A156640 -id:A156640 - cp's OEIS frontend

A156718 Numbers k such that k^2 == -1 (mod 13^2).

70, 99, 239, 268, 408, 437, 577, 606, 746, 775, 915, 944, 1084, 1113, 1253, 1282, 1422, 1451, 1591, 1620, 1760, 1789, 1929, 1958, 2098, 2127, 2267, 2296, 2436, 2465, 2605, 2634, 2774, 2803, 2943, 2972, 3112, 3141, 3281, 3310, 3450, 3479, 3619, 3648, 3788

Offset: 1

Vincenzo Librandi, Feb 14 2009

Also, numbers of the form 169k +- 70.

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

Cf. A156636, A156640, A156627, A156639.

[Floor(n/2)*169-70*(-1)^n: n in [1..50]];

LinearRecurrence[{1,1,-1},{70,99,239},50]

a(n)=n\2*169-70*(-1)^n \\ Charles R Greathouse IV, Dec 23 2011

a(n) = a(n-1) + a(n-2) - a(n-3).
G.f.: x*(70 + 29*x + 70*x^2) / ( (1+x)*(x-1)^2 ). - Alexander R. Povolotsky, Feb 15 2009
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(29*Pi/338)*Pi/169. - Amiram Eldar, Feb 26 2023

A156636 a(n) = 4394*n + 1820.

Original entry on oeis.org

1820, 6214, 10608, 15002, 19396, 23790, 28184, 32578, 36972, 41366, 45760, 50154, 54548, 58942, 63336, 67730, 72124, 76518, 80912, 85306, 89700, 94094, 98488, 102882, 107276, 111670, 116064, 120458, 124852, 129246, 133640, 138034, 142428, 146822, 151216, 155610

Offset: 0

Vincenzo Librandi, Feb 15 2009

The identity (57122*n^2 + 47320*n + 9801)^2 - (169*n^2 + 140*n + 29)*(4394*n + 1820)^2 = 1 can be written as A156735(n)^2 - A156640(n)*a(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A156640, A156718, A156735.

I:=[1820, 6214]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];

Mathematica
```
LinearRecurrence[{2,-1},{1820,6214},50]
```

a(n)=4394*n+1820 \\ Charles R Greathouse IV, Dec 23 2011

G.f.: 26*(70+99*x)/(x-1)^2. - R. J. Mathar, Jan 05 2011
a(n) = 2*a(n-1) - a(n-2).
E.g.f.: 26*exp(x)*(70 + 169*x). - Elmo R. Oliveira, Dec 08 2024

Offset corrected by R. J. Mathar, Jan 05 2011

A156735 a(n) = 57122n^2 + 47320n + 9801.

Original entry on oeis.org

9801, 114243, 332929, 665859, 1113033, 1674451, 2350113, 3140019, 4044169, 5062563, 6195201, 7442083, 8803209, 10278579, 11868193, 13572051, 15390153, 17322499, 19369089, 21529923, 23805001, 26194323, 28697889, 31315699

Offset: 0

Vincenzo Librandi, Feb 15 2009

The identity (57122*n^2 +47320*n +9801)^2 - (169*n^2 +140*n +29)*(4394*n +1820)^2 = 1 can be written as a(n)^2 - A156640(n)*A156636(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A156636, A156640, A156718.

I:=[9801, 114243, 332929]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

A156735:= n-> 57122*n^2 + 47320*n + 9801; seq(A156735(n), n=0..50); # G. C. Greubel, Feb 28 2021

LinearRecurrence[{3,-3,1},{9801,114243,332929},50]
CoefficientList[Series[(9801 +84840x +19603x^2)/(1-x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 03 2014 *)
Table[57122n^2+47320n+9801,{n,0,30}] (* Harvey P. Dale, Jan 30 2024 *)

a(n)= 57122*n^2+47320*n+9801 \\ Charles R Greathouse IV, Dec 23 2011

[57122*n^2 + 47320*n + 9801 for n in (0..50)] # G. C. Greubel, Feb 28 2021

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) for n>2.
G.f.: (9801 + 84840*x + 19603*x^2)/(1 - x)^3. -  Vincenzo Librandi, May 03 2014
E.g.f.: (9801 +104442*x +57122*x^2)*exp(x). - G. C. Greubel, Feb 28 2021

Edited by Charles R Greathouse IV, Jul 25 2010

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

A156718 Numbers k such that k^2 == -1 (mod 13^2).

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A156636 a(n) = 4394*n + 1820.

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A156735 a(n) = 57122*n^2 + 47320*n + 9801.

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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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A156735 a(n) = 57122n^2 + 47320n + 9801.