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A054269
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Length of period of continued fraction for sqrt(prime(n)).
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18
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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
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1,2
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COMMENTS
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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
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LINKS
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MAPLE
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with(numtheory): for i from 1 to 150 do cfr := cfrac(ithprime(i)^(1/2), 'periodic', 'quotients'); printf(`%d, `, nops(cfr[2])) od:
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MATHEMATICA
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Table[p=Prime[n]; Length[Last[ContinuedFraction[Sqrt[p]]]], {n, 100}] (* T. D. Noe, May 22 2007 *)
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CROSSREFS
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Cf. A003285, A130272 (primes at which the period length sets a new record).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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