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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 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 September 24 20:09 EDT 2018. Contains 315356 sequences. (Running on oeis4.)