 and in A285770). By A284967, the first three positions of zeros are {4,59,434} and, according to the last calculations by _Michel Marcus_, no more positions up to 5*10^7.
%C A285813 There are many more terms in A284919 than zeros in this sequence. The reason of this phenomenon is the following. In A284919, if n is not divisible by 3 and 2*n-3 is composite then 2*n+p is composite for every prime for which 2*n-p is prime. Indeed, for these 2*n all such primes p are in the interval (3, 2*n-3). Then either 2*n-p or 2*n+p should be divisible by 3, but 2*n-p is prime >3. So, 2*n+p is composite.
%t A285813 Flatten@ Table[FirstPosition[#, p_ /; PrimeQ@ p] /. k_ /; MissingQ@ k -> {0} &@ Map[2 n + NestWhile[#/2 &, # - 1, EvenQ@ # &] &, Select[Prime@ Range@ PrimePi[2 n - 2], PrimeQ[2 n - #] &]], {n, 86}] (* _Michael De Vlieger_, Apr 27 2017, Version 10.2 *)
%o A285813 (PARI) oddp(n) = n/2^valuation(n,2);
%o A285813 a(n) = {i = 0; forprime(p=2, 2*n, if (isprime(2*n-p), i++; if (isprime(2*n+oddp(p-1)), return(i)););); return(0);} \\ _Michel Marcus_, Apr 29 2017
%Y A285813 Cf. A284919, A284967, A285770.
%K A285813 nonn
%O A285813 1,7
%A A285813 _Vladimir Shevelev_, Apr 27 2017
%E A285813 More terms from _Michael De Vlieger_, Apr 27 2017