

A213859


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


3



1, 2, 0, 3, 4, 4, 0, 2, 6, 6, 4, 7, 8, 2, 0, 9, 16, 10, 4, 2, 12, 12, 16, 8, 14, 20, 4, 15, 16, 16, 0, 2, 18, 22, 16, 19, 20, 2, 24, 21, 16, 22, 4, 38, 24, 24, 16, 32, 6, 2, 4, 27, 34, 52, 8, 2, 30, 30, 4, 31, 32, 2, 0, 8, 16, 34, 4, 2, 46, 36, 16, 37, 38, 17
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OFFSET

0,2


COMMENTS

Conjectures:
1. Indices of zeros: 2^(x+2)2, x >= 0.
2. There are infinitely many n's such that a(n)=n.
3. Every integer k >= 0 appears in a(n) at least once.
4. Every k >= 0 appears in a(n) infinitely many times.


LINKS

Table of n, a(n) for n=0..73.


FORMULA

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


MATHEMATICA

Table[PowerMod[2, n, n+2], {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)


PROG

(Python)
print([2**n % (n+2) for n in range(222)])


CROSSREFS

Cf. A015910, A066606, A062173.
Sequence in context: A287016 A285722 A274441 * A330492 A350534 A101336
Adjacent sequences: A213856 A213857 A213858 * A213860 A213861 A213862


KEYWORD

nonn,easy


AUTHOR

Alex Ratushnyak, Jun 22 2012


STATUS

approved



