A136035 Remainder when dividing 2^q - 1 by q + 1 where q is the n-th prime.

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, 71, 31, 127, 67, 31, 127, 61, 127, 1, 7, 31, 31, 67, 1, 127, 1, 193, 87, 7, 127, 223, 51

Offset

1,2

Comments

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.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

N. M. Stephens, On the Feit-Thompson Conjecture, Mathematics of Computation, Vol. 25 (1971), p. 625.

Formula

a(n) = A001348(n) mod A008864(n). - Michel Marcus, Jun 07 2023

Example

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.

Mathematica

Table[Mod[2^Prime[n] - 1, Prime[n] + 1], {n, 50}]
Mod[2^#-1, #+1]&/@Prime[Range[50]] (* Harvey P. Dale, Feb 21 2012 *)

Prog

(PARI) a(n) = my(q=prime(n)); lift(Mod(2, q+1)^q - 1); \\ Michel Marcus, Jun 07 2023

Crossrefs

Cf. A001348, A008864.

Sequence in context: A245684 A082053 A322753 * A132307 A372968 A188463

Adjacent sequences: A136032 A136033 A136034 * A136036 A136037 A136038

Keyword

easy,nonn

Author

Alonso del Arte, Mar 21 2008

Status

approved