A003285 -id:A003285 - cp's OEIS frontend

A091449 Array T(n,k) read by antidiagonals, where row n is the increasing sequence of numbers m for which the simple continued fraction of sqrt(m) has period n, n >= 0, k >= 1.

1, 2, 4, 3, 5, 9, 41, 6, 10, 16, 7, 130, 8, 17, 25, 13, 14, 269, 11, 26, 36, 19, 29, 23, 370, 12, 37, 49, 58, 21, 53, 28, 458, 15, 50, 64, 31, 73, 22, 74, 32, 697, 18, 65, 81, 106, 44, 202, 45, 85, 33, 986, 20, 82, 100, 43, 113, 69, 250, 52, 89, 34, 1313, 24, 101, 121

Offset: 0

Clark Kimberling, Feb 03 2004

A permutation of the positive integers.

			Array begins:
  n\k|   1   2   3   4   5   6   7    8    9   10   11
  ---+------------------------------------------------
   0 |   1   4   9  16  25  36  49   64   81  100  121
   1 |   2   5  10  17  26  37  50   65   82  101  122
   2 |   3   6   8  11  12  15  18   20   24   27   30
   3 |  41 130 269 370 458 697 986 1313 1325 1613 1714
   4 |   7  14  23  28  32  33  34   47   55   60   62
   5 |  13  29  53  74  85  89 125  173  185  218  229
   6 |  19  21  22  45  52  54  57   59   70   77   88
   7 |  58  73 202 250 274 314 349  425  538  761 1010
   8 |  31  44  69  71  91  92 108  135  153  158  160
   9 | 106 113 137 149 265 389 493  610  698  754  970
  10 |  43  67  86  93 115 116 118  129  154  159  161
The least n for which CF(sqrt(n)) has period of length 4 is n=7, with CF=[2;1,1,1,4,1,1,1,4,1,1,1,4,...]; thus T(4,1)=7.
[The array T(n,k) is indexed by n=0,1,2,3,..., k=1,2,3... .]
Row 0 consists of squares: 1,4,9,...

Cf. A002522, A003285, A013642, A091450, A091451, A091453.

Rows 0-100 are: A000290 (except the initial 0), A002522 (except the initial 1), A013642, A013643, A013644, A010337, A020347, A010338, A020348, A010339, A020349-A020439.

a(17) = T(3,3) corrected by Pontus von Brömssen, Nov 23 2024

A091451 Array T(n,k) read by antidiagonals: (row 0)=squares, (row 1)={numbers m for which the simple continued fraction (CF) of sqrt(m) has period length 1}; once (row n) is defined, let (row n+1) begin with the least positive integer not already in a row and let the rest of (row n+1) be the other m's for which CF(sqrt(m)) has the same period length.

Original entry on oeis.org

1, 2, 4, 3, 5, 9, 7, 6, 10, 16, 13, 14, 8, 17, 25, 19, 29, 23, 11, 26, 36, 31, 21, 53, 28, 12, 37, 49, 41, 44, 22, 74, 32, 15, 50, 64, 43, 130, 69, 45, 85, 33, 18, 65, 81, 46, 67, 269, 71, 52, 89, 34, 20, 82, 100, 58, 76, 86, 370, 91, 54, 125, 47, 24, 101, 121

Offset: 0

Clark Kimberling, Feb 03 2004

A permutation of the positive integers.
From Pontus von Brömssen, Nov 23 2024: (Start)
Rows of A091449 sorted by the first term.
First column gives indices of new terms of A003285.
(End)

			7 is the least positive integer not in rows 0,1,2, so 7=T(3,0); the period length of sqrt(7) is 4, as is the case with T(3,1)=14, T(3,2)=23, etc.
Corner:
  1    4    9    16    25    36   49   64
  2    5   10    17    26    37   50   65
  3    6    8    11    12    15   18   20
  7   14   23    28    32    33   34   47
 13   29   53    74    85    89  125  173
 19   21   22    45    52    54   57   59

Cf. A003285, A091449, A091453.

Map[Map[#[[1]] &, #] &,
 GatherBy[Map[{#, Flatten[ContinuedFraction[Sqrt[#]]]} &, Range[500]],
   Length[#[[2]]] &]]  (* Peter J. C. Moses, May 11 2023 *)

a(47) = T(7,2) corrected by Clark Kimberling, May 20 2023

More terms from Pontus von Brömssen, Nov 23 2024

A096495 Number of distinct terms in the periodic part of the continued fraction for sqrt(prime(n)).

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 1, 4, 3, 3, 4, 1, 2, 4, 3, 3, 4, 5, 5, 4, 3, 3, 2, 3, 3, 1, 5, 4, 6, 3, 6, 4, 3, 6, 5, 7, 5, 6, 3, 3, 6, 6, 6, 5, 1, 7, 8, 3, 2, 3, 3, 6, 5, 5, 1, 4, 2, 7, 7, 5, 6, 3, 6, 6, 6, 5, 8, 6, 5, 4, 4, 3, 7, 3, 9, 4, 3, 7, 1, 6, 6, 8, 7, 6, 3, 2, 5, 7, 5, 9, 4, 6, 9, 8, 4, 4, 6, 6, 8, 9, 8, 2, 4, 6, 10

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) = 6.

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

Cf. A003285, A028832, A054269, A005980, A096491, A096492, A096493, A096494, A096496.

{te=Table[0, {m}], u=1}; Do[s=Length[Union[Last[ContinuedFraction[Prime[n]^(1/2)]]]]; te[[u]]=s;u=u+1, {n, 1, m}];te
Table[Length[Union[ContinuedFraction[Sqrt[Prime[n]]][[2]]]],{n,110}] (* Harvey P. Dale, Jun 22 2017 *)

a(n) = A028832(A000040(n)). - Amiram Eldar, Nov 10 2021

A096496 Number of distinct primes in the periodic part of the continued fraction for sqrt(prime(n)).

Original entry on oeis.org

1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 2, 0, 1, 2, 1, 1, 2, 2, 3, 2, 1, 1, 0, 2, 1, 0, 1, 1, 2, 1, 4, 2, 1, 4, 2, 4, 3, 4, 1, 0, 4, 1, 3, 2, 0, 3, 4, 1, 0, 1, 1, 2, 2, 2, 0, 0, 1, 1, 3, 1, 1, 0, 4, 3, 3, 1, 5, 3, 2, 2, 2, 1, 3, 2, 4, 2, 1, 2, 0, 3, 4, 5, 5, 3, 1, 0, 3, 4, 1, 4, 1, 3, 3, 2, 1, 1, 2, 2, 2, 4, 4, 0, 2, 3, 4

Offset: 1

Labos Elemer, Jun 29 2004

nonn

			n=31: prime(31) = 127, and the periodic part of the continued fraction of sqrt(127) is {3,1,2,2,7,11,7,2,2,1,3,22}, so a(31) = 4.

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

Cf. A003285, A054269, A005980, A096491, A096492, A096493, A096494, A096495.

{te=Table[0, {m}], u=1}; Do[s=Count[PrimeQ[Union[Last[ContinuedFraction[f[n]^(1/2)]]]], True]; te[[u]]=s;u=u+1, {n, 1, m}];te
Count[Union[ContinuedFraction[Sqrt[#]][[2]]],?PrimeQ]&/@Prime[ Range[ 110]] (* _Harvey P. Dale, Apr 27 2016 *)

A130272 Primes p for which the period of the continued fraction of sqrt(p) increases.

Original entry on oeis.org

2, 3, 7, 13, 19, 31, 43, 61, 103, 109, 139, 151, 181, 211, 331, 421, 541, 571, 631, 751, 919, 1291, 1381, 1549, 1579, 1621, 1759, 1831, 2011, 2311, 2671, 3019, 3469, 3691, 3931, 4909, 4951, 4999, 5119, 6211, 6451, 6679, 8269, 8719, 8779, 8941, 9739, 9949

Offset: 1

T. D. Noe, May 19 2007

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..344 (terms n = 1..200 from T. D. Noe)

Cf. A003285, A013645.

mx=0; n=0; t=Table[n++; While[p=Prime[n]; len=Length[Last[ContinuedFraction[Sqrt[p]]]]; len<=mx, n++ ]; mx=len; p, {50}]

Where records occur in A054269.

A288186 Numbers k such that the continued fractions for sqrt(k) and sqrt(k+1) have the same period.

Original entry on oeis.org

11, 21, 32, 33, 38, 39, 78, 83, 91, 95, 104, 111, 115, 140, 141, 146, 147, 161, 164, 204, 205, 206, 219, 222, 227, 230, 242, 245, 299, 310, 320, 321, 326, 327, 340, 344, 371, 383, 395, 404, 413, 428, 434, 438, 443, 447, 451, 452, 464, 471, 498, 504, 515, 539, 545, 572, 573, 578, 579, 594, 596, 644, 654, 659, 695

Offset: 1

Ilya Gutkovskiy, Jun 06 2017

nonn

Numbers k such that A003285(k) = A003285(k+1).

			11 is in the sequence because sqrt(11) = 3 + 1/(3 + 1/(6 + 1/(3 + 1/(6 + 1/...)))), period 2: [3, 6] and sqrt(12) = 3 + 1/(2 + 1/(6 + 1/(2 + 1/(6 + 1/...)))), period  2: [2, 6].

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

Cf. A003285, A013943, A031400, A097853.

A020406 Numbers k such that the continued fraction for sqrt(k) has period 67.

Original entry on oeis.org

1381, 1789, 2521, 3754, 3994, 5953, 6389, 7349, 7765, 7801, 8297, 8842, 9274, 10253, 12413, 14458, 15349, 15749, 16073, 16229, 19073, 21485, 22433, 24106, 24317, 24506, 24538, 24929, 25577, 25706, 26170, 26773, 26954, 28018, 28345, 30197, 32218, 32290

Offset: 1

David W. Wilson

nonn

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

Cf. A003285.

cf67Q[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==67]; Select[Range[33000],cf67Q] (* Harvey P. Dale, Sep 21 2018 *)

A020640 Successive record periods of continued fraction for sqrt(k) (period of continued fraction for sqrt(A013645(n+1))).

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 10, 12, 16, 18, 20, 22, 26, 34, 37, 40, 42, 44, 48, 52, 54, 60, 62, 64, 70, 76, 79, 88, 94, 96, 102, 104, 108, 114, 118, 122, 130, 136, 152, 156, 158, 170, 172, 174, 194, 196, 202, 207, 210, 217, 228, 234, 238, 239, 248, 262, 268, 280, 281

Offset: 1

David W. Wilson

Patrick McKinley, Table of n, a(n) for n=1..512

Cf. A003285, A013645.

per[n_] := ContinuedFraction[Sqrt[n]][[2]] // Length; record = 0; t = Reap[For[n = 1, n < 2*10^4, n++, If[ !IntegerQ[Sqrt[n]], p = per[n]; If[p > record, record = p; Print[{n, p}]; Sow[{n, p}]]]]][[2, 1]]; A020640 = t[[All, 2]] (* Jean-François Alcover, Dec 27 2012 *)
DeleteDuplicates[Table[If[IntegerQ[Sqrt[n]],Nothing,Length[ContinuedFraction[Sqrt[n]][[2]]]],{n,20000}],GreaterEqual] (* Harvey P. Dale, Dec 28 2023 *)

a(n) = A003285(A013645(n+1)). - Pontus von Brömssen, Nov 24 2024

A059904 Periodic part of continued fraction for sqrt(n), encoded by recursively interleaving the bits in the binary expansions of the repeating terms.

Original entry on oeis.org

0, 4, 33, 0, 16, 516, 549755813899, 513, 0, 20, 549, 548, 604462909948052075708555, 549764202537, 545, 0, 64, 8208, 13479973333575319897333507543509815336818572211270286381289293482126

Offset: 1

Marc LeBrun, Feb 07 2001

Could be made less gigantic by omitting final terms in continued fraction, which are always 2*c0.

			sqrt(3)=1+[1,2] so a(3) is encoded as:
.....0 0 0 0 1 -> 1
....... 1 . 0 .-> 2
--------------
.....000100001 = 33.

A059903, A059884, A003285.

a(n) = A059884(A059903(n)).

A064025 Length of period of the continued fraction for sqrt(n!).

Original entry on oeis.org

1, 2, 2, 2, 4, 2, 16, 48, 8, 4, 56, 180, 44, 156, 300, 7936, 10388, 11516, 9104, 13469268, 2684084, 2418800, 28468692, 143007944, 85509116, 402570696, 2287868888, 204306960, 48715166536, 147160740856, 317585614148

Offset: 2

Labos Elemer, Sep 18 2001

			Quotients for 10! are [[1904], [1, 15, 1, 13, 1, 15, 1, 3808]], so period length of 10! is 8.

Cf. A000142, A003285.

with(numtheory): [seq(nops(cfrac(sqrt(k!),'periodic','quotients')[2]),k=2..16)];

Do[ Print[ Length[ Last[ ContinuedFraction[ Sqrt[ n! ]]]]], {n, 2, 24} ]

a(n) = A003285(A000142(n)). - Michel Marcus, Sep 25 2019

More terms from Robert G. Wilson v, Oct 01 2001
a(25)-a(28) from Daniel Suteu, Jan 24 2019
a(29) from Chai Wah Wu, Sep 23 2019
a(30) from Chai Wah Wu, Sep 25 2019
a(31) from Chai Wah Wu, Jan 27 2021
a(32) from Chai Wah Wu, Feb 02 2021

cp's OEIS Frontend

A091449 Array T(n,k) read by antidiagonals, where row n is the increasing sequence of numbers m for which the simple continued fraction of sqrt(m) has period n, n >= 0, k >= 1.

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A096495 Number of distinct terms in the periodic part of the continued fraction for sqrt(prime(n)).

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A096496 Number of distinct primes in the periodic part of the continued fraction for sqrt(prime(n)).

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A130272 Primes p for which the period of the continued fraction of sqrt(p) increases.

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A288186 Numbers k such that the continued fractions for sqrt(k) and sqrt(k+1) have the same period.

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A020406 Numbers k such that the continued fraction for sqrt(k) has period 67.

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A020640 Successive record periods of continued fraction for sqrt(k) (period of continued fraction for sqrt(A013645(n+1))).

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A059904 Periodic part of continued fraction for sqrt(n), encoded by recursively interleaving the bits in the binary expansions of the repeating terms.

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A064025 Length of period of the continued fraction for sqrt(n!).