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A101208
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Smallest odd prime p such that n = (p - 1) / ord_p(2).
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9
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3, 7, 43, 113, 251, 31, 1163, 73, 397, 151, 331, 1753, 4421, 631, 3061, 257, 1429, 127, 6043, 3121, 29611, 1321, 18539, 601, 15451, 14327, 2971, 2857, 72269, 3391, 683, 2593, 17029, 2687, 42701, 11161, 13099, 1103, 71293, 13121, 17467, 2143, 83077, 25609, 5581
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OFFSET
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1,1
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COMMENTS
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First time n appears is given in A001917.
Smallest p (let it be the k-th prime) such that A001917(k) = n, or the smallest prime which has ratio n in base 2.
First cyclic number (in base 2) of n-th degree (or n-th order): the reciprocals of these numbers belong to one of n different cycles. Each cycle has (a(n) - 1)/n digits.
Conjecture: a(n) is defined for all n.
Recursive by indices: (See A054471)
1, 3, 43, 83077, ...
2, 7, 1163, ...
4, 113, 257189, ...
5, 251, 6846277, ...
6, 31, 683, ...
8, 73, 472019, ...
9, 397, 13619483, ...
10, 151, 349717, ...
...
The records for the ratio in base 2 are: 1, 2, 6, 8, 18, 24, 31, 38, 72, 105, 129, 630, 1285, 1542, 2048, ..., the primes are: 3, 7, 31, 73, 127, 601, 683, 1103, 1801, 2731, 5419, 8191, 43691, 61681, 65537, ...
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LINKS
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MATHEMATICA
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f[n_Integer] := Block[{k = 1, p}, While[p = k*n + 1; ! PrimeQ[p] || p != 1 + n*MultiplicativeOrder[2, p] || p = 2, k++]; p]; Array[f, 128] (* Eric Chen, Jun 01 2015 *)
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PROG
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(PARI) a(n) = {p=3; ok = 0; until(ok, if (n == (p-1)/znorder(Mod(2, p)), ok = 1, p = nextprime(p+1)); ); return (p); } \\ Michel Marcus, Jun 27 2013
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CROSSREFS
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Cf. A001122, A115591, A001133, A001134, A001135, A001136, A152307, A152308, A152309, A152310, A152311, which are sequences of primes p where the period of the reciprocal in base 2 is (p-1)/n for n=1 to 11.
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KEYWORD
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nonn,nice,base
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AUTHOR
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Leigh Ellison (le(AT)maths.gla.ac.uk), Dec 14 2004
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STATUS
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approved
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