

A101208


Smallest odd prime p such that n = (p  1) / ord_p(2).


9



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

1,1


COMMENTS

First time n appears is given in A001917.
Smallest p (let it be the kth prime) such that A001917(k) = n, or the smallest prime which has ratio n in base 2.
First cyclic number (in base 2) of nth degree (or nth 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, ...


LINKS



MATHEMATICA

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 *)


PROG

(PARI) a(n) = {p=3; ok = 0; until(ok, if (n == (p1)/znorder(Mod(2, p)), ok = 1, p = nextprime(p+1)); ); return (p); } \\ Michel Marcus, Jun 27 2013


CROSSREFS

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 (p1)/n for n=1 to 11.


KEYWORD

nonn,nice,base


AUTHOR

Leigh Ellison (le(AT)maths.gla.ac.uk), Dec 14 2004


STATUS

approved



