A003285 -id:A003285 - cp's OEIS frontend

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.

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

A035015 Period of continued fraction for square root of n-th squarefree integer.

Original entry on oeis.org

1, 2, 1, 2, 4, 1, 2, 5, 4, 2, 1, 6, 6, 6, 4, 1, 5, 2, 8, 4, 4, 2, 1, 2, 2, 3, 2, 10, 12, 4, 2, 5, 4, 6, 7, 6, 11, 4, 1, 2, 10, 8, 6, 8, 7, 5, 6, 4, 4, 1, 2, 5, 10, 2, 5, 8, 10, 16, 4, 11, 1, 2, 12, 2, 9, 6, 15, 2, 6, 9, 6, 10, 10, 4, 1, 2, 12, 10, 3, 6, 16, 14, 9, 4, 18, 4, 4, 2, 1, 2, 9, 20, 10, 4

Offset: 2

David L. Treumann (alewifepurswest(AT)yahoo.com)

Friesen proved that each value appears infinitely often. - Michel Marcus, Apr 12 2019

			a(2)=1 because 2 is the 2nd smallest squarefree integer and sqrt 2 = [ 1,2,2,2,2,... ] thus has an eventual period of 1.

David W. Wilson, Table of n, a(n) for n = 2..10000
S. R. Finch, Class number theory [broken link]
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Christian Friesen, On continued fractions of given period, Proc. Amer. Math. Soc. 103 (1988), 9-14.
Ron Knott, An Introduction to Continued Fractions

Cf. A003285, A005117 (squarefree numbers), A013943.

sqf:= select(numtheory:-issqrfree,[$2..1000]):
map(n->nops(numtheory:-cfrac(sqrt(n),'periodic','quotients')[2]),sqf); # Robert Israel, Dec 21 2014

Length[ContinuedFraction[Sqrt[#]][[2]]]&/@Select[ Range[ 2,200], SquareFreeQ] (* Harvey P. Dale, Jul 17 2011 *)

a(n) = A003285(A005117(n)). - Michel Marcus, Dec 29 2014

Corrected and extended by James Sellers

A059866 Period of the continued fraction for sqrt(2^n-1).

Original entry on oeis.org

2, 4, 2, 8, 2, 12, 2, 20, 2, 12, 2, 164, 2, 40, 2, 40, 2, 1208, 2, 660, 2, 1304, 2, 3056, 2, 2492, 2, 1080, 2, 13004, 2, 10232, 2, 11296, 2, 148736, 2, 56576, 2, 615482, 2, 44448, 2, 64, 2, 2628524, 2, 28219952, 2, 139558, 2, 3067080, 2, 2683626, 2, 90740360, 2, 103050292, 2

Offset: 2

Labos Elemer, Feb 28 2001

nonn

			For n=7 and n=8 the continued fractions are [[11], [3, 1, 2, 2, 7, 11, 7, 2, 2, 1, 3, 22]] and [[15], [1, 30]] with periods 12 and 2, respectively.

Chai Wah Wu, Table of n, a(n) for n = 2..78

Cf. A003285, A059926, A059927, A064932.

with(numtheory): [seq(nops(cfrac(sqrt(2^k-1),'periodic','quotients')[2]),k=2..30)];

Table[Length@ Last@ ContinuedFraction[Sqrt[2^n - 1]], {n, 2, 56}] (* Michael De Vlieger, Mar 21 2015 *)

Corrected and extended by Naohiro Nomoto, Nov 09 2001

a(57)-a(60) from Daniel Suteu, Jan 25 2019

A097853 Period of continued fraction for square root of n (or 1 if n is a square).

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 4, 2, 1, 1, 2, 2, 5, 4, 2, 1, 1, 2, 6, 2, 6, 6, 4, 2, 1, 1, 2, 4, 5, 2, 8, 4, 4, 4, 2, 1, 1, 2, 2, 2, 3, 2, 10, 8, 6, 12, 4, 2, 1, 1, 2, 6, 5, 6, 4, 2, 6, 7, 6, 4, 11, 4, 2, 1, 1, 2, 10, 2, 8, 6, 8, 2, 7, 5, 4, 12, 6, 4, 4, 2, 1, 1, 2, 2, 5, 10, 2, 6, 5, 2, 8, 8, 10, 16, 4, 4, 11, 4, 2, 1, 1, 2, 12

Offset: 1

N. J. A. Sloane, Sep 01 2004

			1 is a square. 2 has continued fraction [1;2,2,2...], 3 has [1;1,2,1,2,1,2...]. - _Georg Fischer_, Jun 14 2019

See A003285, which is the main entry for this sequence.

a:= n-> `if`(issqr(n), 1, nops(numtheory[cfrac](
         sqrt(n), 'periodic', 'quotients')[2])):
seq(a(n), n=1..120);  # Alois P. Heinz, Jun 14 2019

a[n_] := If[IntegerQ[Sqrt[n]], 1, Length[ContinuedFraction[Sqrt[n]][[2]]]];
Table[a[n], {n, 1, 103}] (* Jean-François Alcover, Jan 09 2025 *)

A013644 Numbers k such that the continued fraction for sqrt(k) has period 4.

Original entry on oeis.org

7, 14, 23, 28, 32, 33, 34, 47, 55, 60, 62, 75, 78, 79, 95, 96, 98, 119, 126, 128, 136, 138, 140, 141, 142, 155, 167, 174, 176, 180, 189, 192, 194, 215, 219, 220, 222, 223, 248, 252, 254, 266, 287, 299, 300, 305, 312, 315, 318, 320, 321, 322, 335, 359, 368, 377, 390, 392

Offset: 1

Clark Kimberling

nonn

			The continued fraction for sqrt(7) is [2;1,1,1,4,...] with period 4, so 7 is in the sequence.  The continued fractions sqrt(3) = [1;1,2,...] with period 2 and sqrt(13) = [3;1,1,1,1,6,...] with period 5 do not have period 4, so 3 and 13 are not in the sequence. - _Michael B. Porter_, Sep 20 2016

Kenneth H. Rosen, Elementary Number Theory and Its Applications, Addison-Wesley, 1984, page 426 (but beware of errors!).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Austin Mack and Timothy Sawicki, Pell's Equations Through Dynamical Systems, 2012. [Broken link?]

Cf. A003285.

cfp4Q[n_]:=Module[{sr=Sqrt[n]},!IntegerQ[sr]&&Length[ ContinuedFraction[ sr][[2]]]==4]; Select[Range[500],cfp4Q] (* Harvey P. Dale, Jul 29 2014 *)

See Austin Mack and Timothy Sawicki(2012).

A013645 Values of k at which the period of the continued fraction for sqrt(k) sets a new record.

Original entry on oeis.org

1, 2, 3, 7, 13, 19, 31, 43, 46, 94, 139, 151, 166, 211, 331, 421, 526, 571, 604, 631, 751, 886, 919, 1291, 1324, 1366, 1516, 1621, 1726, 2011, 2311, 2566, 2671, 3004, 3019, 3334, 3691, 3931, 4174, 4846, 5119, 6211, 6451, 6679, 6694, 7606, 8254, 8779, 8941, 9739

Offset: 1

Clark Kimberling

Periods of the fractions (sequence offset by one term) are given by A020640.

For n = 1 to 513 (the range of the b-file), the class number of the field Q(sqrt(a(n))) is 1 (computed with Mathematica). - Emmanuel Vantieghem, Mar 16 2017

			The continued fraction for sqrt(31) is {5; 1, 1, 3, 5, 3, 1, 1, 10}, the continued fraction for sqrt(43) is {6; 1, 1, 3, 1, 5, 1, 3, 1, 1, 12}, and there is no number between 31 and 43 whose square root produces a continued fraction whose period exceeds that of 31.

Kenneth H. Rosen, Elementary Number Theory and Its Applications, Addison-Wesley, 1984, page 426 (but beware of errors!).

Patrick McKinley, Table of n, a(n) for n = 1..513 (first 200 terms from T. D. Noe)
H. C. Williams, A Numerical Investigation into the Length of the Period of the Continued Fraction Expansion of sqrt(D), Mathematics of Computation 36:154 (1981), 593-601 (see especially Tables 1 through 5 of this paper).

Cf. A003285, A020640.

mx = -1; t = {}; Do[len = Length[ Last[ ContinuedFraction[ Sqrt[ n]]]]; If[len > mx, mx = len; AppendTo[t, n]], {n, 10^4}]; t

More terms from David W. Wilson

A020439 Numbers k such that the continued fraction for sqrt(k) has period 100.

Original entry on oeis.org

2629, 3646, 4924, 5692, 5833, 5836, 6172, 6703, 6801, 7389, 7438, 8158, 8287, 8551, 8654, 9103, 10041, 10079, 10096, 10629, 10936, 11038, 11068, 11116, 11335, 11383, 11519, 11824, 11863, 11995, 12016, 12044, 12494, 12751, 12811, 12895, 13372, 13569

Offset: 1

David W. Wilson

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A003285.

nn=15000;With[{nonsq=Complement[Range[nn],Range[ Floor[Sqrt[nn]]]^2]}, Select[ nonsq,Length[ ContinuedFraction[Sqrt[#]][[2]]]==100&]] (* Harvey P. Dale, May 22 2012 *)

A059926 Length of period of the continued fraction expansion of sqrt(2^n+1).

Original entry on oeis.org

1, 4, 1, 10, 1, 16, 1, 44, 1, 74, 1, 46, 1, 204, 1, 714, 1, 702, 1, 908, 1, 404, 1, 7754, 1, 1136, 1, 9886, 1, 8154, 1, 23578, 1, 65096, 1, 404762, 1, 23992, 1, 3514774, 1, 110124, 1, 4802160, 1, 6490450, 1, 180832, 1, 115972, 1, 770304, 1, 62665998, 1, 133093360, 1, 1019300318, 1, 60079334

Offset: 4

Labos Elemer, Mar 01 2001

For n=1,2 a(1)=2, a(2)=1; for n=3 it is not a quadratic surd.

			For n=7 and n=8 the periods after the transient are as follows: cfrac(sqrt(2^7+1),'periodic','quotients'); gives [[11], [2, 1, 3, 1, 6, 1, 3, 1, 2, 22]] cfrac(sqrt(2^8+1),'periodic','quotients'); gives [[16], [32]]

Chai Wah Wu, Table of n, a(n) for n = 4..84

Cf. A000051, A003285, A059866, A061682.

with(numtheory): [seq(nops(cfrac(sqrt(2^k+1),'periodic','quotients')[2]),k=4..28)];

Table[Length[ContinuedFraction[Sqrt[2^n+1]][[2]]],{n,4,60}] (* Harvey P. Dale, Feb 05 2012 *)

a(n) = A003285(A000051(n)). - Michel Marcus, Sep 27 2019

Two more terms from David W. Wilson, Jun 18 2001
Corrected and extended by Naohiro Nomoto, Nov 09 2001
a(58)-a(63) from Daniel Suteu, Jan 25 2019

A059927 Period of the continued fraction for sqrt(2^(2n+1)).

Original entry on oeis.org

1, 2, 4, 4, 12, 24, 48, 96, 196, 368, 760, 1524, 3064, 6068, 12168, 24360, 48668, 97160, 194952, 389416, 778832, 1557780, 3116216, 6229836, 12462296, 24923320, 49849604, 99694536, 199394616, 398783628, 797556364, 1595117676, 3190297400, 6380517544, 12761088588, 25522110948, 51044281208, 102088450460, 204177067944, 408353857832, 816708255152

Offset: 0

Labos Elemer, Mar 01 2001

nonn

K. R. Matthews (Feb 2007) showed that lim_{n -> oo} a(n)/2^n = 0.7427.... - A.H.M. Smeets, Nov 14 2017

			For n=5 we look at the square root of 2^11 = 2048, and find that the cycle has length 24. Here is Maple's calculation:  cfrac(sqrt(2048),'periodic','quotients') = [[45],[3,1,12,5,1,1,2,1,2,4,1,21,1,4,2,1,2,1,1,5,12,1,3,90]], the periodic part having length 24.

K. R. Matthews, On the continued fraction expansion of sqrt(2^(2n+1)), Feb 2007.

Cf. A003285, A004171, A059866 (for sqrt(2^n-1)).

Cf. A064932 (for sqrt(3^(2n+1))), A293028 (for sqrt(5^(2n+1))).

with(numtheory): [seq(nops(cfrac(sqrt(2^(2*k-1)),'periodic','quotients')[2]),k=1..15)];

Array[Length@ ContinuedFraction[Sqrt[2^(2 # + 1)]][[-1]] &, 15, 0] (* Michael De Vlieger, Oct 09 2017 *)

a(n) = A003285(A004171(n)). - Michel Marcus, Sep 27 2019

More terms from Don Reble, Oct 31 2001
a(32) = 3190297400 from Don Reble, Feb 10 2007
a(33)-a(35) from Keith Matthews (keithmatt(AT)gmail.com), Feb 16 2007, Feb 28 2007
Name clarified by Joerg Arndt, Oct 09 2017
a(36)-a(37) from Chai Wah Wu, Sep 26 2019
a(38)-a(40) from Chai Wah Wu, Sep 30 2019

A067280 Number of terms in continued fraction for sqrt(n), excl. 2nd and higher periods.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 5, 3, 1, 2, 3, 3, 6, 5, 3, 1, 2, 3, 7, 3, 7, 7, 5, 3, 1, 2, 3, 5, 6, 3, 9, 5, 5, 5, 3, 1, 2, 3, 3, 3, 4, 3, 11, 9, 7, 13, 5, 3, 1, 2, 3, 7, 6, 7, 5, 3, 7, 8, 7, 5, 12, 5, 3, 1, 2, 3, 11, 3, 9, 7, 9, 3, 8, 6, 5, 13, 7, 5, 5, 3, 1, 2, 3, 3, 6, 11, 3, 7, 6, 3, 9, 9, 11, 17, 5, 5, 12, 5

Offset: 1

Frank Ellermann, Feb 23 2002

			a(2)=2: [1,(2)+ ]; a(3)=3: [1,(1,2)+ ]; a(4)=1: [2]; a(5)=2: [2,(4)+ ].

H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th edition, 1999, table 1.

Related sequences: 2 : A040000, ..., 44: A040037, 48: A040041, ..., 51: A040043, 56: A040048, 60: A040052, 63: A040055, ..., 66: A040057. 68: A040059, 72: A040063, 80: A040071.
Related sequences: 45: A010135, ..., 47: A010137, 52: A010138, ..., 55: A010141, 57: A010142, ..., 59: A010144. 61: A010145, 62: A010146. 67: A010147, 69: A010148, ..., 71: A010150.
Cf. A003285.

from sympy import continued_fraction_periodic
def A067280(n): return len((a := continued_fraction_periodic(0,1,n))[:1]+(a[1] if a[1:] else [])) # Chai Wah Wu, Jun 14 2022

a(n) = A003285(n) + 1. - Andrey Zabolotskiy, Jun 23 2020

Name clarified by Michel Marcus, Jun 22 2020

cp's OEIS Frontend

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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A035015 Period of continued fraction for square root of n-th squarefree integer.

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A059866 Period of the continued fraction for sqrt(2^n-1).

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A097853 Period of continued fraction for square root of n (or 1 if n is a square).

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A013644 Numbers k such that the continued fraction for sqrt(k) has period 4.

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A013645 Values of k at which the period of the continued fraction for sqrt(k) sets a new record.

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Mathematica

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A020439 Numbers k such that the continued fraction for sqrt(k) has period 100.

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Mathematica

A059926 Length of period of the continued fraction expansion of sqrt(2^n+1).

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