A213381 a(n) = n^n mod (n+2).

1, 1, 0, 2, 4, 3, 0, 7, 6, 5, 4, 6, 8, 13, 0, 8, 16, 9, 4, 19, 12, 11, 16, 17, 14, 7, 4, 14, 16, 15, 0, 31, 18, 13, 16, 18, 20, 37, 24, 20, 16, 21, 4, 7, 24, 23, 16, 17, 6, 49, 4, 26, 34, 3, 8, 55, 30, 29, 4, 30, 32, 61, 0, 57, 16, 33, 4, 67, 46, 35, 16, 36, 38

Offset

0,4

Comments

Conjectures:

1. Indices of zeros: 2^(x+2)-2, x >= 0.

2. a(n)=n if n is in A176003.

3. Every integer k >= 0 appears in a(n) at least once.

4. Every k >= 0 appears in a(n) infinitely many times.

From Robert Israel, May 05 2015: (Start)

Conjecture 1) is true: with m = n+2, a(n) = (-2)^(m-2) mod m = 0 iff m divides 2^(m-2), i.e., m = 2^k for some k with k <= m-2 (which is true for k >= 2).

Conjecture 2) is true: if n = 3*p-2 where p is prime, then n == 1 (mod 3) so n^n == n (mod 3), and n^(p-1) == 1 (mod p) so n^n == n (mod p), and therefore (if p <> 3) n^n == n (mod 3*p). A separate computation verifies the case p=3.

If p is an odd prime, then a(p+2) = (p-1)/2. (End)

Links

Paolo P. Lava, Table of n, a(n) for n = 0..1000

Formula

a(n) = (n^n) mod (n+2).

a(n) = (-2)^n mod (n+2). - Robert Israel, May 05 2015

Example

a(5) = 5^5 mod 7 = 3125 mod 7 = 3.

Mathematica

a[n_] := PowerMod[-2, n, n+2];
a /@ Range[0, 100] (* Jean-François Alcover, Jun 04 2020 *)

Prog

(PARI) a(n) = lift(Mod(n, n+2)^n); \\ Michel Marcus, Jun 04 2020

Crossrefs

Cf. A000312.

Sequence in context: A048644 A246713 A106137 * A127651 A019641 A085008

Adjacent sequences: A213378 A213379 A213380 * A213382 A213383 A213384

Keyword

nonn

Author

Alex Ratushnyak, Jun 10 2012

Status

approved