%I A136035
%S A136035 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 A136035 71,31,127,67,31,127,61,127,1,7,31,31,67,1,127,1,193,87,7,127,223,51
%N A136035 Remainder when dividing 2^q - 1 by q + 1 where q is the n-th prime.
%C A136035 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.
%D A136035 N. M. Stephens, On the Feit-Thompson Conjecture, Mathematics of Computation, 
               Vol. 25 (1971), p. 625
%e A136035 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 A136035 Table[Mod[2^Prime[n] - 1, Prime[n] + 1], {n, 50}]
%Y A136035 Sequence in context: A046913 A118228 A082053 this_sequence A132307 A101748 
               A058606
%Y A136035 Adjacent sequences: A136032 A136033 A136034 this_sequence A136036 A136037 
               A136038
%K A136035 easy,nonn
%O A136035 1,2
%A A136035 Alonso Delarte (alonso.delarte(AT)gmail.com), Mar 21 2008