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 A072982 Primes p for which the period length 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS All Fermat primes>5 (A019434) are in the sequence, since it can be shown that the period length 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 length 1048576 = 2^20; 46179488366593 has period length 2199023255552 = 2^41; 101702694862849 has period length 8388608 = 2^23; 171523813933057 has period length 4398046511104 = 2^42; 505775348776961 has period length 2199023255552 = 2^41; 834427406578561 has period length 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 length 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

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Last modified February 23 16:14 EST 2020. Contains 332173 sequences. (Running on oeis4.)