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.

A036411 9-gonal square numbers.

Original entry on oeis.org

1, 9, 1089, 8281, 978121, 7436529, 878351769, 6677994961, 788758910641, 5996832038649, 708304623404049, 5385148492712041, 636056763057925561, 4835857349623374369, 571178264921393749929, 4342594514813297471521
Offset: 1

Views

Author

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

Keywords

Comments

From Ant King, Nov 17 2011: (Start)
lim_{n -> infinity} a(2n+1)/a(2n) = 1/625 * (36913 + 9864 * sqrt(14));
lim_{n -> infinity} a(2n)/a(2n-1) = 1/625 * (2417 + 624 * sqrt(14)).
(End)

Crossrefs

Programs

  • Magma
    I:=[1, 9, 1089, 8281]; [n le 4 select I[n] else 898*Self(n-2)-Self(n-4)+200: n in [1..20]]; // Vincenzo Librandi, Nov 18 2011
  • Maple
    a(0):=1:a(1):=9:a(2):=1089:a(3):=8281: a(4):=978121:for n from 0 to 20 do a(n+5):=a(n+4)+898*a(n+3)-898*a(n+2)-a(n+1)+a(n):od:seq(a(n),n=0..20); # Richard Choulet, May 08 2009
  • Mathematica
    LinearRecurrence[ {1, 898, - 898, - 1, 1 }, { 1, 9, 1089, 8281, 978121 }, 16] (* Ant King, Nov 17 2011 *)

Formula

O.g.f. f(z) = 1 + 9*z + ... = ((1 + 8*z + 182*z^2 + 8*z^3 + z^4)/((1-z)*(1 - 898*z^2 + z^4))). With the first values, for n > 0: a(n+5) = a(n+4) + 898*a(n+3) - 898*a(n+2) - a(n+1) + a(n). On every bisection modulo 2: a(n+2) = 30*a(n+1) - a(n) + 200. On every bisection modulo 2: a(n+1) = 449*a(n) + 100 + 60*sqrt(56*a(n)^2 + 25*a(n)). a(n) = -25/112 + (11/28 + (11/112)*sqrt(14))*(15 + 4*sqrt(14))^n + (11/28 - (11/112)*sqrt(14))*(15 - 4*sqrt(14))^n + (7/32 - (1/16)*sqrt(14))*(-15 + 4*sqrt(14))^n + (7/32 + (1/16)*sqrt(14))*(-15 - 4*sqrt(14))^n. - Richard Choulet, May 08 2009
a(n) = 898 * a(n-2) - a(n-4) + 200. - Ant King, Nov 17 2011

Extensions

More terms from Eric W. Weisstein
More terms from Richard Choulet, May 08 2009

A048920 Indices of heptagonal numbers (A000566) which are also 9-gonal.

Original entry on oeis.org

1, 104, 14725, 2090804, 296879401, 42154784096, 5985682462189, 849924754846700, 120683329505769169, 17136182865064375256, 2433217283509635517141, 345499718075503179058724
Offset: 1

Views

Author

Keywords

Comments

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1) = (6 + sqrt(35))^2 = 71 + 12*sqrt(35). - Ant King, Dec 31 2011

Crossrefs

Programs

  • Magma
    I:=[1, 104, 14725]; [n le 3 select I[n] else 143*Self(n-1)-143*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 21 2011
  • Mathematica
    LinearRecurrence[{143,-143,1},{1,104,14725},30] (* Vincenzo Librandi, Dec 21 2011 *)
  • Maxima
    makelist(expand((42+(-21+5*sqrt(35))*(6+sqrt(35))^(2*n-1)-(21+5*sqrt(35))*(6-sqrt(35))^(2*n-1))/140), n, 1, 12); /* Bruno Berselli, Dec 20 2011 */
    

Formula

From Bruno Berselli, Dec 20 2011: (Start)
G.f.: x*(1 - 39*x - 4*x^2)/((1-x)*(1 - 142*x + x^2)).
a(n) = (42 + (-21+5r)*(6+r)^(2n-1) - (21+5r)*(6-r)^(2n-1))/140, where r=sqrt(35). (End)
From Ant King, Dec 31 2011: (Start)
a(n) = 142*a(n-1) - a(n-2) - 42.
a(n) = ceiling(1/140*(49+9*sqrt(35))*(6+sqrt(35))^(2*n-2)).
(End)

A048921 9-gonal heptagonal numbers (A000566).

Original entry on oeis.org

1, 26884, 542041975, 10928650279834, 220343446399977901, 4442564555387704166896, 89570986345383445012986019, 1805930222253056462964119954950, 36411165051495138060899141518722649, 734121907962314751330792028336366100956
Offset: 1

Views

Author

Keywords

Comments

As n increases, this sequence is approximately geometric with common ratio r = lim(n->Infinity,a(n)/a(n-1)) = (6+sqrt(35))^4 = 10081+1704*sqrt(35). - Ant King, Dec 31 2011

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{20163, -20163, 1}, {1, 26884, 542041975}, 9]; (* Ant King, Dec 31 2011 *)
  • PARI
    Vec(x*(1+6721*x+46*x^2)/((1-x)*(1-20162*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 22 2015

Formula

From Ant King, Dec 31 2011: (Start)
a(n) = 20163*a(n-1)-20163*a(n-2)+a(n-3).
a(n) = 20162*a(n-1)-a(n-2)+6768.
a(n) = 1/560*((39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)+(39-4*sqrt(35))*(6-sqrt(35))^(4*n-3)-188).
a(n) = floor(1/560*(39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)).
G.f.: x(1+6721*x+46*x^2) / ((1-x)(1-20162*x+x^2)).
(End)
Showing 1-3 of 3 results.