cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489
Offset: 1

Views

Author

Paul Weisenhorn, May 23 2009

Keywords

Comments

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014
    
  • Mathematica
    LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
    CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
  • PARI
    a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

Formula

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Extensions

Edited, extended by R. J. Mathar, Sep 02 2009
First formula corrected by Harvey P. Dale, Oct 08 2012

A245031 Numbers m such that 3*m+1 and 8*m+1 are both squares.

Original entry on oeis.org

0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600
Offset: 1

Views

Author

Bruno Berselli, Jul 15 2014

Keywords

Comments

Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=1: A006454;
k=3: this sequence;
k=4: A029549;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
k=6: A200999;
k=7: A157879.
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=1: A045899;
k=2: A045502;
k=4: A059989;
k=5: A159683;
k=6: 8*A029546;
k=7: A160695;
k=8: this sequence.

Crossrefs

Programs

  • Magma
    I:=[0,1,21,120,2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];
  • Mathematica
    LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]
  • Maxima
    a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);
    
  • PARI
    a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a
    

Formula

G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1): 40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).

Extensions

Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016

A334673 a(n) = 23*a(n-1) - a(n-2) + 1 for n > 1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 24, 552, 12673, 290928, 6678672, 153318529, 3519647496, 80798573880, 1854847551745, 42580695116256, 977501140122144, 22439945527693057, 515141245996818168, 11825808712399124808, 271478459139183052417, 6232178751488811080784, 143068632825103471805616
Offset: 0

Views

Author

Francesca Arici, Sep 11 2020

Keywords

Crossrefs

Cf. A004253, A004254, A030221, A097778 (first differences).
Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Programs

  • Mathematica
    CoefficientList[Series[x/((1 - x) (x^2 - 23 x + 1)), {x, 0, 18}], x] (* Michael De Vlieger, Apr 07 2021 *)

Formula

a(n) = A004254(n)*A004254(n+1)/5 = A160695(n+1)/5.
G.f.: x/((1-x)*(x^2-23*x+1)). - Alois P. Heinz, Sep 11 2020
From Klaus Purath, Jun 18 2025: (Start)
a(n) = (A004253(n+1)^2 - 1) / 15.
a(n) = (A030221(n)^2 - 1) / 35.
a(n) + a(n+1) = A004253(n+1)^2. (End)

Extensions

a(13)-a(14) corrected and more terms added by Alois P. Heinz, Sep 11 2020
Showing 1-3 of 3 results.