A154377 -id:A154377 - cp's OEIS frontend

A154375 a(n) = 1250n^2 + 100n + 1.

1351, 5201, 11551, 20401, 31751, 45601, 61951, 80801, 102151, 126001, 152351, 181201, 212551, 246401, 282751, 321601, 362951, 406801, 453151, 502001, 553351, 607201, 663551, 722401, 783751, 847601, 913951, 982801, 1054151, 1128001

Offset: 1

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 + 100*n + 1)^2 - (25*n^2 + 2*n)*(250*n + 10)^2 = 1 can be written as a(n)^2 - A154377(n)*A154379(n)^2 = 1. - Vincenzo Librandi, Jan 30 2012

This is the case s=5 of the identity (2*s^4*n^2 + 4*s^2*n + 1)^2 - (s^2*n^2 + 2*n)*(2*s^3*n + 2*s)^2 = 1. - Bruno Berselli, Jan 30 2012

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

Cf. A154377, A154379.

Table[1250n^2+100n+1,{n,30}] (* or *) LinearRecurrence[{3,-3,1},{1351, 5201, 11551},30] (* Harvey P. Dale, Apr 25 2011 *)

a(n)=1250*n^2+100*n+1 \\ Charles R Greathouse IV, Dec 27 2011

a(1)=1351, a(2)=5201, a(3)=11551, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Apr 25 2011
G.f.: x*(x^2 + 1148*x + 1351)/(1-x)^3. - Vincenzo Librandi, Jan 30 2012
E.g.f.: (1250*x^2 + 1350*x + 1)*exp(x) - 1. - G. C. Greubel, Sep 15 2016

Minor corrections by M. F. Hasler, Oct 08 2014

A154379 a(n) = 250*n + 10.

Original entry on oeis.org

260, 510, 760, 1010, 1260, 1510, 1760, 2010, 2260, 2510, 2760, 3010, 3260, 3510, 3760, 4010, 4260, 4510, 4760, 5010, 5260, 5510, 5760, 6010, 6260, 6510, 6760, 7010, 7260, 7510, 7760, 8010, 8260, 8510, 8760, 9010, 9260, 9510, 9760, 10010, 10260

Offset: 1

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 + 100*n + 1)^2 - (25*n^2 + 2*n)*(250*n + 10)^2 = 1 can be written as A154375(n)^2 - A154377(n)*a(n)^2 = 1 (see also the second comment in A154375). - Vincenzo Librandi, Jan 30 2012

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

Cf. A154375, A154377.

LinearRecurrence[{2, -1}, {260, 510}, 50] (* Vincenzo Librandi, Jan 30 2012 *)

a(n)=250*n+10 \\ Charles R Greathouse IV, Dec 27 2011

a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jan 30 2012
G.f.: 10*x*(26 - x)/(1-x)^2. - Vincenzo Librandi, Jan 30 2012 [corrected by Georg Fischer, May 12 2019]
E.g.f.: 10*( (25*x + 1)*exp(x) - 1). - G. C. Greubel, Sep 15 2016

Definition corrected by Paolo P. Lava, Jan 14 2009

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

A154375 a(n) = 1250n^2 + 100n + 1.

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A154379 a(n) = 250*n + 10.

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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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cp's OEIS Frontend

A154375 a(n) = 1250*n^2 + 100*n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A154379 a(n) = 250*n + 10.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maxima

Python

A154375 a(n) = 1250n^2 + 100n + 1.