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A072982
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Primes p for which the period of 1/p is a power of 2.
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12
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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
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1,1
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COMMENTS
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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
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LINKS
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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)
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EXAMPLE
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15361 has a period of 256 = 2^8, hence 15361 is in the sequence.
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MAPLE
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filter:= proc(p) local k;
if not isprime(p) then return false fi;
k:=igcd(p-1, 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)}):
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MATHEMATICA
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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 *)
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PROG
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(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')
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CROSSREFS
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Cf. A197224 (power of 2 which is the period of the decimal 1/a(n)).
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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