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 A237419 4^(n+1)^2 - 3. 1

%I

%S 1,253,262141,4294967293,1125899906842621,4722366482869645213693,

%T 316912650057057350374175801341,

%U 340282366920938463463374607431768211453,5846006549323611672814739330865132078623730171901

%N 4^(n+1)^2 - 3.

%C There are no primes of this form.

%C Subsequence of A141725.

%C From _Michel Lagneau_, Feb 24 2014: (Start)

%C If n == 1, 2, 6, 7 (mod 10) then a(n) is divisible by 11.

%C Proof: (n+1)^2 == 4, 9 (mod 10) and the property 4^(10k) == 1 (mod 11) gives 4^(4+10k) = 4^4*4^(10k) = 256*4^(10k) == 3*1 (mod 11) and 4^(9+10k) = 4^9*4^(10k) = 262144*4^(10k) == 3*1 (mod 11). Hence 4^(n+1)^2 - 3 == 0 (mod 11). (End)

%C From _Michel Lagneau_, Feb 25 2014: (Start)

%C Generalization:

%C If n == r1,r2,r3,r4 (mod p-1) then a(n) is divisible by p if (n+1)^2 == u, v (mod p-1) with the property 4^u and 4^v == 3 mod p. (It is possible to find u = v, for example with p = 37.)

%C The sequence of the primes p having this property is {11, 23, 37, 59, 83, 263, 359, 383, 467, 479, 503, 563, 587, 839, 853, 887, 983, 1019, 1187, 1223, 1319, 1523, 1823, 1871, 2027, 2039, 2063, ...}.

%C Examples:

%C If n == 1,8,12,19 (mod 22) then a(n) is divisible by 23

%C If n == 6,10,24,28 (mod 36) then a(n) is divisible by 37

%C If n == 4,23,33,52 (mod 58) then a(n) is divisible by 59. (End)

%H Vincenzo Librandi, <a href="/A237419/b237419.txt">Table of n, a(n) for n = 0..40</a>

%e a(0) = 4^(0+1)^2 - 3 = 1.

%p A237419:=n->4^(n + 1)^2 - 3; seq(A237419(n), n=0..20); # _Wesley Ivan Hurt_, Feb 27 2014

%t Table[(4^(n + 1)^2 - 3), {n, 0, 20}]

%o (MAGMA) [4^(n+1)^2-3: n in [0..10]];

%Y Cf. A141725.

%K nonn

%O 0,2

%A _Vincenzo Librandi_, Feb 22 2014

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Last modified January 19 15:32 EST 2021. Contains 340270 sequences. (Running on oeis4.)