

A136035


Remainder when dividing 2^q  1 by q + 1 where q is the nth prime.


1



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
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OFFSET

1,2


COMMENTS

The FeitThompson 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 FeitThompson conjecture would have to be a pair of odd primes.


REFERENCES

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


LINKS

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


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 *)


CROSSREFS

Sequence in context: A245684 A082053 A322753 * A132307 A188463 A319298
Adjacent sequences: A136032 A136033 A136034 * A136036 A136037 A136038


KEYWORD

easy,nonn


AUTHOR

Alonso del Arte, Mar 21 2008


STATUS

approved



