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.

A080987 Ratios of successive terms of A080985.

Original entry on oeis.org

3, 11, 145, 24721, 706521601, 568754681712768961, 364030550787463437509470123011290881, 147562413614008475146723669284672702440197339884722672618106791371480321
Offset: 1

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Author

Hugo Pfoertner, Feb 26 2003

Keywords

Crossrefs

A079269 Define b by b(1) = 1 and for n>1, b(n) = b(n-1)+1/(1+1/b(n-1)); sequence gives numerator of b(n).

Original entry on oeis.org

1, 3, 21, 861, 1275141, 2551762438701, 9546380157472159016030421, 126857284256055227389078067834858327568823447932861
Offset: 1

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Author

N. J. A. Sloane, Feb 16 2003

Keywords

Comments

The next term is too large to include.

Examples

			The b sequence begins 1, 3/2, 21/10, 861/310, 1275141/363010, 2551762438701/594665194510, ... = a(n)/A079278.
		

Crossrefs

Programs

  • Maple
    b := proc(n) option remember; if n=1 then 1 else b(n-1)+1/(1+1/b(n-1)); fi; end;
  • Mathematica
    nxt[n_]:=n+1/(1+1/n); Numerator/@Nest[Append[#,nxt[Last[#]]]&,{1},10]  (* Harvey P. Dale, Apr 21 2011 *)

Formula

Conjecture: a(m+1) = a(m)^2 + a(m)^3 /(2a(m-1)^2) - a(m)a(m-1)^2/2 for m >= 2. - Leroy Quet

A080984 Define b by b(1) = 1 and for n > 1, b(n) = b(n-1) + 1/(2 + 1/b(n-1)); sequence gives numerator of b(n).

Original entry on oeis.org

1, 4, 56, 9968, 294115808, 242590126064151488, 158248601344912132157178428071499648, 65129411362626329768830076910903417752818896343320137665280356705971968
Offset: 1

Views

Author

Hugo Pfoertner, Feb 26 2003

Keywords

Comments

The next term has 285 digits. - Harvey P. Dale, Jul 07 2011

Examples

			The sequence {b(n)} begins 1, 4/3, 56/33, 9968/4785, 294115808/118289985, ...
		

Crossrefs

Programs

  • Mathematica
    Numerator/@NestList[#+1/(2+1/#)&,1,9] (* Harvey P. Dale, Jul 07 2011 *)
  • Reduce
    a := 1; for i := 1:8 do write a := a+1/(2+1/a);

Formula

b(k) = n(k)/d(k); n(1)=1, d(1)=1, m=2; for k >= 2: n(k+1) = n(k) *(m*n(k) + 2*d(k)), d(k+1) = d(k)*(m*n(k) + d(k)). - Leroy Quet
Showing 1-3 of 3 results.