
COMMENTS

There are no primes of this form.
Subsequence of A141725.
From Michel Lagneau, Feb 24 2014: (Start)
If n == 1, 2, 6, 7 (mod 10) then a(n) is divisible by 11.
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)
From Michel Lagneau, Feb 25 2014: (Start)
Generalization:
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.)
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, ...}.
Examples:
If n == 1,8,12,19 (mod 22) then a(n) is divisible by 23
If n == 6,10,24,28 (mod 36) then a(n) is divisible by 37
If n == 4,23,33,52 (mod 58) then a(n) is divisible by 59. (End)
