A003814 -id:A003814 - cp's OEIS frontend

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

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

A284601 Numbers k such that the decimal representation of 1/k does not terminate and has odd period.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 24, 27, 30, 31, 36, 37, 41, 43, 45, 48, 53, 54, 60, 62, 67, 71, 72, 74, 75, 79, 81, 82, 83, 86, 90, 93, 96, 106, 107, 108, 111, 120, 123, 124, 129, 134, 135, 142, 144, 148, 150, 151, 155, 158, 159, 162, 163, 164, 166, 172, 173, 180, 185, 186, 191, 192, 199, 201, 205, 212, 213, 214, 215

Offset: 1

Ilya Gutkovskiy, Mar 30 2017

From Robert G. Wilson v, Apr 02 2017: (Start)
If k is in the sequence, then so are 2k and 5k.
The complement of A284602.
Primitives: 3, 9, 27, 31, 37, 41, 43, 53, 67, 71, 79, 81, 83, 93, 107, 111, 123, ..., .
(End)
From Robert Israel, Apr 03 2017: (Start)
Numbers of the form 2^j * 5^k * m where m > 1, gcd(m,10)=1 and the multiplicative order of 10 (mod m) is odd.
Complement of A003592 in the multiplicative semigroup generated by A186635, i.e., numbers whose prime factors are in A186635 with at least one prime factor not 2 or 5. (End)

			27 is in the sequence because 1/27 = 0.0370(370)... period is 3, 3 is odd.
2 and 5 are not in the sequence because 1/2 = 0.5 and 1/5 = 0.2 are terminating expansions. See also comments in A051626 and A284602.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Index entries for sequences related to decimal expansion of 1/n

Cf. A002371, A003592, A003814, A051626, A186635, A284602.

filter:= proc(n) local m;
  m:= n/2^padic:-ordp(n,2);
  m:= m/5^padic:-ordp(m,5);
  m > 1 and numtheory:-order(10,m)::odd
end proc:
select(filter, [$1..1000]); # Robert Israel, Apr 03 2017

Select[Range[215], Mod[Length[RealDigits[1/#][[1, -1]]], 2] == 1 & ]

A031422 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 9.

Original entry on oeis.org

601, 1073, 1930, 2017, 2621, 2825, 3037, 3533, 3769, 4013, 4714, 5701, 6218, 6373, 6689, 7013, 7757, 8461, 8825, 9197, 9277, 12629, 13394, 13621, 14081, 14549, 15613, 15754, 18265, 18797, 20005, 20282, 20441, 21410, 22277, 22993, 23762, 24065, 24370, 25114

Offset: 1

David W. Wilson

nonn

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

Cf. A031404-A031423.

Subsequence of A003814.

n = 1; t = {}; While[Length[t] < 50, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 9, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 04 2014 *)

a(1) removed by T. D. Noe, Apr 04 2014

A206587 Numbers k such that the periodic part of the continued fraction of sqrt(k) has even length.

Original entry on oeis.org

1, 3, 4, 6, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 59, 60, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 75, 76, 77, 78, 79, 80, 81, 83, 84

Offset: 1

T. D. Noe, Mar 20 2012

nonn

This is A206586 and the squares, A000290.

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

Cf. A003814 (period has odd length).

Cf. A000290, A206586.

Select[Range[100], IntegerQ[Sqrt[#]] || EvenQ[Length[ContinuedFraction[Sqrt[#]][[2]]]] &]

cyc(cf) = {
  if(#cf==1, return([])); \\ There is no cycle
  my(s=[]);
  for(k=2, #cf,
    s=concat(s, cf[k]);
    if(cf[k]==2*cf[1], return(s)) \\ Cycle found
  );
  0 \\ Cycle not found
}
select(n->#cyc(contfrac(sqrt(n)))%2==0, vector(400, n, n)) \\ Colin Barker, Oct 19 2014

A307303 Triangle T(n, k) read as upwards antidiagonals of array A, where A(n, k) is the number of families (also called classes) of proper solutions of the Pell equation x^2 - D(n)*y^2 = -k, for k >= 1.

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, Apr 20 2019

For details see A324252 which gives the array for the numbers of families of proper solutions of x^2 - D(n)*y^2 = k for positive integers k. See also the W. Lang link in A324251, section 3.
The D(n) values for nonzero entries in column k = 1 are given in A003814 (representation of -1).
The position list for nonzero entries in row n = 1 is A057126 (conjecture).

			The array A(n, k) begins:
n,  D(n) \k  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...
-------------------------------------------------------------------
1,   2:      1  1  0  0  0  0  2  0  0  0  0  0  0  2  0
2,   3:      0  1  1  0  0  0  0  0  0  0  2  0  0  0  0
3,   5:      1  0  0  2  1  0  0  0  0  0  2  0  0  0  0
4,   6:      0  1  0  0  2  1  0  0  0  0  0  0  0  0  2
5,   7:      0  0  2  0  0  2  1  0  0  0  0  0  0  1  0
6,   8:      0  0  0  1  0  0  2  1  0  0  0  0  0  0  0
7,  10:      1  0  0  0  0  2  0  0  2  1  0  0  0  0  2
8,  11:      0  1  0  0  0  0  2  0  0  2  1  0  0  0  0
9,  12:      0  0  1  0  0  0  0  2  0  0  2  1  0  0  0
10, 13:      1  0  2  2  0  0  0  0  2  0  0  4  1  0  0
11, 14:      0  0  0  0  2  0  1  0  0  2  0  0  2  1  0
12, 15:      0  0  0  0  0  1  0  0  0  0  2  0  0  2  1
13, 17:      1  0  0  0  0  0  0  2  0  0  0  0  2  0  0
14, 18:      0  1  0  0  0  0  0  0  2  0  0  0  0  2  0
15, 19:      0  1  2  0  0  0  0  0  0  2  0  0  0  0  4
16, 20:      0  0  0  1  0  0  0  0  0  0  2  0  0  0  0
17, 21:      0  0  1  0  2  0  0  0  0  0  0  2  0  0  0
18, 22:      0  1  0  0  0  0  2  0  0  0  0  0  2  0  0
19, 23:      0  0  0  0  0  0  0  0  0  0  2  0  0  2  0
20, 24:      0  0  0  0  0  0  0  1  0  0  0  0  0  0  2
-------------------------------------------------------------------
The triangle T(n, k) begins:
n\k   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 ..
1:    1
2:    0  1
3:    1  1  0
4:    0  0  1  0
5:    0  1  0  0  0
6:    0  0  0  2  0  0
7:    1  0  2  0  1  0  2
8:    0  0  0  0  2  0  0  0
9:    0  1  0  1  0  1  0  0  0
10:   1  0  0  0  0  2  0  0  0  0
11:   0  0  1  0  0  0  1  0  0  0  0
12:   0  0  2  0  0  2  2  0  0  0  2  0
13:   1  0  0  2  0  0  0  1  0  0  2  0  0
14:   0  0  0  0  0  0  2  0  0  0  0  0  0  2
15:   0  1  0  0  2  0  0  0  2  0  0  0  0  0  0
16:   0  1  0  0  0  0  0  2  0  1  0  0  0  0  0  0
17:   0  0  2  0  0  1  1  0  0  2  0  0  0  0  0  0  2
18:   0  0  0  0  0  0  0  0  2  0  1  0  0  1  2  0  0  0
19:   0  1  1  1  0  0  0  0  0  0  2  0  0  0  0  0  0  0  0
20:   0  0  0  0  0  0  0  2  0  2  0  1  0  0  0  0  0  0  0  0
...
For this triangle more than the shown columns of the array have been used.
----------------------------------------------------------------------------
A(5, 6) = 2 = T(10, 6)  because D(5) =  7, and the Pell form F(5) with disc(F(5)) = 4*7 = 28 representing k = -6 has 2 families (classes) of proper solutions generated from the two positive fundamental positive solutions (x10, y10) = (13, 5) and  (x20, y20) = (1, 1). They are obtained from the trivial solutions of the parallel forms [-6, 2, 1] and [-6, 10, -3], respectively.

D. A. Buell, Binary Quadratic Forms, Springer, 1989.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973.

Cf. A000037, A000194, A003814, A057126, A324252 (positive k), A324251.

T(n, k) = A(n-k+1, k) for 1 <= k <= n, with A(n,k) the number of proper (positive) fundamental solutions of the Pell equation x^2 - D(n)*y^2 = -k for k >= 1, with D(n) = A000037(n), for n >= 1. Each such fundamental solution generates a family of proper solutions.

A031414 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 1.

Original entry on oeis.org

13, 29, 53, 58, 74, 85, 97, 106, 125, 137, 157, 173, 185, 229, 233, 241, 293, 298, 314, 338, 346, 353, 365, 389, 397, 425, 433, 445, 457, 461, 533, 538, 541, 554, 557, 593, 629, 634, 641, 661, 673, 698, 733, 746, 754, 769, 794, 818, 821, 829, 845, 857, 877

Offset: 1

David W. Wilson

nonn

			The continued fraction of sqrt[29] is {5; 2, 1, 1, 2, 10}. The center number in the periodic part is 1.

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

Cf. A031404-A031423.

Subsequence of A003814.

n = 1; t = {}; While[Length[t] < 60, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 1, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 03 2014 *)

Initial 2 removed by T. D. Noe, Apr 03 2014

Definitions of A031414-A031423 clarified by N. J. A. Sloane, Aug 18 2021

A031415 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 2.

Original entry on oeis.org

41, 61, 113, 130, 181, 202, 265, 269, 313, 317, 394, 421, 458, 586, 613, 617, 685, 697, 761, 773, 853, 925, 929, 937, 986, 1013, 1066, 1109, 1117, 1201, 1213, 1301, 1325, 1354, 1409, 1417, 1429, 1466, 1586, 1625, 1637, 1649, 1714, 1741, 1745, 1753, 1861

Offset: 1

David W. Wilson

nonn

In general, the simple continued fraction expansion of sqrt(m) is a periodic palindromic sequence. That is, contfrac( sqrt(m) ) = [c(0); c(1), c(2), ..., c(p), c(p+1), ...] where p is the period. c(p) = 2*c(0), c(k) = c(p+k) for k>0, c(k) = c(p-k) for p>k>0. If the period p is odd, then p = 2*k+1 and c(k) = c(k+1) can be considered a pair of equal central terms. If the period is even, then p = 2*k and the unique central term is c(k). - Michael Somos, Apr 04 2014

			The simple continued fraction expansion of sqrt(41) = [6; 2, 2, 12,  2, 2, 12, 2, 2, 12, ...] with odd period 3 and two terms equal to 2. Another example is sqrt(202) = [14; 4, 1, 2, 2, 1, 4, 28, 4, 1, 2, 2, 1, 4, 28, 4, 1, 2, 2, 1, 4, 28,  ...] with odd period 7 and two terms equal to 2. - _Michael Somos_, Apr 03 2014

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..999 from T. D. Noe)

Cf. A031404-A031423.

Subsequence of A003814.

n = 1; t = {}; While[Length[t] < 50, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 2, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 03 2014 *)

from sympy.ntheory.continued_fraction import continued_fraction_periodic
A031415_list = []
for n in range(1,10**3):
    cf = continued_fraction_periodic(0,1,n)
    if len(cf) > 1 and len(cf[1]) > 1 and len(cf[1]) % 2 and cf[1][len(cf[1])//2] == 2:
        A031415_list.append(n) # Chai Wah Wu, Sep 16 2021

a(1) corrected by T. D. Noe, Apr 03 2014

A031416 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 3.

Original entry on oeis.org

89, 149, 193, 218, 250, 277, 337, 493, 521, 569, 653, 709, 914, 1009, 1018, 1037, 1385, 1465, 1553, 1597, 1618, 1754, 1781, 1898, 1921, 1973, 1994, 2069, 2129, 2146, 2293, 2378, 2389, 2441, 2474, 2561, 2725, 2741, 2777, 2897, 2957, 2986, 3170, 3229, 3265

Offset: 1

David W. Wilson

nonn

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

Cf. A031404-A031423.

Subsequence of A003814.

n = 1; t = {}; While[Length[t] < 60, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 3, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 03 2014 *)
cfo3Q[n_]:=Module[{s=Sqrt[n],cf,len},cf=If[IntegerQ[s],{0,0}, ContinuedFraction[ s ][[2]]];len=Length[cf];OddQ[len]&&cf[[ (len+1)/2]] == cf[[(len-1)/2]]==3]; Select[Range[3300],cfo3Q] (* Harvey P. Dale, Sep 25 2019 *)

Initial 10 removed by T. D. Noe, Apr 03 2014

A031417 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 4.

Original entry on oeis.org

274, 370, 481, 797, 953, 1069, 1249, 1313, 1378, 1381, 1514, 1657, 1658, 1733, 1889, 2125, 2297, 2377, 2554, 2557, 2833, 2834, 2929, 2941, 3226, 3329, 3338, 3433, 3541, 3761, 3874, 3989, 4093, 4106, 4441, 4442, 4561, 4682, 4685, 4933, 4937, 5197, 5450

Offset: 1

David W. Wilson

nonn

			The simple continued fraction for sqrt(274) = [16; 1, 1, 4, 4, 1, 1, 32, ...] with odd period 7 and central term 4. Another example is sqrt(481) = [21; 1, 13, 1, 1, 1, 4, 4, 1, 1, 1, 13, 1, 42, ...] with odd period 13 and central term 4. - _Michael Somos_, Apr 03 2014

T. D. Noe, Table of n, a(n) for n = 1..999

Cf. A031404-A031423.

Subsequence of A003814.

n = 1; t = {}; While[Length[t] < 50, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 4, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 03 2014 *)
cf4Q[n_]:=Module[{s=Sqrt[n],cf,len},cf=If[IntegerQ[s],{1,1},ContinuedFraction[ s][[2]]];len=Length[cf];OddQ[len]&&cf[[(len+1)/2]] == cf[[(len-1)/2]]==4]; Select[Range[5500],cf4Q] (* Harvey P. Dale, Jul 28 2021 *)

a(1) corrected by T. D. Noe, Apr 03 2014

A031418 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 5.

Original entry on oeis.org

73, 373, 449, 565, 610, 757, 1021, 1145, 1193, 1594, 1669, 1906, 2053, 2074, 2138, 2314, 2477, 2593, 2861, 3065, 3145, 4129, 4346, 4373, 4469, 4498, 4721, 5018, 5114, 5386, 5741, 6025, 6317, 6617, 6737, 6925, 7241, 7489, 7522, 7897, 7978, 8017, 8186, 8314

Offset: 1

David W. Wilson

nonn

			The simple continued fraction expansion of sqrt(73) = [8, 1, 1, 5, 5, 1, 1, 16, ...] of odd period 7 with a pair of central terms both equal to 5. Another example is sqrt(373) = [19, 3, 5, 5, 3, 38, ...] of odd period 5 with a pair of central terms both equal to 5. - _Michael Somos_, Apr 03 2014

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

Cf. A031404-A031423.

Subsequence of A003814.

opct5Q[n_]:=Module[{s=Sqrt[n],cf,len},If[IntegerQ[s],cf={1,1}, cf= ContinuedFraction[s][[2]]];len=Length[cf];OddQ[len] && cf[[Floor[len/2]]] == cf[[Ceiling[len/2]]]==5]; Select[Range[10000],opct5Q] (* Harvey P. Dale, Feb 22 2013 *)

Corrected and extended by Harvey P. Dale, Feb 22 2013

cp's OEIS Frontend

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

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A284601 Numbers k such that the decimal representation of 1/k does not terminate and has odd period.

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A031422 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 9.

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A206587 Numbers k such that the periodic part of the continued fraction of sqrt(k) has even length.

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A307303 Triangle T(n, k) read as upwards antidiagonals of array A, where A(n, k) is the number of families (also called classes) of proper solutions of the Pell equation x^2 - D(n)*y^2 = -k, for k >= 1.

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A031414 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 1.

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A031415 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 2.

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A031416 Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 3.

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