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 A237419 4^(n+1)^2 - 3. 1
 1, 253, 262141, 4294967293, 1125899906842621, 4722366482869645213693, 316912650057057350374175801341, 340282366920938463463374607431768211453, 5846006549323611672814739330865132078623730171901 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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 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.) 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) LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..40 EXAMPLE a(0) = 4^(0+1)^2 - 3 = 1. MAPLE A237419:=n->4^(n + 1)^2 - 3; seq(A237419(n), n=0..20); # Wesley Ivan Hurt, Feb 27 2014 MATHEMATICA Table[(4^(n + 1)^2 - 3), {n, 0, 20}] PROG (MAGMA) [4^(n+1)^2-3: n in [0..10]]; CROSSREFS Cf. A141725. Sequence in context: A077695 A253880 A145628 * A243666 A243687 A263380 Adjacent sequences:  A237416 A237417 A237418 * A237420 A237421 A237422 KEYWORD nonn AUTHOR Vincenzo Librandi, Feb 22 2014 STATUS approved

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Last modified November 25 11:32 EST 2020. Contains 338623 sequences. (Running on oeis4.)