

A054269


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


15



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
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OFFSET

1,2


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 2n1 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^22 (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


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


CROSSREFS

Cf. A003285, A130272 (primes at which the period length sets a new record).
Sequence in context: A120988 A095979 A276133 * A086450 A270439 A106044
Adjacent sequences: A054266 A054267 A054268 * A054270 A054271 A054272


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane, May 05 2000


EXTENSIONS

More terms from James A. Sellers, May 05 2000


STATUS

approved



