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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)
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OFFSET
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1,3
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COMMENTS
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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
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, F10.
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LINKS
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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.
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MAPLE
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A015910 := n-> modp(2 &^ n, n) ; # Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 15 2008
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MATHEMATICA
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Table[PowerMod[2, n, n], {n, 85} ]
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PROG
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(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
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CROSSREFS
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Cf. A036236, A015911, A015921, A001567.
Sequence in context: A144182 A037036 A055947 * A182256 A164993 A223487
Adjacent sequences: A015907 A015908 A015909 * A015911 A015912 A015913
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KEYWORD
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nonn,changed
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AUTHOR
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Robert G. Wilson v
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STATUS
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approved
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