A323741 a(n) = m-p where m = (2n+1)^2 and p is the largest prime < m.

2, 2, 2, 2, 8, 2, 2, 6, 2, 2, 6, 6, 2, 2, 8, 2, 2, 2, 10, 12, 2, 8, 2, 2, 8, 6, 2, 20, 12, 2, 2, 6, 6, 2, 2, 6, 2, 2, 12, 8, 6, 6, 8, 2, 8, 2, 12, 6, 10, 8, 2, 22, 2, 14, 20, 6, 6, 2, 2, 2, 8, 6, 2, 8, 2, 6, 2, 12, 2, 14, 6, 2, 8, 8, 14, 10, 2, 18, 20, 2, 8, 14, 6, 2, 10, 2, 32, 2, 12, 12, 2, 8, 6, 44, 2, 6, 14, 6, 20, 14

Offset: 1

Author

Ali Sada, Sep 03 2019

Keywords

• nonn

Comments

a(n) cannot be a square: suppose a(n) = k^2; then p=m-a(n) could be factored as (2n+k-1)*(2n-k-1); hence it would not be a prime.

Legendre's conjecture implies a(n) <= 4*n. Oppermann's conjecture implies a(n) <= 2*n. - Robert Israel, Sep 04 2019

All terms are even. - Alois P. Heinz, Sep 04 2019

Examples

			When n=4, m=81, p=79, so a(4) = 81-79 = 2.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A016754, A049711, A088572, A089166, A151799.

Programs

Maple

seq((2*n+1)^2-prevprime((2*n+1)^2),n=1..100); # Robert Israel, Sep 04 2019

Mathematica

mp[n_]:=Module[{m=(2n+1)^2},m-NextPrime[m,-1]]; Array[mp, 100] (* Harvey P. Dale, Feb 03 2022 *)

PARI

a(n) = (2*n+1)^2 - precprime((2*n+1)^2 - 1); \\ Michel Marcus, Sep 05 2019

Formula

a(n) = A049711(A016754(n)).