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 1-2 of 2 results.

A010709 Constant sequence: the all 4's sequence.

Original entry on oeis.org

4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
Offset: 0

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Author

Keywords

Comments

From Klaus Brockhaus, May 25 2010: (Start)
Continued fraction expansion of 2+sqrt(5).
Decimal expansion of 4/9.
Inverse binomial transform of A020707. (End)

Crossrefs

From Klaus Brockhaus, May 25 2010: (Start)
Equals 4*A000012, 2*A007395, A010731/2, A010855/4, A010871/8.
Cf. A098317 (decimal expansion of 2+sqrt(5)), A020707 (2^(n+2)). (End)

Programs

Formula

From Klaus Brockhaus, May 25 2010: (Start)
a(n) = 4.
G.f.: 4/(1-x). (End)
E.g.f.: 4*e^x. - Vincenzo Librandi, Jan 29 2012

A174927 Periodic sequence: Repeat 1, 64.

Original entry on oeis.org

1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64, 1, 64
Offset: 0

Views

Author

Klaus Brockhaus, Apr 02 2010

Keywords

Comments

Interleaving of A000012 and 2*A010871.
Also continued fraction expansion of (4+sqrt(17))/8.
First differences of A174928.

Crossrefs

Cf. A000012 (all 1's sequence), A010871 (all 32's sequence), A010689 (repeat 1, 8), A174930 (decimal expansion of (4+sqrt(17))/8), A174928.

Programs

  • Magma
    &cat[ [1, 64]: n in [0..41] ];
    [ (65-63*(-1)^n)/2: n in [0..83] ];
  • Mathematica
    PadRight[{},100,{1,64}] (* Harvey P. Dale, Jun 16 2013 *)

Formula

a(n) = (65-63*(-1)^n)/2.
a(n) = a(n-2) for n > 1; a(0) = 0, a(1) = 64.
a(n) = -a(n-1)+65 for n > 0; a(0) = 1.
a(n) = ((n+1) mod 2)+64*(n mod 2).
G.f.: (1+64*x)/((1-x)*(1+x)).
Showing 1-2 of 2 results.