A154357 -id:A154357 - cp's OEIS frontend

A154355 a(n) = 25n^2 - 36n + 13.

13, 2, 41, 130, 269, 458, 697, 986, 1325, 1714, 2153, 2642, 3181, 3770, 4409, 5098, 5837, 6626, 7465, 8354, 9293, 10282, 11321, 12410, 13549, 14738, 15977, 17266, 18605, 19994, 21433, 22922, 24461, 26050, 27689, 29378

Offset: 0

Vincenzo Librandi, Jan 07 2009

The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as A154358(n)^2 - a(n)*A154360(n)^2 = 1. See also the third comment in A154357.
Numbers of the form (3n-2)^2 + (4n-3)^2. - Bruno Berselli, Dec 12 2011
From Klaus Purath, May 06 2025: (Start)
25*a(n)-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*(5*y)^2 = -1 for any integer n where a(1-n) = A154357(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*5^2 - 1), x(0) = 1, x(1) = 4*D*5^2 - 1, y(0) = 1, y(1) = 4*D*5^2 - 3. The two recurrences are of the form (4*D*5^2 - 2, -1).
It follows from the above that the terms of this sequence and of A154357 belong to A031396. (End)

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

Cf. A154354, A154358, A154359, A154360, A154361, A202141.

Essentially a duplicate of A007533.

[25*n^2-36*n+13: n in [0..40]]; // Bruno Berselli, Sep 15 2016

Table[25n^2-36n+13,{n,0,40}]  (* Harvey P. Dale, Apr 02 2011 *)
LinearRecurrence[{3, -3, 1}, {13, 2, 41}, 50] (* Vincenzo Librandi, Feb 21 2012 *)

for(n=0, 40, print1(25*n^2 - 36*n + 13", ")); \\ Vincenzo Librandi, Feb 21 2012

a(n) = A007533(n-1), n>0. - R. J. Mathar, Jan 14 2009
G.f.: (13 - 37*x + 74*x^2) / (1-x)^3. - R. J. Mathar, Jan 05 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Feb 21 2012
E.g.f.: (13 - 11*x + 25*x^2) * exp(x). - G. C. Greubel, Sep 14 2016
From Klaus Purath, May 06 2025: (Start)
a(n) = (3*n-2)^2 + (4*n-3)^2.
25*a(n) - 1 = (25*n - 18)^2. (End)

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

First comment rewritten by Bruno Berselli, Dec 12 2011

A154359 a(n) = 1250n^2 - 700n + 99.

Original entry on oeis.org

99, 649, 3699, 9249, 17299, 27849, 40899, 56449, 74499, 95049, 118099, 143649, 171699, 202249, 235299, 270849, 308899, 349449, 392499, 438049, 486099, 536649, 589699, 645249, 703299, 763849, 826899, 892449, 960499, 1031049

Offset: 0

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as a(n)^2 - A154357(n)*A154361(n)^2 = 1. See also the third comment in A154357.

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

[1250*n^2-700*n+99: n in [0..40]]; // Bruno Berselli, Sep 15 2016

LinearRecurrence[{3, -3, 1}, {99, 649, 3699}, 50] (* Vincenzo Librandi, Feb 21 2012 *)

for(n=0, 40, print1(1250*n^2-700*n+99", ")); \\ Vincenzo Librandi, Feb 21 2012

G.f.: (99 + 352*x + 2049*x^2)/(1-x)^3. - Bruno Berselli, Dec 13 2011
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Feb 21 2012
E.g.f.: (99 + 550*x + 1250*x^2)*exp(x). - G. C. Greubel, Sep 15 2016

Edited by Charles R Greathouse IV, Jul 29 2010

Librandi's comment rewritten by Bruno Berselli, Dec 13 2011

A154361 a(n) = 250*n - 70.

Original entry on oeis.org

-70, 180, 430, 680, 930, 1180, 1430, 1680, 1930, 2180, 2430, 2680, 2930, 3180, 3430, 3680, 3930, 4180, 4430, 4680, 4930, 5180, 5430, 5680, 5930, 6180, 6430, 6680, 6930, 7180, 7430, 7680, 7930, 8180, 8430, 8680, 8930, 9180

Offset: 0

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as A154359(n)^2 - A154357(n)*a(n)^2 = 1. See also the third comment in A154357.

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

Cf. A154360, A154359, A154358, A154357, A154355.

[250*n-70: n in [0..50]]; // Bruno Berselli, Sep 15 2016

LinearRecurrence[{2, -1}, {-70, 180}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
250*Range[0,50]-70 (* Harvey P. Dale, Apr 09 2020 *)

for(n=0, 50, print1(250*n - 70", ")); \\ Vincenzo Librandi, Feb 21 2012

G.f.: -10*(7 - 32*x)/(1-x)^2. - Bruno Berselli, Dec 13 2011
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 21 2012
E.g.f.: 10*(-7 + 25*x)*exp(x). - G. C. Greubel, Sep 15 2016

Offset changed and Librandi's comment rewritten by Bruno Berselli, Dec 13 2011

A154358 a(n) = 1250n^2 - 1800n + 649.

Original entry on oeis.org

649, 99, 2049, 6499, 13449, 22899, 34849, 49299, 66249, 85699, 107649, 132099, 159049, 188499, 220449, 254899, 291849, 331299, 373249, 417699, 464649, 514099, 566049, 620499, 677449, 736899, 798849, 863299, 930249

Offset: 0

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as a(n)^2 - A154355(n)*A154360(n)^2 = 1. See also the third comment in A154357.

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

Cf. A154359, A154360, A154361, A154355, A154357.

[1250*n^2-1800*n+649: n in [0..40]]; // Bruno Berselli, Sep 15 2016

LinearRecurrence[{3, -3, 1}, {649, 99, 2049}, 50] (* Vincenzo Librandi, Feb 21 2012 *)

for(n=0, 40, print1(1250*n^2 - 1800*n + 649", ")); \\ Vincenzo Librandi, Feb 21 2012

G.f.: (649 - 1848*x + 3699*x^2)/(1-x)^3. - R. J. Mathar, Jan 05 2011
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
E.g.f.: (649 - 550*x + 1250*x^2)*exp(x). - G. C. Greubel, Sep 14 2016

Offset and one entry corrected by R. J. Mathar, Jan 05 2011

Librandi's comment rewritten by Bruno Berselli, Dec 13 2011

A154360 a(n) = 250*n - 180.

Original entry on oeis.org

-180, 70, 320, 570, 820, 1070, 1320, 1570, 1820, 2070, 2320, 2570, 2820, 3070, 3320, 3570, 3820, 4070, 4320, 4570, 4820, 5070, 5320, 5570, 5820, 6070, 6320, 6570, 6820, 7070, 7320, 7570, 7820, 8070, 8320, 8570, 8820, 9070, 9320, 9570, 9820, 10070, 10320

Offset: 0

Vincenzo Librandi, Jan 08 2009

The identity (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1 can be written as A154358(n)^2 - A154355(n)*a(n)^2 = 1. See also the third comment in A154357.

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

Cf. A154361, A154359, A154358, A154357, A154355.

[250*n-180: n in [0..50]]; // Bruno Berselli, Sep 15 2016

LinearRecurrence[{2, -1}, {-180, 70}, 50] (* Vincenzo Librandi, Feb 21 2012 *)

for(n=0, 50, print1(250n - 180", ")); \\ Vincenzo Librandi, Feb 21 2012

G.f.: -10*(18 - 43*x)/(1-x)^2. - Bruno Berselli, Dec 13 2011
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 21 2012
E.g.f.: 10*(-18 + 25*x)*exp(x). - G. C. Greubel, Sep 15 2016

Offset changed and Librandi's comment rewritten by Bruno Berselli, Dec 13 2011

A154356 Primes of the form 25n^2-14n+2 for n >= 0.

Original entry on oeis.org

2, 13, 557, 1129, 1901, 5417, 8761, 10733, 15277, 23593, 30137, 41453, 59341, 74857, 80429, 86201, 92173, 104717, 118061, 179437, 253613, 284729, 306473, 352361, 364333, 414221, 453737, 523597, 598457, 798521, 1068329, 1217933, 1285049

Offset: 1

Vincenzo Librandi, Jan 07 2009

			a(8)=10733 corresponds to n=21, therefore 10733=(3*21-1)^2+(4*21-1)^2=62^2+83^2 (see second comment in A154357). - _Bruno Berselli_, Dec 14 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A154357.

[ a: n in [0..400] | IsPrime(a) where a is 25*n^2-14*n+2];

Join[{2},Select[Table[25n^2-14n+2,{n,500}],PrimeQ]] (* Harvey P. Dale, May 15 2011 *)

Edited by Robert Hochberg, Jun 21 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

A154355 a(n) = 25*n^2 - 36*n + 13.

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A154359 a(n) = 1250*n^2 - 700*n + 99.

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A154361 a(n) = 250*n - 70.

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A154358 a(n) = 1250*n^2 - 1800*n + 649.

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A154360 a(n) = 250*n - 180.

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A154356 Primes of the form 25n^2-14n+2 for n >= 0.

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Magma

Mathematica

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.

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A154355 a(n) = 25n^2 - 36n + 13.

A154359 a(n) = 1250n^2 - 700n + 99.

A154358 a(n) = 1250n^2 - 1800n + 649.