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A054269
Length of period of continued fraction for sqrt(prime(n)).
18
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
1,2
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
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 *)
CROSSREFS
Cf. A003285, A130272 (primes at which the period length sets a new record).
Sequence in context: A377781 A276133 A307602 * A373399 A086450 A370153
KEYWORD
nonn,easy,nice
AUTHOR
N. J. A. Sloane, May 05 2000
EXTENSIONS
More terms from James A. Sellers, May 05 2000
STATUS
approved