|
Search: id:A136035
|
|
|
| A136035 |
|
Remainder when dividing 2^q - 1 by q + 1 where q is the n-th prime. |
|
+0
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
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
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.
|
|
REFERENCES
|
N. M. Stephens, On the Feit-Thompson Conjecture, Mathematics of Computation, Vol. 25 (1971), p. 625
|
|
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}]
|
|
CROSSREFS
|
Sequence in context: A046913 A118228 A082053 this_sequence A132307 A101748 A058606
Adjacent sequences: A136032 A136033 A136034 this_sequence A136036 A136037 A136038
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Alonso Delarte (alonso.delarte(AT)gmail.com), Mar 21 2008
|
|
|
Search completed in 0.002 seconds
|
