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%I #35 May 13 2020 06:44:16
%S 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,
%T 4,16,69,16,36,4,100,4,36,36,16,36,36,4,100,4,16,4,144,144,16,4,16,36,
%U 4,100,36,36,36,4,36,16,4,100,196,16,4,16,196,36,100,4,16,36,64,36,36,16
%N a(n) = prime(n)^2 modulo prime(n+1).
%C 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):
%C n = 4: a(4) = 5 < ((p(5)-p(4))^2 = (11-7)^2 = 16) and 16 > 11.
%C n = 9: a(9) = 7 < ((p(10)-p(9))^2 = (29-23)^2 = 36) and 36 > 29.
%C n = 30: a(30) = 69 < ((p(31)-p(30))^2 = (127-113)^2 = 196) and 196 > 127.
%C In all other cases, a(n) = A076821(n) = (prime(n+1)-prime(n))^2, is highly probable but not proved conjecture.
%H Alois P. Heinz, <a href="/A167770/b167770.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = prime(n)^2 modulo prime(n+1).
%F a(n) == A001223(n)^2 (mod A000040(n+1)). - _L. Edson Jeffery_, Oct 01 2014
%p A167770:=n->ithprime(n)^2 mod ithprime(n+1): seq(A167770(n), n=1..70); # _Wesley Ivan Hurt_, Oct 01 2014
%t Table[PowerMod[Prime[n], 2, Prime[n+1]], {n, 221265}]
%o (PARI) a(n)=prime(n)^2%prime(n+1) \\ _M. F. Hasler_, Oct 04 2014
%Y Cf. A076821 (squares of the differences between consecutive primes).
%Y Cf. A001223 (modular square roots of this sequence).
%Y Cf. A000040 (primes), A001248 (squares of primes).
%K nonn
%O 1,2
%A _Zak Seidov_, Nov 11 2009
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