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A230579
a(n) = 2^n mod 341.
1
1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171, 1, 2, 4, 8, 16, 32, 64, 128, 256, 171
OFFSET
0,2
COMMENTS
Jeans asserts that it would have been impossible for the ancient Chinese to have discovered a case of failure for the converse of Fermat's little theorem because the smallest counterexample "(n = 341) consists of 103 figures" in base 10.
Granted that without a computer, the task of calculating 2^340 - 1 and dividing by 341 is tedious and error-prone, thus discouraging the discovery of that number as a counterexample to the so-called Chinese hypothesis.
But by instead computing just a few dozen powers of 2 modulo 341, it becomes readily apparent that the sequence of powers of 2 modulo 341 has a period of length 10 and therefore 2^340 = 1 mod 341, yet 341 = 11 * 31, which is not a prime number.
LINKS
L. Halbeisen and N. Hungerbühler, On generalised Carmichael numbers, Hardy-Ramanujan Society, 1999, 22 (2), pp. 8-22. (hal-01109575). See p. 8.
J. H. Jeans, The converse of Fermat's theorem, Messenger of Mathematics 27 (1898), p. 174.
FORMULA
a(0) = 1, a(n) = 2*a(n-1) mod 341.
EXAMPLE
a(8) = 256 because 2^8 = 256.
a(9) = 171 because 2^9 = 512 and 512 - 341 = 171.
a(10) = 1 because 2 * 171 = 342 and 342 - 341 = 1.
MATHEMATICA
PowerMod[2, Range[0, 79], 341]
LinearRecurrence[{1, -1, 1, -1, 1, -1, 1, -1, 1}, {1, 2, 4, 8, 16, 32, 64, 128, 256}, 70] (* Ray Chandler, Jul 12 2015 *)
PROG
(PARI) a(n)=lift(Mod(2, 341)^n) \\ Charles R Greathouse IV, Mar 22 2016
CROSSREFS
Cf. A206786.
Sequence in context: A243086 A087079 A252757 * A361937 A009694 A275816
KEYWORD
nonn,easy
AUTHOR
Alonso del Arte, Oct 23 2013
STATUS
approved