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

1, 253, 262141, 4294967293, 1125899906842621, 4722366482869645213693, 316912650057057350374175801341, 340282366920938463463374607431768211453, 5846006549323611672814739330865132078623730171901

Offset: 0

Author

Vincenzo Librandi, Feb 22 2014

Keywords

• nonn

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)

Examples

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..40

Crossrefs

Cf. A141725.

Programs

Magma

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

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}]