



0, 0, 2, 0, 2, 4, 2, 0, 8, 4, 2, 4, 2, 4, 8, 0, 2, 10, 2, 16, 8, 4, 2, 16, 7, 4, 26, 16, 2, 4, 2, 0, 8, 4, 18, 28, 2, 4, 8, 16, 2, 22, 2, 16, 17, 4, 2, 16, 30, 24, 8, 16, 2, 28, 43, 32, 8, 4, 2, 16, 2, 4, 8, 0, 32, 64, 2, 16, 8, 44, 2, 64, 2, 4, 68, 16, 18, 64, 2, 16, 80, 4, 2, 64
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OFFSET

1,3


COMMENTS

2^n == 2 mod n if and only if n is a prime or a member of A001567 [Guy].  N. J. A. Sloane, Mar 22 2012
Known solutions to 2^n == 3 (mod n) are given in A050259.
This sequence is conjectured to include every integer k >= 0 except k = 1. A036236 includes a proof that k = 1 is not in this sequence, and n = A036236(k) solves a(n) = k for all other 0 <= k <= 1000.  David W. Wilson, Oct 11 2011


REFERENCES

Richard K. Guy, Unsolved Problems in Number Theory, F10.


LINKS

T. D. Noe, Table of n, a(n) for n=1..10000
Albert Frank, International Contest Of Logical Sequences, 2002  2003. Item 4
Albert Frank, Solutions of International Contest Of Logical Sequences, 2002  2003.
Peter L. Montgomery, 65digit solution.


FORMULA

a(2^k) = 0.  Alonso del Arte, Nov 10 2014


EXAMPLE

a(7) = 2 because 2^7 = 128 = 2 mod 7.
a(8) = 0 because 2^8 = 256 = 0 mod 8.
a(9) = 8 because 2^9 = 512 = 8 mod 9.


MAPLE

A015910 := n> modp(2 &^ n, n) ; # Zerinvary Lajos, Feb 15 2008


MATHEMATICA

Table[PowerMod[2, n, n], {n, 85}]


PROG

(Maxima) makelist(power_mod (2, n, n), n, 1, 84); /* Bruno Berselli, May 20 2011 */
(PARI) a(n)=lift(Mod(2, n)^n) \\ Charles R Greathouse IV, Jul 15 2011


CROSSREFS

Cf. A036236, A015911, A015921, A001567.
Sequence in context: A144182 A037036 A055947 * A182256 A164993 A223487
Adjacent sequences: A015907 A015908 A015909 * A015911 A015912 A015913


KEYWORD

nonn


AUTHOR

Robert G. Wilson v


STATUS

approved



