A003285 -id:A003285 - cp's OEIS frontend

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

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 *)

A096494 Largest value in the periodic part of the continued fraction of sqrt(prime(n)).

Original entry on oeis.org

2, 2, 4, 4, 6, 6, 8, 8, 8, 10, 10, 12, 12, 12, 12, 14, 14, 14, 16, 16, 16, 16, 18, 18, 18, 20, 20, 20, 20, 20, 22, 22, 22, 22, 24, 24, 24, 24, 24, 26, 26, 26, 26, 26, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 32, 32, 32, 32, 32, 32, 32, 34, 34, 34, 34, 34, 36, 36, 36, 36, 36, 36

Offset: 1

Labos Elemer, Jun 29 2004

nonn

			n=31: prime(31) = 127, and the periodic part is {3,1,2,2,7,11,7,2,2,1,3,22}, so a(31)=22.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000006, A003285, A005980, A054269.
Cf. A096491, A096492, A096493, A096495, A096496.
Cf. A117767.

a096494 = (* 2) . a000006  -- Reinhard Zumkeller, Sep 20 2014

A096491 := proc(n)
if issqr(n) then
sqrt(n) ;
else
numtheory[cfrac](sqrt(n),'periodic','quotients') ;
%[2] ;
max(op(%)) ;
end if;
end proc:
A096494 := proc(n)
option remember ;
A096491(ithprime(n)) ;
end proc: # R. J. Mathar, Mar 18 2010

{te=Table[0, {m}], u=1}; Do[s=Max[Last[ContinuedFraction[Prime[n]^(1/2)]]]; te[[u]]=s;u=u+1, {n, 1, m}];te
a[n_]:=IntegerPart[Sqrt[Prime[n]]] 2 IntegerPart[Sqrt[#]]&/@Prime[Range[90]] (* Vincenzo Librandi, Aug 09 2015 *)

It seems that lim_{n->infinity} a(n)/n = 0. - Benoit Cloitre, Apr 19 2003

a(n) = 2*A000006(n). - Benoit Cloitre, Apr 19 2003

A064932 Period of the continued fraction for sqrt(3^(2n+1)).

Original entry on oeis.org

2, 10, 30, 98, 270, 818, 2382, 7282, 21818, 65650, 196406, 589982, 1768938, 5309294, 15924930, 47779238, 143322850, 429998586, 1289970842, 3869957114, 11609762666, 34829416842, 104488103446

Offset: 1

Wouter Meeussen, Oct 26 2001

Limit_{n->oo} a(n)/3^n = 1.11... - A.H.M. Smeets, Oct 25 2017

R. K. Guy, personal communication.

Cf. A003285, A013708, A059927.

Table[Length[Last[ContinuedFraction[Sqrt[3^(2n+1)] ]]], {n, 10}]

a(n) = A003285(A013708(n)). - Michel Marcus, Sep 25 2019

a(11)-a(13) from A.H.M. Smeets, Sep 28 2017
a(14) from A.H.M. Smeets, Oct 08 2017
a(15)-a(19) from Daniel Suteu, Jan 24 2019
a(20) from Chai Wah Wu, Sep 23 2019
a(21)-a(23) from Chai Wah Wu, Sep 25 2019

A065938 Position of sqrt(n) in the mapping N2QuQR1 given in A065936.

Original entry on oeis.org

1, 6, 14, 7, 120, 248, 16160, 1019, 127, 32640, 65408, 16373, 8386032, 4194056, 4194239, 32767, 2147450880, 4294934528, 4611672824287851743, 268435343, 8796091842564, 1125899889968159, 70368744112268, 70368744161279

Offset: 1

Antti Karttunen, Dec 07 2001

nonn

Eric Weisstein's World of Mathematics, Continued Fraction.

Cf. A003285. N2QuQR1(a[n])^2 = n, see A065936. For frac2position_in_0_1_SB_tree see A065658. Cf. also A065939.

[seq(frac2position_in_0_1_SB_tree(sqrt_n_confrac2binfrac(j)),j=1..40)];
sqrt_n_confrac2binfrac := proc(n) local c,t; c := CONFRACS_FOR_sqrt_N[n]; t := `if`((1 = nops(c)),[],`if`((0 = (nops(c) mod 2)),[op(c[2..nops(c)]),op(c[2..nops(c)])],c[2..nops(c)])); RETURN( (((2^c[1])-1) + `if`(1 = nops(c),0,(runcounts2binexp0(t) / ((2^(convert(t,`+`)))-1)))) / (2^c[1])); end;
runcounts2binexp0 := proc(c) local i,e,n; n := 0; for i from 0 to nops(c)-1 do e := c[i+1]; n := ((2^e)*n) + ((i mod 2)*((2^e)-1)); od; RETURN(n); end;
CONFRACS_FOR_sqrt_N := [[1], [1, 2], [1, 1, 2], [2], [2, 4], [2, 2, 4], [2, 1, 1, 1, 4], [2, 1, 4], [3], [3, 6], etc., adapted from Weisstein's encyclopedia entry for Continued Fractions]

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

A121339 Periodic part of continued fraction for square roots of integers.

Original entry on oeis.org

2, 1, 2, 4, 2, 4, 1, 1, 1, 4, 1, 4, 6, 3, 6, 2, 6, 1, 1, 1, 1, 6, 1, 2, 1, 6, 1, 6, 8, 4, 8, 2, 1, 3, 1, 2, 8, 2, 8, 1, 1, 2, 1, 1, 8, 1, 2, 4, 2, 1, 8, 1, 3, 1, 8, 1, 8, 10, 5, 10, 3, 2, 3, 10, 2, 1, 1, 2, 10, 2, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, 1, 1, 10, 1, 2, 1, 10, 1, 4, 1, 10, 1, 10, 12, 6, 12, 4, 12

Offset: 2

Franklin T. Adams-Watters, Aug 28 2006

			The table starts:
  2
  1 2
  <empty>
  4
  2 4
  1 1 1 4
  1 4

T. D. Noe, Rows n = 2..1000 of triangle, flattened

Cf. A003285 (row lengths), A013943 (row lengths for nonempty rows).

a[n_] := If[ IntegerQ[ Sqrt[n] ], {}, ContinuedFraction[ Sqrt[n] ] // Last]; Table[a[n], {n, 2, 39}] // Flatten (* Jean-François Alcover, Sep 10 2012 *)

A059853 Period of continued fraction for sqrt(n^2+3), n >= 2.

Original entry on oeis.org

4, 2, 6, 4, 2, 6, 10, 2, 12, 16, 2, 16, 20, 2, 10, 10, 2, 12, 10, 2, 28, 10, 2, 26, 16, 2, 18, 48, 2, 34, 12, 2, 26, 32, 2, 32, 32, 2, 20, 70, 2, 56, 34, 2, 24, 22, 2, 54, 52, 2, 70, 16, 2, 18, 38, 2, 16, 36, 2, 12, 72, 2, 114, 30, 2, 64, 32, 2, 52, 54, 2, 22, 92, 2, 154, 88, 2, 56

Offset: 2

Labos Elemer, Feb 27 2001

nonn

The old name was "Quotient cycle length of sqrt(n^2+3)." - Jianing Song, May 01 2021

			sqrt(35^2+3) = [35; 23, 2, 1, 7, 8, 1, 1, 1, 2, 2, 1, 1, 5, 3, 1, 16, 1, 3, 5, 1, 1, 2, 2, 1, 1, 1, 8, 7, 1, 2, 23, 70], so a(35) = 32.
sqrt(36^2+3) = [36; 24, 72], so a(36) = 2.
sqrt(37^2+3) = [37; 24, 1, 2, 7, 1, 8, 2, 1, 1, 1, 2, 2, 5, 1, 3, 18, 3, 1, 5, 2, 2, 1, 1, 1, 2, 8, 1, 7, 2, 1, 24, 74], so a(37) = 32.

Amiram Eldar, Table of n, a(n) for n = 2..10000

Cf. A003285.

Period of continued fraction for sqrt(n^2+k): this sequence (k=3), A059855 (k=4), A059854 (k=5).

with(numtheory): [seq(nops(cfrac(sqrt(k^2+3),'periodic','quotients')[2]),k=2..256)];

a[n_] := Length[ContinuedFraction[Sqrt[n^2 + 3]][[2]]]; Array[a, 100, 2] (* Amiram Eldar, Jul 10 2024 *)

If n is a multiple of 3 then a(n) = 2.

a(n) = A003285(n^2+3). - Jianing Song, May 01 2021

New name by Jianing Song, May 01 2021

A059854 Period of continued fraction for sqrt(n^2+5), n >= 3.

Original entry on oeis.org

4, 6, 2, 3, 6, 8, 10, 2, 4, 9, 4, 14, 2, 16, 6, 12, 12, 2, 16, 22, 10, 24, 2, 24, 12, 24, 16, 2, 6, 26, 30, 26, 2, 7, 20, 12, 18, 2, 18, 11, 20, 64, 2, 20, 30, 19, 22, 2, 40, 20, 10, 50, 2, 10, 38, 74, 14, 2, 22, 64, 50, 72, 2, 48, 10, 30, 48, 2, 22, 51, 10, 36, 2, 34, 12, 47, 46, 2

Offset: 3

Labos Elemer, Feb 27 2001

nonn

The old name was "Quotient cycle length of sqrt(n^2+5) for n>=3." - Jianing Song, May 01 2021

			sqrt(13^2+5) = [13; 5, 4, 5, 26], so a(13) = 4.
sqrt(14^2+5) = [14; 5, 1, 1, 1, 2, 1, 8, 1, 2, 1, 1, 1, 5, 28], so a(14) = 14.
sqrt(15^2+5) = [15; 6, 30], so a(15) = 2.
sqrt(16^2+5) = [16; 6, 2, 3, 7, 1, 3, 1, 2, 1, 3, 1, 7, 3, 2, 6, 32], so a(16) = 16.

Amiram Eldar, Table of n, a(n) for n = 3..10000

Cf. A003285.

Period of continued fraction for sqrt(n^2+k): A059853 (k=3), A059855 (k=4), this sequence (k=5).

with(numtheory): [seq(nops(cfrac(sqrt(k^2+5), 'periodic', 'quotients')[2]), k=3..256)];

a[n_] := Length[ContinuedFraction[Sqrt[n^2 + 5]][[2]]]; Array[a, 100, 3] (* Amiram Eldar, Jul 10 2024 *)

If n is a multiple of 5 then a(n) = 2.

a(n) = A003285(n^2+5). - Jianing Song, May 01 2021

New name by Jianing Song, May 01 2021

A059855 Period of continued fraction for sqrt(n^2+4), n >= 1.

Original entry on oeis.org

1, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2, 5, 2

Offset: 1

Labos Elemer, Feb 27 2001

From Jianing Song, May 01 2021: (Start)
The old name was "Quotient cycle length of sqrt(n^2+4)."
Essentially the same as A010695 and A021400. (End)

			For even n, sqrt(n^2+4) = [n; n/2, 2*n], hence a(n) = 2.
For odd n > 1, sqrt(n^2+4) = [n; (n-1)/2, 1, 1, (n-1)/2, 2*n], hence a(n) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A003285, A010695, A021400.

Period of continued fraction for sqrt(n^2+k): A059853 (k=3), this sequence (k=4), A059854 (k=5).

with(numtheory): [seq(nops(cfrac(sqrt(k^2+4), 'periodic', 'quotients')[2]), k=1..100)];

a[n_] := Length @ ContinuedFraction[Sqrt[n^2 + 4]][[2]]; Array[a, 100] (* Amiram Eldar, May 13 2020 *)

a(n) = 2 for even n, a(n) = 5 for odd n > 1.

a(n) = A003285(n^2+4). - Jianing Song, May 01 2021

A077636 Length of periodic part of continued fraction expansion of square root of A051451(n), i.e., sqrt(lcm(1..x)) where x is a prime power from A000961.

Original entry on oeis.org

0, 1, 2, 2, 4, 2, 2, 2, 4, 8, 18, 14, 36, 38, 232, 268, 110, 280, 4348, 3244, 32684, 148184, 207616, 9988, 1946132, 2154482, 13319736, 8971624, 12345748, 69705504, 159413696, 1184191340, 1183672188, 23656693528, 28963250020, 701296434876, 754283490078

Offset: 1

Labos Elemer, Nov 13 2002

			For A051451(10) = 360360, the periodic part is {3,2,1,132,1,2,3,1200} with 8 terms, so a(10) = 8.

Cf. A000961, A003285, A051451.

pp = Join[{1}, Select[Range[2, 50], Mod[ #, # - EulerPhi[ # ]] == 0 &]]; Table[ Length[ Last[ ContinuedFraction[ Sqrt[ Apply[ LCM, Table[i, {i, 1, pp[[n]]}]]]]]], {n, 1, 31}]

a(n) = A003285(A051451(n)). - Michel Marcus, Sep 30 2019

Edited and extended by Robert G. Wilson v, Nov 14 2002
a(31) from Ray Chandler, Jan 16 2009
a(32)-a(35) from Chai Wah Wu, Sep 26 2019
a(36) from Chai Wah Wu, Sep 29 2019
a(37) from Chai Wah Wu, Sep 26 2021

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A096494 Largest value in the periodic part of the continued fraction of sqrt(prime(n)).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

A064932 Period of the continued fraction for sqrt(3^(2n+1)).

Views

Author

Keywords

Comments

References

Crossrefs

Programs

Mathematica

Formula

Extensions

A065938 Position of sqrt(n) in the mapping N2QuQR1 given in A065936.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A121339 Periodic part of continued fraction for square roots of integers.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A059853 Period of continued fraction for sqrt(n^2+3), n >= 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A059854 Period of continued fraction for sqrt(n^2+5), n >= 3.

Views

Author

Keywords

Comments

Examples

Links