A293621 Numbers k such that (2*k)^2 + 1 and (2*k+2)^2 + 1 are both primes.

1, 2, 7, 12, 27, 62, 102, 192, 232, 317, 322, 357, 547, 572, 587, 622, 637, 657, 687, 782, 807, 837, 842, 982, 1027, 1042, 1047, 1202, 1227, 1267, 1332, 1417, 1462, 1567, 1652, 1767, 1877, 1887, 2012, 2077, 2087, 2182, 2302, 2307, 2367, 2392, 2397, 2477, 2507

Offset: 1

Author

Amiram Eldar, Oct 13 2017

Keywords

• nonn

Comments

Sierpiński proved that under Schinzel's hypothesis H this sequence is infinite. He gives the 24 terms below 10^3.

Sierpiński noted that the only triple of consecutive primes of the form (2n)^2 + 1 are for n = 1 (i.e., 1 and 2 are the only consecutive terms in this sequence), since every triple of consecutive terms contains at least one term which is divisible by 5.

Examples

			1 is in the sequence since (2*1)^2 + 1 = 5 and (2*1+2)^2 + 1 = 17 are both primes.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Wacław Sierpiński, Remarque sur la distribution de nombres premiers, Matematički Vesnik, Vol. 2(17), Issue 31 (1965), pp. 77-78.
• Eric Weisstein's World of Mathematics, Near-Square Prime.
• Wikipedia, Schinzel's hypothesis H.

Crossrefs

Subsequence of A001912.

Cf. A002496, A002522, A005574, A053755, A096012.

Programs

Mathematica

Select[Range[10^4], AllTrue[{(2#)^2+1, (2#+2)^2+1}, PrimeQ] &]

PARI

isok(n) = isprime((2*n)^2 + 1) && isprime((2*n+2)^2 + 1); \\ Michel Marcus, Oct 13 2017

Formula

a(n) = A096012(n)/2. - Amiram Eldar, Feb 24 2020