A157265 -id:A157265 - cp's OEIS frontend

A020347 Numbers k such that the continued fraction for sqrt(k) has period 6.

19, 21, 22, 45, 52, 54, 57, 59, 70, 77, 88, 107, 111, 112, 114, 117, 131, 164, 165, 175, 178, 183, 187, 208, 216, 221, 232, 267, 270, 273, 275, 278, 280, 285, 294, 296, 303, 308, 350, 357, 371, 372, 374, 381, 387, 407, 418, 437, 456, 470, 498, 499, 507, 510, 514, 518

Offset: 1

David W. Wilson

nonn

Includes A157265, corresponding to continued fractions [6*k+4,1,1,2,1,1,12*k+8,1,1,2,1,1,12*k+8,...]. - Robert Israel, Nov 21 2019

			The continued fraction for sqrt(19) is 4 + 1/(2 + 1/(1 + 1/(3 + 1/(1 + 1/(2 + 1/(8 + 1/(2 + 1/(1 + 1/(3 + 1/(1 + 1/(2 + 1/(8 + ..., which has period 6, so 19 is in the sequence.
The continued fraction for sqrt(20) is 4 + 1/(2 + 1/(8 + 1/(2 + 1/(8 + 1/(2 + 1/(8 + ..., which has a period of 2, so 20 is not in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..2500 from Robert Israel)

Cf. A157265.

filter:= proc(n)
  not issqr(n) and nops(numtheory:-cfrac(sqrt(n),periodic,quotients)[2])=6
end proc:
select(filter, [$1..1000]); # Robert Israel, Nov 21 2019

Select[Range[500], Length[Last[ContinuedFraction[Sqrt[#]]]] == 6 &] (* Alonso del Arte, Mar 04 2018 *)

A157266 a(n) = 1728*n - 408.

Original entry on oeis.org

1320, 3048, 4776, 6504, 8232, 9960, 11688, 13416, 15144, 16872, 18600, 20328, 22056, 23784, 25512, 27240, 28968, 30696, 32424, 34152, 35880, 37608, 39336, 41064, 42792, 44520, 46248, 47976, 49704, 51432, 53160, 54888, 56616, 58344

Offset: 1

Vincenzo Librandi, Feb 26 2009

The identity (10368*n^2-4896*n+577)^2-(36*n^2-17*n+2)*(1728*n-408)^2=1 can be written as A157267(n)^2-A157265(n)*a(n)^2=1 (see also the second comment in A157267). - Vincenzo Librandi, Jan 27 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A157265, A157267.

I:=[1320, 3048]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 27 2012

LinearRecurrence[{2,-1},{1320,3048},40] (* Vincenzo Librandi, Jan 27 2012 *)
Table[1728n-408,{n,40}] (* Harvey P. Dale, Apr 18 2020 *)

for(n=1, 40, print1(1728*n - 408", ")); \\ Vincenzo Librandi, Jan 27 2012

a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jan 27 2012
G.f.: x*(1320+408*x)/(x-1)^2. - Vincenzo Librandi, Jan 27 2012
E.g.f.: 24*((72*x - 17)*exp(x) + 17). - G. C. Greubel, Feb 04 2018

A157267 a(n) = 10368n^2 - 4896n + 577.

Original entry on oeis.org

6049, 32257, 79201, 146881, 235297, 344449, 474337, 624961, 796321, 988417, 1201249, 1434817, 1689121, 1964161, 2259937, 2576449, 2913697, 3271681, 3650401, 4049857, 4470049, 4910977, 5372641, 5855041, 6358177, 6882049, 7426657, 7992001, 8578081, 9184897, 9812449

Offset: 1

Vincenzo Librandi, Feb 26 2009

The identity (10368*n^2 - 4896*n + 577)^2 - (36*n^2 - 17*n + 2)*(1728*n - 408)^2 = 1 can be written as a(n)^2 - A157265(n)*A157266(n)^2 = 1. - Vincenzo Librandi, Jan 27 2012

This is the case s = 4*n - 1 of the identity (2*r^2 - 1)^2 - ((r^2 - 1)/144)*(24*r)^2 = 1, where r = 18*s + 9*i^(s*(s+1)) - (-1)^s - 9 and i = sqrt(-1). - Bruno Berselli, Jan 29 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A157265, A157266.

I:=[6049, 32257, 79201]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 27 2012

LinearRecurrence[{3,-3,1},{6049,32257,79201},40] (* Vincenzo Librandi, Jan 27 2012 *)

for(n=1, 40, print1(10368*n^2 - 4896*n + 577", ")); \\ Vincenzo Librandi, Jan 27 2012

From Vincenzo Librandi, Jan 27 2012: (Start)
G.f.: x*(6049 + 14110*x + 577*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(10368*x^2 + 5472*x + 577) - 577. - Elmo R. Oliveira, Nov 09 2024

A334116 a(n) is the least number k greater than n such that the square roots of both k and n have continuous fractions with the same period p and, if p > 1, the same periodic terms except for the last term.

Original entry on oeis.org

1, 5, 8, 4, 10, 12, 32, 15, 9, 17, 40, 20, 74, 33, 24, 16, 26, 39, 1880, 30, 112, 660, 96, 35, 25, 37, 104, 299, 338, 42, 77600, 75, 60, 78, 48, 36, 50, 84, 68, 87, 130, 56, 288968, 468, 350, 3242817, 192, 63, 49, 65, 200, 2726, 1042, 1628, 180, 72, 308, 425, 5880, 95

Offset: 1

Gerhard Kirchner, Apr 14 2020

nonn

Note that a(n)=n if n is a square. The square root of a squarefree integer n has a continued fraction of the form [e(0);[e(1),...,e(p)]] with e(p)=2e(0) and e(i)=e(p-i) for 0 < i < p, see reference. The symmetric part [e(1),...,e(p-1)] of the continued fraction [m;[e(1),...,e(p-1), 2m]] will be called the pattern of n. 2 has the empty pattern (sqrt(2)=[1,[2]]), 3 has the pattern [1] (sqrt(3)=[1,[1,2]]) and so on. In this sense, the description of the sequence can be simplified as "Least number greater than n with the same pattern".
It can be can proved (see link) that integers with the same pattern are terms of a quadratic sequence.
An ambiguity has to be fixed: sqrt(2)=[1,[2]] = [1,[2,2]] = [1,[2,2,2]] and so on. We define that the shortest pattern is correct, here it is empty. Comment on the third subsequence (2),6,12,... below: The second term 6 has the pattern [2], but the first term 2 in brackets has the "wrong" pattern, after fixing the ambiguity.

			1) p=1: f(1)=2, f(2)=a(2)=5, f(3)=a(5)=10, f(4)=a(10)=17,..
sqrt(2)=[1,[2]], sqrt(5)=[2,[4]], sqrt(10)=[3,[6]], sqrt(17)=[4,[8]],..
2) p=2: f(1)=3, f(2)=a(3)=8, f(3)=a(8)=15, f(4)=a(15)=24,..
sqrt(3)=[1,[1,2]], sqrt(8)=[2,[1,4]], sqrt(15)=[3,[1,6]], sqrt(24)=[4,[1,8]],..
3) p=3: f(1)=41, f(2)=a(41)=130, f(3)=a(130)=269,..
sqrt(41)=[6,[2,2,12]], sqrt(130)=[11,[2,2,121]], sqrt(269)=[16,[2,2,256]],..
4) p=4: f(1)=33, f(2)=a(33)=60, f(3)=a(60)=95,..
sqrt(33)=[5,[1,2,1,10]], sqrt(60)=[7,[1,2,1,49]], sqrt(95)=[9,[1,2,1,81]],..
Several subsequences f(k) with f(k+1)=a(f(k)).
k>1 if first term in brackets, k>0 otherwise.
First terms  Period  Formula           Example
1) 2,5,10,17   1  A002522(k)=k^2+1           1
2) 3,8,15,24   2  A005563(k)=(k+1)^2-1       2
3)(2),6,12     2  A002378(k)=k*(k+1)
4) 7,32,75     4  A013656(k)=k*(9*k-2)
5) 11,40,87    2  A147296(k)=k*(9*k+2)
6) 13,74,185   5  A154357(k)=25*k^2-14*k+2
7) (3),14,33   4  A033991(k)=k*(4*k-1)       4
8) (5),18,39   2  A007742(k)=k*(4*k+1)
9) 21,112,275  6  A157265(k)=36*k^2-17*k+2
10)23,96,219   4  A154376(k)=25*k^2-2*k
11)27,104,231  2  A154377(k)=25*k^2+2*k
12)28,299,858  4  A156711(k)=144*k^2-161*k+45
13)29,338,985  5  A156640(k)=169*k^2+140*k+29
14)(8),34,78   4  A154516(k)=9*k^2-k
15)(10),38,84  2  A154517(k)=9*k^2+k
16)(2),41,130  3  A154355(k)=25*k^2-36*k+13  3
17)47,192,435  4  A157362(k)=49*k^2-2*k

Kenneth H. Rosen, Elementary number theory and its applications, Addison-Wesley, 3rd ed. 1993, page 428.

Chai Wah Wu, Table of n, a(n) for n = 1..138
Gerhard Kirchner, Continued fractions with the same pattern

Cf. A002522, A005563, A002378, A013656, A147296, A154357, A033991, A007742, A157265, A154376, A154377, A156711, A156640, A154516, A154517, A154355, A157362.

block([nmax: 100],
/*saves the first nmax terms in the current directory*/
algebraic: true, local(coeff), showtime: true,
fl: openw(sconcat("terms",nmax, ".txt")),
coeff(w,m):=
  block(a: m, p: 0, s: w, vv:[],
   while a<2*m do
    (p: p+1, s: ratsimp(1/(s-floor(s))), a: floor(s),
     if a<2*m then vv: append(vv, [a])),
   j: floor((p-1)/2),
   if mod(p,2)=0 then v: [1,0,vv[j+1]] else v: [0,1,1],
   for i from j thru 1 step(-1) do
    (h: vv[i], u: [v[1]+h*v[3], v[3], 2*h*v[1]+v[2]+h^2*v[3]], v: u),
   return(v)),
   for n from 1 thru nmax do
    (w: sqrt(n), m: floor(w),
     if w=m then  b: n else
      (v: coeff(w,m),  x: v[1], y: v[2], z: v[3], q: mod(z,2),
       if q=0 then (z: z/2, y: y/2) else x: 2*x,
       fr: (x*m+y)/z, m: m+z, fr: fr+x, b: m^2+fr),
      printf( fl, "~d, ", b)),
      close(fl));

from sympy import floor, S, sqrt
def coeff(w,m):
    a, p, s, vv = m, 0, w, []
    while a < 2*m:
        p += 1
        s = S.One/(s-floor(s))
        a = floor(s)
        if a < 2*m:
            vv.append(a)
    j = (p-1)//2
    v = [0,1,1] if p % 2 else [1, 0, vv[j]]
    for i in range(j-1,-1,-1):
        h = vv[i]
        v = [v[0]+h*v[2], v[2], 2*h*v[0]+v[1]+h**2*v[2]]
    return v
def A334116(n):
    w = sqrt(n)
    m = floor(w)
    if w == m:
        return n
    else:
        x, y, z = coeff(w,m)
        if z % 2:
            x *= 2
        else:
            z //= 2
            y //= 2
        return (m+z)**2+x+(x*m+y)//z # Chai Wah Wu, Sep 30 2021, after Maxima code

cp's OEIS Frontend

A020347 Numbers k such that the continued fraction for sqrt(k) has period 6.

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Maple

Mathematica

A157266 a(n) = 1728*n - 408.

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Magma

Mathematica

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A157267 a(n) = 10368*n^2 - 4896*n + 577.

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Magma

Mathematica

PARI

Formula

A334116 a(n) is the least number k greater than n such that the square roots of both k and n have continuous fractions with the same period p and, if p > 1, the same periodic terms except for the last term.

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Maxima

Python

A157267 a(n) = 10368n^2 - 4896n + 577.