

A213381


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


6



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
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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)^(m2) mod m = 0 iff m divides 2^(m2), i.e., m = 2^k for some k with k <= m2 (which is true for k >= 2).
Conjecture 2) is true: if n = 3*p2 where p is prime, then n == 1 (mod 3) so n^n == n (mod 3), and n^(p1) == 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) = (p1)/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.


MAPLE

List213381:=proc(q) local a, n; print(1);
for n from 1 to q do print(n^n mod (n+2)); od; end:
List213381(1000); # Paolo P. Lava, Apr 10 2013


MATHEMATICA

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


PROG

(Python)
print([n**n % (n+2) for n in range(99)])
(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



