A212844 a(n) = 2^(n+2) mod n.

0, 0, 2, 0, 3, 4, 1, 0, 5, 6, 8, 4, 8, 2, 2, 0, 8, 4, 8, 4, 11, 16, 8, 16, 3, 16, 23, 8, 8, 16, 8, 0, 32, 16, 2, 4, 8, 16, 32, 24, 8, 4, 8, 20, 23, 16, 8, 16, 22, 46, 32, 12, 8, 4, 7, 16, 32, 16, 8, 4, 8, 16, 32, 0, 63, 58, 8, 64, 32, 36, 8, 40, 8, 16, 47

Offset: 1

Author

Alex Ratushnyak, Jul 22 2012

Keywords

• nonn

Comments

Also a(n) = x^x mod (x-2), where x = n+2.

Indices of 0's: 2^k, k>=0.

Indices of 1's: 7, 511, 713, 11023, 15553, 43873, 81079, 95263, 323593, 628153, 2275183, 6520633, 6955513, 7947583, 10817233, 12627943, 14223823, 15346303, 19852423, 27923663, 28529473, ...

Conjecture: every integer k >= 0 appears in a(n) at least once.

Each number below 69 appears at least once. Some large first occurrences: a(39806401) = 25, a(259274569) = 33, a(10571927) = 55, a(18039353) = 81. - Charles R Greathouse IV, Jul 21 2015

Examples

			a(3) = 2^5 mod 3 = 32 mod 3 = 2.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000312, A015910, A062173, A112983, A213381, A213859.

Programs

Maple

A212844 := proc(n)
    modp( 2&^ (n+2),n) ;
end proc: # R. J. Mathar, Jul 24 2012

Mathematica

Table[PowerMod[2, n+2, n], {n, 79}] (* Alonso del Arte, Jul 22 2012 *)

PARI

A212844(n)=lift(Mod(2,n)^(n+2)) \\ M. F. Hasler, Jul 23 2012

Python

for n in range(1,99):
    print(2**(n+2) % n, end=',')

Formula

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