A003285 -id:A003285 - cp's OEIS frontend

A267857 Length of the period of the continued fraction for the square root of D, the discriminant of indefinite binary quadratic forms. D is given in A079896.

1, 2, 2, 5, 1, 2, 6, 2, 4, 5, 4, 4, 1, 2, 3, 8, 6, 2, 6, 5, 2, 6, 4, 11, 1, 2, 8, 2, 7, 12, 6, 2, 2, 5, 6, 5, 8, 10, 4, 11, 1, 2, 2, 8, 15, 6, 9, 10, 6, 2, 16, 5, 4, 10, 2, 16, 4, 9, 4, 4, 1, 2, 9, 2, 8, 2, 17, 8, 10, 6, 6, 2, 16, 5, 4, 8, 4, 21

Offset: 1

Wolfdieter Lang, Feb 03 2016

This is a subsequence of A003285.
If a(n) is even then the smallest positive integer solution of the Pell equation x^2 - D(n)*y^2 = +1 with D(n) = A079896(n) is given by (x0, y0) = (P,Q) with P/Q = [a,b[1], ..., b[a(n)-1]]. If a(n) is odd then the smallest positive integer solution of the Pell equation x^2 - D(n)*y^2 = +1 is given by (x0, y0) = (P^2 + D(n)*Q^2, 2*P*Q). See e.g., the Silverman reference Theorem 40.4 on p. 351.
For positive integer d, d not a square, the Pell equations X^2 - d*Y^2 = +4 and X^2 - d*Y^2 = -4 have no proper solutions. For D(n) = A079896(n) there are solutions for X^2 - D(n)*Y^2 = +4 or -4 (inclusive or). See the Wolfdieter Lang link under A225953 for Pell +4 or -4 solutions.

			a(1)  = 1  because sqrt(5)  = [2,repeat(4)].
a(2)  = 2  because sqrt(8)  = [2,repeat(1,4)].
a(24) = 11 because sqrt(61) = [7,repeat(1,4,3,1,2,2,1,3,4,1,14)].
Pell +1 equation: n = 24 with D = 61 has odd a(24)
  P/Q = [7,1,4,3,1,2,2,1,3,4,1] = 29718/3805 (in lowest terms). Therefore (x0, y0) = (1766319049, 226153980), see A174762 (Of course, (1, 0) is the smallest nonnegative solution.)

J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 351.

Robin Visser, Table of n, a(n) for n = 1..10000

Cf. A003285, A033313, A033317, A079896, A225953.

See the Robert Israel program under A003285, adapted to n -> a(n).

Length[Last@ #] & /@ ContinuedFraction@ Sqrt@ Select[Range@ 200, And[MemberQ[{0, 1}, Mod[#, 4]], ! IntegerQ@ Sqrt@ #] &] (* Michael De Vlieger, Feb 11 2016, after A079896 *)

def a(n):
    i, D = 1, Integer(5)
    while(i < n):
        D += 1; i += 1*(((D%4) in [0, 1]) and (not D.is_square()))
    K. = QuadraticField(D)
    return continued_fraction(a).period_length()  # Robin Visser, Jun 06 2025

Offset corrected by Robin Visser, Jun 06 2025

A276689 Least term in the periodic part of the continued fraction expansion of sqrt(n) or 0 if n is square.

Original entry on oeis.org

0, 0, 2, 1, 0, 4, 2, 1, 1, 0, 6, 3, 2, 1, 1, 1, 0, 8, 4, 1, 2, 1, 1, 1, 1, 0, 10, 5, 2, 1, 2, 1, 1, 1, 1, 1, 0, 12, 6, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 0, 14, 7, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 0, 16, 8, 1, 4, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 18, 9, 6, 1

Offset: 0

Chai Wah Wu, Sep 28 2016

nonn

If r > 0 is even, then a((rm/2)^2+m) = r for all m >= 1 and a((r^2-2)^2/4 + (r+1)^3) = r.

If r is odd, then a((rm)^2+2m) = r for all m >= 1 and a(r^4 + r^3 + 5(r+1)^2/4) = r.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A031424, A003285, A091453, A096491.

from sympy import continued_fraction_periodic
def A276689(n):
    x = continued_fraction_periodic(0,1,n)
    return min(x[1]) if len(x) > 1 else 0

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.

A293028 Period of the continued fraction for sqrt(5^(2n+1)).

Original entry on oeis.org

1, 5, 31, 153, 745, 3779, 19011, 95121, 475251, 2375759, 11880177, 59406541, 297033919, 1485165459, 7425797777, 37129022423

Offset: 0

A.H.M. Smeets, Sep 28 2017

Cf. A003285, A013710, A064932.

Table[Length[Last[ContinuedFraction[Sqrt[5^(2n+1)] ]]],{n,0,11}] (* Metin Sariyar, Sep 25 2019 *)

a(n) = A003285(A013710(n)). - Michel Marcus, Sep 25 2019

a(11)-a(13) from Daniel Suteu, Jan 24 2019
Better name from Daniel Suteu, Jan 24 2019
a(14)-a(15) from Chai Wah Wu, Sep 25 2019

A384923 a(n) is the smallest number of leading significant digits of the square root of the n-th nonsquare that includes all decimal digits.

Original entry on oeis.org

19, 23, 37, 39, 45, 36, 27, 17, 25, 15, 36, 19, 20, 36, 25, 37, 28, 13, 27, 52, 39, 17, 38, 27, 26, 17, 23, 24, 37, 19, 25, 26, 26, 41, 58, 57, 25, 12, 25, 22, 24, 19, 33, 48, 23, 41, 49, 23, 32, 32, 23, 30, 19, 17, 31, 27, 24, 47, 24, 26, 18, 22, 19, 48, 31, 22

Offset: 1

Felix Huber, Jun 26 2025

Squares are excluded by definition because a(n) would only exist for positive integers s that include all decimal digits. The smallest square s^2 for which a(n) would exist is 1023456789^2 = 1047463798950190521.

			The leading 19 significant digits of sqrt(2) are [1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8]. These digits include all decimal digits, with the digit '8' appearing for the first time at position 19. Since 2 is the first nonsquare, it follows that a(1) = 19.

Felix Huber, Table of n, a(n) for n = 1..10000
Wikipedia, Significant Figures

Cf. A000037, A000196, A003285, A061845, A113507, A384924.

A384923:=proc(n)
    local m,b,k;
    m:=n+floor(1/2+sqrt(n));
    b:=floor(log10(sqrt(m)));
    k:=9-b;
    while nops(convert(ListTools:-Reverse(convert(floor(10^k*sqrt(m)),'base',10)),set))<10 do
        k:=k+1
    od;
    return k+b+1
end proc;
seq(A384923(n),n=1..66);

from itertools import count
from math import isqrt
def A384923(n):
    m = n+(k:=isqrt(n))+(n>k*(k+1))
    return 1+next(n for n in count(9) if len(set(str(isqrt(10**(n<<1)*m))))==10) # Chai Wah Wu, Jul 01 2025

a(n) >= max(10, A384924(n)).

a(A113507(k) - floor(sqrt(A113507(k)))) = 10 for positive integers k.

A384924 a(n) is the position of the first occurrence of the digit 0 among the leading significant decimal digits of the square root of the n-th nonsquare.

Original entry on oeis.org

14, 5, 5, 17, 11, 16, 10, 10, 6, 3, 36, 12, 6, 7, 13, 37, 16, 4, 26, 52, 2, 12, 6, 9, 11, 13, 16, 14, 4, 5, 2, 8, 18, 10, 3, 4, 12, 10, 3, 20, 9, 6, 2, 48, 6, 4, 49, 11, 32, 13, 9, 15, 19, 4, 5, 21, 2, 5, 24, 17, 3, 6, 19, 16, 5, 3, 4, 11, 17, 7, 19, 9, 2, 4, 16

Offset: 1

Felix Huber, Jun 26 2025

			The leading 14 significant digits of sqrt(2) are [1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0], with the digit '0' appearing for the first time at position 14. Since 2 is the first nonsquare, it follows that a(1) = 14.

Felix Huber, Table of n, a(n) for n = 1..10000
Wikipedia, Significant Figures

Cf. A000037, A000196, A003285, A113507, A384923.

A384924:=proc(n)
    local m,b,k;
    m:=n+floor(1/2+sqrt(n));
    b:=floor(log10(sqrt(m)));
    k:=1-b;
    while not member(0,ListTools:-Reverse(convert(floor(10^k*sqrt(m)),'base',10))) do
        k:=k+1
    od;
    return k+b+1
end proc;
seq(A384924(n),n=1..75);

b[n_] := (n + Floor[Sqrt[n + Floor[Sqrt[n]]]]);a[n_]:=Position[RealDigits[N[Sqrt[b[n]],100]][[1]],0][[1]];Array[a,75]//Flatten (* Increase precision for n>23000 *) (* James C. McMahon, Jul 05 2025 *)

from itertools import count
from math import isqrt
def A384924(n):
    m = n+(k:=isqrt(n))+(n>k*(k+1))
    return 1+next(n for n in count(1) if not isqrt(10**(n<<1)*m)%10) # Chai Wah Wu, Jul 01 2025

2 <= a(n) <= A384923(n).

A020376 Numbers k such that the continued fraction for sqrt(k) has period 37.

Original entry on oeis.org

421, 769, 1409, 1481, 2237, 2609, 2977, 3109, 3194, 3257, 3301, 3517, 4282, 5617, 6154, 6637, 7474, 8573, 8770, 9181, 9685, 9893, 9965, 10042, 10085, 10949, 11674, 12185, 12301, 12473, 12658, 12953, 13061, 13697, 13885, 15025, 15146, 15889, 16042

Offset: 1

David W. Wilson

nonn

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

Cf. A003285.

cf37Q[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==37]; Select[Range[17000],cf37Q] (* Harvey P. Dale, Jan 03 2019 *)

A020377 Numbers k such that the continued fraction for sqrt(k) has period 38.

Original entry on oeis.org

619, 669, 691, 811, 1057, 1108, 1357, 1461, 1629, 1723, 1809, 1816, 1819, 1857, 2084, 2169, 2182, 2185, 2201, 2246, 2266, 2356, 2391, 2515, 2571, 2578, 2634, 2950, 2959, 2979, 2983, 3139, 3235, 3254, 3382, 3411, 3441, 3460, 3473, 3558, 3574, 3621, 3714

Offset: 1

David W. Wilson

nonn

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

Cf. A003285.

q[k_] := Module[{c = ContinuedFraction[Sqrt[k]]}, Length[c] == 2 && Length[c[[2]]] == 38]; Select[Range[4000], q] (* Amiram Eldar, Aug 22 2025 *)

A020378 Numbers k such that the continued fraction for sqrt(k) has period 39.

Original entry on oeis.org

541, 661, 3809, 4637, 4874, 6250, 6625, 6922, 7033, 7633, 7642, 7706, 8017, 8693, 9497, 9533, 10337, 10553, 10585, 10874, 11773, 12337, 12370, 12401, 13037, 13282, 13781, 13810, 13898, 14281, 14290, 14741, 15425, 15970, 16153, 16762, 17362, 18365

Offset: 1

David W. Wilson

nonn

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

Cf. A003285.

q[k_] := Module[{c = ContinuedFraction[Sqrt[k]]}, Length[c] == 2 && Length[c[[2]]] == 39]; Select[Range[19000], q] (* Amiram Eldar, Aug 22 2025 *)

cp's OEIS Frontend

A267857 Length of the period of the continued fraction for the square root of D, the discriminant of indefinite binary quadratic forms. D is given in A079896.

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A276689 Least term in the periodic part of the continued fraction expansion of sqrt(n) or 0 if n is square.

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

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

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A293028 Period of the continued fraction for sqrt(5^(2n+1)).

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A384923 a(n) is the smallest number of leading significant digits of the square root of the n-th nonsquare that includes all decimal digits.

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A384924 a(n) is the position of the first occurrence of the digit 0 among the leading significant decimal digits of the square root of the n-th nonsquare.

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Maple

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A020376 Numbers k such that the continued fraction for sqrt(k) has period 37.

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