

A072982


Primes p for which the period of 1/p is a power of 2.


12



3, 11, 17, 73, 101, 137, 257, 353, 449, 641, 1409, 10753, 15361, 19841, 65537, 69857, 453377, 976193, 1514497, 5767169, 5882353, 6187457, 8253953, 8257537, 70254593, 167772161, 175636481, 302078977, 458924033, 639631361, 1265011073
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OFFSET

1,1


COMMENTS

All Fermat primes > 5 (A019434) are in the sequence, since it can be shown that the period of 1/(2^(2^n)+1) is 2^(2^n) whenever 2^(2^n)+1 is prime.  Benoit Cloitre, Jun 13 2007
Take all the terms from row 2^k of triangle in A046107 for k >= 0 and sort to arrive at this sequence.  Ray Chandler, Nov 04 2011
Additional terms, but not necessarily the next in sequence: 13462517317633 has period 1048576 = 2^20; 46179488366593 has period 2199023255552 = 2^41; 101702694862849 has period 8388608 = 2^23; 171523813933057 has period 4398046511104 = 2^42; 505775348776961 has period 2199023255552 = 2^41; 834427406578561 has period 64 = 2^6  Ray Chandler, Nov 09 2011
Furthermore (excluding the initial term 3) this sequence is also the ascending sequence of primes dividing 10^(2^k)+1 for some nonnegative integer k. For a prime dividing 10^(2^k)+1, the period of 1/p is 2^(k+1). Thus for the prime p = 558711876337536212257947750090161313464308422534640474631571587847325442162307811\
65223702155223678309562822667655169, a factor of 10^(2^7)+1, the period of 1/p is only 2^8. This large prime then also belongs to the sequence.  Christopher J. Smyth, Mar 13 2014
For any m, every term that is not a factor of 10^(2^k)1 for some k < m is congruent to 1 (mod 2^m). Thus all terms except 3, 11, 17, 73, 101, 137, 353, 449, 69857, 976193, 5882353, 6187457 are congruent to 1 (mod 128).  Robert Israel, Jun 17 2016
Additional terms listed earlier confirmed as next terms in sequence.  Arkadiusz Wesolowski, Jun 17 2016


LINKS

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..45 (first 33 terms from Ray Chandler, to 36 terms from Robert G. Wilson v, to 39 terms from Ray Chandler)


EXAMPLE

15361 has a period of 256 = 2^8, hence 15361 is in the sequence.


MAPLE

filter:= proc(p) local k;
if not isprime(p) then return false fi;
k:=igcd(p1, 2^ilog2(p));
evalb(10 &^ k mod p = 1)
end proc:
r:= select(`<=`, `union`(seq(numtheory:factorset(10^(2^k)1), k=1..6)), 10^9):
b:= select(filter, {seq(i, i=129..10^9, 128)}):


MATHEMATICA

Do[ If[ IntegerQ[ Log[2, Length[ RealDigits[ 1/Prime[n]] [[1, 1]]]]], Print[ Prime[n]]], {n, 1, 47500}] (* Robert G. Wilson v, May 09 2007 *)
pmax = 10^10; p = 1; While[p < pmax, p = NextPrime[p]; If[ IntegerQ[Log[2, MultiplicativeOrder[10, p] ] ], Print[ p]; ]; ]; (* Ray Chandler, May 14 2007 *)


PROG

(PARI) ? a(n)=if(n<4, n==2, znorder(Mod(10, prime(n)))) ? for(n=1, 20000, if(gcd(a(n), 2^1000)==a(n), print1(prime(n), ", ")))
(Python)
from itertools import count
from sympy import prime, n_order
def A072982_gen(): return (p for p in (prime(n) for n in count(2)) if p != 5 and bin(n_order(10, p))[2:].rstrip('0') == '1')


CROSSREFS

Cf. A197224 (power of 2 which is the period of the decimal 1/a(n)).


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



