A054269 Length of period of continued fraction for sqrt(prime(n)).

1, 2, 1, 4, 2, 5, 1, 6, 4, 5, 8, 1, 3, 10, 4, 5, 6, 11, 10, 8, 7, 4, 2, 5, 11, 1, 12, 6, 15, 9, 12, 6, 9, 18, 9, 20, 17, 18, 4, 5, 14, 21, 16, 13, 1, 20, 26, 4, 2, 5, 11, 12, 17, 14, 1, 12, 3, 24, 21, 13, 18, 5, 14, 16, 17, 11, 34, 19, 14, 7, 15, 4, 20, 5, 30, 8, 9, 21, 1, 21, 18, 37, 16

Offset: 1

Author

N. J. A. Sloane, May 05 2000

Keywords

• nonn
• easy
• nice

Comments

The following sequences (allowing offset of first term) all appear to have the same parity: A034953, triangular numbers with prime indices; A054269, length of period of continued fraction for sqrt(p), p prime; A082749, difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254, numbers n such that 2n-1 is prime; A067076, 2n+3 is a prime. - Jeremy Gardiner, Sep 10 2004

Note that primes of the form n^2+1 (A002496) have a continued fraction whose period length is 1; odd primes of the form n^2+2 (A056899) have length 2; odd primes of the form n^2-2 (A028871) have length 4. - T. D. Noe, Nov 03 2006

For an odd prime p, the length of the period is odd if p=1 (mod 4) or even if p=3 (mod 4). - T. D. Noe, May 22 2007

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• A. I. Gliga, On continued fractions of the square root of prime numbers

Crossrefs

Cf. A003285, A130272 (primes at which the period length sets a new record).

Programs

Maple

with(numtheory): for i from 1 to 150 do cfr := cfrac(ithprime(i)^(1/2), 'periodic','quotients'); printf(`%d,`, nops(cfr[2])) od:

Mathematica

Table[p=Prime[n]; Length[Last[ContinuedFraction[Sqrt[p]]]],{n,100}] (* T. D. Noe, May 22 2007 *)
Length[ContinuedFraction[Sqrt[#]][[2]]]&/@Prime[Range[100]] (* Harvey P. Dale, Sep 28 2024 *)

Extensions

More terms from James Sellers, May 05 2000