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%I #20 Mar 14 2023 03:40:15
%S 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,
%T 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,
%U 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
%N Length of period of continued fraction for sqrt(prime(n)).
%C 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
%C 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
%C 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
%H T. D. Noe, <a href="/A054269/b054269.txt">Table of n, a(n) for n = 1..10000</a>
%H A. I. Gliga, <a href="http://www.math.princeton.edu/mathlab/jr02fall/Periodicity/alexajp.pdf">On continued fractions of the square root of prime numbers</a>
%p with(numtheory): for i from 1 to 150 do cfr := cfrac(ithprime(i)^(1/2), 'periodic','quotients'); printf(`%d,`, nops(cfr[2])) od:
%t Table[p=Prime[n]; Length[Last[ContinuedFraction[Sqrt[p]]]],{n,100}] (* _T. D. Noe_, May 22 2007 *)
%Y Cf. A003285, A130272 (primes at which the period length sets a new record).
%K nonn,easy,nice
%O 1,2
%A _N. J. A. Sloane_, May 05 2000
%E More terms from _James A. Sellers_, May 05 2000
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