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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 (list; graph; refs; listen; history; text; internal format)
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 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

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

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Last modified March 28 21:25 EDT 2017. Contains 284246 sequences.