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a(n) = 2^n mod 341.
1

%I #23 Mar 09 2021 04:15:57

%S 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,

%T 32,64,128,256,171,1,2,4,8,16,32,64,128,256,171,1,2,4,8,16,32,64,128,

%U 256,171,1,2,4,8,16,32,64,128,256,171,1,2,4,8,16,32,64,128,256,171

%N a(n) = 2^n mod 341.

%C 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.

%C 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.

%C 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.

%H L. Halbeisen and N. Hungerbühler, <a href="https://hal.archives-ouvertes.fr/hal-01109575">On generalised Carmichael numbers</a>, Hardy-Ramanujan Society, 1999, 22 (2), pp. 8-22. (hal-01109575). See p. 8.

%H J. H. Jeans, <a href="http://oeis.org/w/images/c/c4/The_converse_of_Fermat%27s_theorem.pdf">The converse of Fermat's theorem</a>, Messenger of Mathematics 27 (1898), p. 174.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,-1,1,-1,1,-1,1).

%F a(0) = 1, a(n) = 2*a(n-1) mod 341.

%e a(8) = 256 because 2^8 = 256.

%e a(9) = 171 because 2^9 = 512 and 512 - 341 = 171.

%e a(10) = 1 because 2 * 171 = 342 and 342 - 341 = 1.

%t PowerMod[2, Range[0, 79], 341]

%t 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 *)

%o (PARI) a(n)=lift(Mod(2,341)^n) \\ _Charles R Greathouse IV_, Mar 22 2016

%Y Cf. A206786.

%K nonn,easy

%O 0,2

%A _Alonso del Arte_, Oct 23 2013