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A015910 2^n mod n. 44
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

2^n == 2 mod n iff 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

R. 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, 65-digit solution.

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

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Last modified April 25 03:36 EDT 2014. Contains 240994 sequences.