A013646 -id:A013646 - cp's OEIS frontend

A215485 Periods of square root continued fractions at which A013646 sets a new record.

0, 1, 2, 3, 7, 9, 13, 19, 23, 27, 35, 41, 43, 45, 53, 55, 71, 77, 101, 127, 129, 135, 147, 163, 169, 189, 199, 201, 247, 283, 335, 353, 367, 459, 465, 503, 537, 587, 625, 637, 643, 739, 767, 827, 1009, 1135, 1325, 1423, 1433, 1543, 1561, 1775, 1781, 1951, 2011

Offset: 0

Patrick McKinley, Aug 12 2012

nonn

Each term of this sequence takes a turn at being the smallest unknown period for a square root continued fraction. Periods 1 and 2 are seen as the periods of sqrt(2) and sqrt(3) respectively, but a period of 3 is not seen until sqrt(41).
By convention, the period for perfect squares (e.g., 1) is 0.
Open question: Are there any more even terms after the 2?

			When a square root continued fraction with a period of 3 is first seen (at sqrt(41)), the lowest period not yet seen is 7, which first occurs as the period of sqrt(58).

Patrick McKinley, Table of n, a(n) for n = 0..254

Cf. A013646.

A215508 Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n); records of A013646.

Original entry on oeis.org

1, 2, 3, 41, 58, 106, 193, 337, 586, 949, 1061, 1117, 1153, 1249, 1669, 2381, 3733, 5857, 6577, 6781, 8389, 11173, 14293, 15817, 17137, 17209, 23017, 37921, 38377, 46261, 47293, 56929, 82561, 90121, 113173, 122401, 148957, 151057, 161149, 163729, 193873, 206209, 225769, 322513, 497473, 576529, 676129, 686893, 706621, 862921, 946489, 992281, 1032649, 1198081, 1597033, 1655677, 1779409, 1930021, 2299489, 2367481, 2584081, 3209281, 3528409, 3933073, 4068241, 4160521, 4283689, 4726009, 4833901

Offset: 0

Patrick McKinley, Aug 13 2012

nonn

The continued fractions of these numbers have the "hard to get" lengths listed in sequence A215485. They fill the last gaps in the table when computing A013646.

			The lengths of the continued fractions of sqrt(1), sqrt(2), sqrt(3) and sqrt(41) are 0, 1, 2 and 3 respectively. The rest of the sequence follows A215485 similarly.

Patrick McKinley, Table of n, a(n) for n = 0..253

Cf. A013646, A215485.

a(n) = A013646(A215485(n)). - Pontus von Brömssen, Nov 24 2024

A096491 a(n) = sqrt(n) of n if n is a perfect square, otherwise a(n) = largest term in period of continued fraction expansion of square root of n.

Original entry on oeis.org

1, 2, 2, 2, 4, 4, 4, 4, 3, 6, 6, 6, 6, 6, 6, 4, 8, 8, 8, 8, 8, 8, 8, 8, 5, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 6, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 7, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16

Offset: 1

Labos Elemer, Jun 29 2004

nonn

			For n=127: the period={3,1,2,2,7,11,7,2,2,1,3,22}, max=a(127)=22.

Cf. A003285, A013646.

A096491 := proc(n)
if issqr(n) then
sqrt(n) ;
else
numtheory[cfrac](sqrt(n),'periodic','quotients') ;
%[2] ;
max(op(%)) ;
end if;
end proc:
# R. J. Mathar, Mar 18 2010

u=1;Do[s=Max[Last[ContinuedFraction[n^(1/2)]]];tc[[u]]=s;u=u+1, {n, 1, m}]

Definition revised by N. J. A. Sloane, Mar 18 2010

A096493 Number of distinct primes in continued fraction period of square root of n.

Original entry on oeis.org

0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 2, 1, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 2, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 0, 0, 0, 0, 0, 3, 0, 1, 1, 2, 1, 1, 0, 0, 2, 2, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 1, 0, 3, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0

Offset: 1

Labos Elemer, Jun 29 2004

nonn

			n=127: the period={3,1,2,2,7,11,7,2,2,1,3,22},
distinct-primes={2,3,7,11}, so a[127]=4;

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A003285, A013646, A096491, A096492.

{te=Table[0, {m}], u=1}; Do[s=Count[PrimeQ[Union[Last[ContinuedFraction[n^(1/2)]]]], True]; te[[u]]=s;u=u+1, {n, 1, m}];te
dpcf[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],0,Count[Union[ ContinuedFraction[ s][[2]]],?PrimeQ]]]; Array[dpcf,110] (* _Harvey P. Dale, Mar 18 2016 *)

A096492 Number of distinct terms in continued fraction period of square root of n.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 4, 2, 3, 4, 3, 2, 1, 1, 2, 3, 3, 2, 4, 2, 3, 3, 2, 1, 1, 2, 2, 2, 2, 2, 4, 3, 3, 5, 3, 2, 1, 1, 2, 4, 3, 4, 2, 2, 3, 2, 4, 3, 5, 3, 2, 1, 1, 2, 5, 2, 4, 3, 4, 2, 3, 2, 2, 5, 4, 3, 3, 2, 1, 1, 2, 2, 3, 4, 2, 3, 3, 2, 3, 4, 4, 6, 3, 3, 3, 3, 2, 1, 1, 2, 5, 2, 2

Offset: 1

Labos Elemer, Jun 29 2004

nonn

Essentially the same as A028832. - Amiram Eldar, Nov 10 2021

			n=127: the period={3,1,2,2,7,11,7,2,2,1,3,22},distinct-terms={1,2,3,7,11,22}, so a[127]=6;

Cf. A003285, A013646, A028832, A096491, A096493.

{tc=Table[0, {m}], u=1}; Do[s=Length[Union[Last[ContinuedFraction[n^(1/2)]]]]; tc[[u]]=s;u=u+1, {n, 1, m}], tc

a(n) = 1 if n is a square and a(n) = A028832(n) otherwise. - Amiram Eldar, Nov 10 2021

A059800 Smallest prime p such that the quotient-cycle length in continued fraction expansion of sqrt(p) is n: smallest prime p(m) for which A054269(m)=n.

Original entry on oeis.org

2, 3, 41, 7, 13, 19, 73, 31, 113, 43, 61, 103, 193, 179, 109, 191, 157, 139, 337, 151, 181, 491, 853, 271, 457, 211, 1109, 487, 821, 379, 601, 463, 613, 331, 1061, 1439, 421, 619, 541, 1399, 1117, 571, 1153, 823, 1249, 739, 1069, 631, 1021, 1051, 1201

Offset: 1

Labos Elemer, Feb 23 2001

nonn

			The quotient-cycle length L=9=A054269(m) first appears for p(30)=113, so a(9)=113 namely, at first A054269(30)=9; a(A054269(30)) = p(30) = 113 = a(9). The quotient cycle with L=16 first emerges for sqrt(191) and it is: cfrac(sqrt(191), 'periodic', 'quotients')= [[13],[1,4,1,1,3,2,2,13,2 2,3,1,1,4,1,26]].

T. D. Noe, Table of n, a(n) for n = 1..2000

Cf. A054269.

Cf. A013646, A130272

a(n) = Min{p|A054269(sequence number of p)=n; p is prime}.

A062769 Smallest number m such that the continued fraction expansion of sqrt(m) has period 2n + 1.

Original entry on oeis.org

2, 41, 13, 58, 106, 61, 193, 109, 157, 337, 181, 586, 457, 949, 821, 601, 613, 1061, 421, 541, 1117, 1153, 1249, 1069, 1021, 1201, 1669, 2381, 1453, 2137, 2053, 1801, 2293, 1381, 1549, 3733, 3541, 3217, 5857, 1621, 3169, 4657, 2689, 3049, 2389, 4057, 4549

Offset: 0

Lekraj Beedassy, Jul 17 2001

nonn

If the continued fraction for sqrt(N) has period (2k + 1) and k-th convergent P(k)/Q(k) [taking P(-1)=1; Q(-1)=0 where necessary], then the i-th positive solution V(i) = [x(i),y(i)] to the Pell equation x^2 - N*y^2 = 1 satisfies the recurrence V(i+2) = 2*A*V(i+1) - V(i) starting with V(0)=(1,0); V(1) = (A,B) where A = 2*S^2 + 1; B = 2*S*T and S = P(k)*Q(k) + P(k-1)*Q(k-1); T = Q(k)^2 + Q(k-1)^2.

			For n = 2, 2n+1 = 5. a(2) = 13 and we indeed have sqrt(13) = [3; 1, 1, 1, 1, 6] with period 5, the first one in the sequence sqrt(29) = [5; 2, 1, 1, 2, 10], sqrt(53) = [7; 3, 1, 1, 3, 14], sqrt(74) = [8; 1, 1, 1, 1, 16], sqrt(85) = [9; 4, 1, 1, 4, 18], sqrt(89) = [9; 2, 3, 3, 2, 18], ...

Michael De Vlieger, Table of n, a(n) for n = 0..1000 (Terms 0..999 from T. D. Noe)
Dario Alpern, Continued Fraction calculator.
Keith Matthews, Calculating the simple continued fraction of a quadratic irrational.
Ulrich Sondermann, Continued Fractions.
Gang Xiao, Contfrac,continued fraction expansion server with k-th convergent calculator.

Cf. A013646, A003814, A031396, A003654.

nn = 50; t = Table[0, {nn}]; n = 1; found = 0; While[found < nn, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && (len + 1)/2 <= nn && t[[(len + 1)/2]] == 0, t[[(len + 1)/2]] = n; found++]]]; t (* T. D. Noe, Apr 04 2014 *)

More terms from Naohiro Nomoto, Jan 01 2002

A240072 Least number k with continued fraction of sqrt(k) having periodic part of length 2*n.

Original entry on oeis.org

3, 7, 19, 31, 43, 46, 134, 94, 139, 151, 166, 271, 211, 334, 379, 463, 331, 478, 619, 526, 571, 604, 694, 631, 1051, 751, 886, 1039, 1141, 919, 1291, 1324, 1699, 1879, 1366, 2476, 2038, 1516, 1894, 1759, 2164, 1831, 2179, 1726, 2851, 2461, 2011, 2311, 4603

Offset: 1

T. D. Noe, Apr 04 2014

nonn

It appears that, in general, these numbers are less than the corresponding numbers for the odd lengths, A062769.

			The continued fractions of sqrt(3), sqrt(7), and sqrt(19) are {1; 1, 2}, {2; 1, 1, 1, 4}, and {4; 2, 1, 3, 1, 2, 8}.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A013646 (even and odd), A062769 (similar, but odd length).

nn = 50; t = Table[0, {nn}]; n = 1; found = 0; While[found < nn, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[EvenQ[len] && len/2 <= nn && t[[len/2]] == 0, t[[len/2]] = n; found++]]]; t

A288184 Least odd number k such that the continued fraction for sqrt(k) has period n.

Original entry on oeis.org

5, 3, 41, 7, 13, 19, 73, 31, 113, 43, 61, 103, 193, 179, 109, 133, 157, 139, 337, 151, 181, 253, 853, 271, 457, 211, 949, 487, 821, 379, 601, 463, 613, 331, 1061, 1177, 421, 619, 541, 589, 1117, 571, 1153, 823, 1249, 739, 1069, 631, 1021, 1051, 1201, 751

Offset: 1

Ilya Gutkovskiy, Jun 06 2017

nonn

			a(2) = 3, sqrt(3) = 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + 1/...)))), period 2: [1, 2].

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Periodic Continued Fraction

Cf. A000035, A003285, A010337-A010339, A013642-A013644, A013646, A013943, A020347-A020439, A059800, A062769, A097853, A288185.

from sympy import continued_fraction_periodic
def A288184(n):
    d = 1
    while True:
        s = continued_fraction_periodic(0,1,d)[-1]
        if isinstance(s, list) and len(s) == n:
            return d
        d += 2 # Chai Wah Wu, Jun 07 2017

A003285(a(n)) = n, A000035(a(n)) = 1.

A288185 Least even number k such that the continued fraction for sqrt(k) has period n.

Original entry on oeis.org

2, 6, 130, 14, 74, 22, 58, 44, 106, 86, 298, 46, 746, 134, 1066, 94, 1018, 424, 922, 268, 394, 166, 586, 382, 1306, 214, 1354, 334, 1642, 436, 2122, 508, 1114, 454, 4138, 478, 3194, 1108, 4874, 526, 3418, 724, 2458, 604, 9914, 694, 4618, 844, 2746, 1318

Offset: 1

Ilya Gutkovskiy, Jun 06 2017

nonn

			a(2) = 6, sqrt(6) = 2 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + 1/...)))), period 2: [2, 4].

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Periodic Continued Fraction

Cf. A000035, A003285, A010337-A010339, A013642-A013644, A013646, A013943, A020347-A020439, A097853, A240072, A288184.

from sympy import continued_fraction_periodic
def A288185(n):
    d = 2
    while True:
        s = continued_fraction_periodic(0,1,d)[-1]
        if isinstance(s, list) and len(s) == n:
            return d
        d += 2 # Chai Wah Wu, Jun 08 2017

A003285(a(n)) = n, A000035(a(n)) = 0.

cp's OEIS Frontend

A215485 Periods of square root continued fractions at which A013646 sets a new record.

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A215508 Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n); records of A013646.

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A096491 a(n) = sqrt(n) of n if n is a perfect square, otherwise a(n) = largest term in period of continued fraction expansion of square root of n.

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A096493 Number of distinct primes in continued fraction period of square root of n.

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Mathematica

A096492 Number of distinct terms in continued fraction period of square root of n.

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Mathematica

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A059800 Smallest prime p such that the quotient-cycle length in continued fraction expansion of sqrt(p) is n: smallest prime p(m) for which A054269(m)=n.

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A062769 Smallest number m such that the continued fraction expansion of sqrt(m) has period 2n + 1.

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Mathematica

Extensions

A240072 Least number k with continued fraction of sqrt(k) having periodic part of length 2*n.

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Mathematica

A288184 Least odd number k such that the continued fraction for sqrt(k) has period n.

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