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%I #14 Apr 11 2024 09:37:22
%S 0,3,1,7,7,1,13,7,7,1,31,1,31,7,31,13,7,1,7,31,1,47,31,31,57,31,23,67,
%T 71,31,127,67,31,127,61,127,1,7,31,31,67,1,127,1,193,87,7,127,223,51
%N Remainder when dividing 2^q - 1 by q + 1 where q is the n-th prime.
%C The Feit-Thompson conjecture states that given primes p and q, (p^q - 1)/(p - 1) is never divisible by (q^p - 1)/(q - 1). Assigning p = 2, the two expressions simplify to 2^q - 1 and q + 1. The former is an odd number and the latter is even, therefore there will always be a remainder when dividing the former by the latter (with the obvious exception of q = 2). This means that any counterexample to the Feit-Thompson conjecture would have to be a pair of odd primes.
%H Harvey P. Dale, <a href="/A136035/b136035.txt">Table of n, a(n) for n = 1..1000</a>
%H N. M. Stephens, <a href="https://doi.org/10.1090/S0025-5718-1971-0297686-1">On the Feit-Thompson Conjecture</a>, Mathematics of Computation, Vol. 25 (1971), p. 625.
%F a(n) = A001348(n) mod A008864(n). - _Michel Marcus_, Jun 07 2023
%e a(7) = 13 because the 7th prime is 17. (2^17 - 1)/(2 - 1) gives the Mersenne prime 131071, which when divided by (17^2 - 1)/(17 - 1) = 18, leaves a remainder of 13.
%t Table[Mod[2^Prime[n] - 1, Prime[n] + 1], {n, 50}]
%t Mod[2^#-1,#+1]&/@Prime[Range[50]] (* _Harvey P. Dale_, Feb 21 2012 *)
%o (PARI) a(n) = my(q=prime(n)); lift(Mod(2, q+1)^q - 1); \\ _Michel Marcus_, Jun 07 2023
%Y Cf. A001348, A008864.
%K easy,nonn,changed
%O 1,2
%A _Alonso del Arte_, Mar 21 2008
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