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A167770
a(n) = prime(n)^2 modulo prime(n+1).
7
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
1,2
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
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
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}]
PROG
(PARI) a(n)=prime(n)^2%prime(n+1) \\ M. F. Hasler, Oct 04 2014
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).
Sequence in context: A195783 A376178 A360997 * A080800 A253443 A140341
KEYWORD
nonn
AUTHOR
Zak Seidov, Nov 11 2009
STATUS
approved