A167770 a(n) = prime(n)^2 modulo prime(n+1).

1, 4, 4, 5, 4, 16, 4, 16, 7, 4, 36, 16, 4, 16, 36, 36, 4, 36, 16, 4, 36, 16, 36, 64, 16, 4, 16, 4, 16, 69, 16, 36, 4, 100, 4, 36, 36, 16, 36, 36, 4, 100, 4, 16, 4, 144, 144, 16, 4, 16, 36, 4, 100, 36, 36, 36, 4, 36, 16, 4, 100, 196, 16, 4, 16, 196, 36, 100, 4, 16, 36, 64, 36, 36, 16

Offset: 1

Author

Zak Seidov, Nov 11 2009

Keywords

• nonn

Comments

Only for three cases n = 4,9,30, a(n) < (prime(n+1)-prime(n))^2 because only in these cases (prime(n+1)-prime(n))^2 > prime(n+1):

n = 4: a(4) = 5 < ((p(5)-p(4))^2 = (11-7)^2 = 16) and 16 > 11.

n = 9: a(9) = 7 < ((p(10)-p(9))^2 = (29-23)^2 = 36) and 36 > 29.

n = 30: a(30) = 69 < ((p(31)-p(30))^2 = (127-113)^2 = 196) and 196 > 127.

In all other cases, a(n) = A076821(n) = (prime(n+1)-prime(n))^2, is highly probable but not proved conjecture.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A076821 (squares of the differences between consecutive primes).

Cf. A001223 (modular square roots of this sequence).

Cf. A000040 (primes), A001248 (squares of primes).

Programs

Maple

A167770:=n->ithprime(n)^2 mod ithprime(n+1): seq(A167770(n), n=1..70); # Wesley Ivan Hurt, Oct 01 2014

Mathematica

Table[PowerMod[Prime[n], 2, Prime[n+1]], {n, 221265}]

PARI

a(n)=prime(n)^2%prime(n+1) \\ M. F. Hasler, Oct 04 2014

Formula

a(n) = prime(n)^2 modulo prime(n+1).

a(n) == A001223(n)^2 (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014