%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 p1) then a(n) is divisible by p if (n+1)^2 == u, v (mod p1) 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)^23: n in [0..10]];
%Y Cf. A141725.
%K nonn
%O 0,2
%A _Vincenzo Librandi_, Feb 22 2014
