A101208 Smallest odd prime p such that n = (p - 1) / ord_p(2).

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

Offset: 1

Author

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

Keywords

• nonn
• nice
• base

Comments

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, ...

(Updated by Eric Chen, Jun 01 2015)

Links

• Eric Chen, Table of n, a(n) for n = 1..612
• V. Papadimitriou, The 1/p ratio of the first hundred million primes

Crossrefs

Cf. A001917, A101209, A054471.

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.

Programs

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 *)

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