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.

A215495 a(4*n) = a(4*n+2) = a(2*n+1) = 2*n + 1.

Original entry on oeis.org

1, 1, 1, 3, 3, 5, 3, 7, 5, 9, 5, 11, 7, 13, 7, 15, 9, 17, 9, 19, 11, 21, 11, 23, 13, 25, 13, 27, 15, 29, 15, 31, 17, 33, 17, 35, 19, 37, 19, 39, 21, 41, 21, 43, 23, 45, 23, 47, 25, 49, 25, 51, 27, 53, 27, 55, 29, 57, 29, 59, 31, 61, 31, 63, 33, 65, 33, 67, 35, 69, 35, 71, 37, 73, 37, 75, 39, 77, 39, 79, 41, 81, 41, 83, 43, 85, 43
Offset: 0

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Author

Paul Curtz, Aug 13 2012

Keywords

Comments

A214282(n) and -A214283(n) are companions. Separately or together, they have many links with the Catalan's numbers A000108(n). Examples:
A214282(n+1) - 2*A214282(n) = -1, -1, 1, 0, -2, -5, 5, 0, -14, -42, 42, 0, -132, ....
2*A214283(n) - A214283(n+1) = 1, 0, -1, -2, 2, 0, -5, -14, 14, 0, -42, -132, 132, ....
A214282(n) + A214283(n) = 1, 0, -1, 0, 2, 0, -5, 0, 14, 0, -42,... (A126120).
The companion to a(n) is b(n) = -A214283(n)/(1,1,1,1,2,2,5,5,...) = 0, 1, 2, 3, 2, 5, 4, 7, 4, 9, 6, ....
a(n) - b(n) = A056594(n).
Discovered as a(n) = A214282(n+1)/A000108([n/2]). See abs(A129996(n-2)).

Crossrefs

Programs

  • Magma
    I:=[1,1,1,3,3,5]; [n le 6 select I[n] else Self(n-2) +Self(n-4) -Self(n-6): n in [1..30]]; // G. C. Greubel, Apr 23 2018
  • Mathematica
    a[n_?EvenQ] := n/2 + Boole[Mod[n, 4] == 0]; a[n_?OddQ] := n; Table[a[n], {n, 0, 86}] (* Jean-François Alcover, Aug 14 2012 *)
    LinearRecurrence[{0,1,0,1,0,-1}, {1,1,1,3,3,5}, 50] (* G. C. Greubel, Apr 23 2018 *)
  • PARI
    x='x+O('x^30); Vec(( 1+x+2*x^3+x^4+x^5 )/( (x^2+1)*(x-1)^2*(1+x)^2 )) \\ G. C. Greubel, Apr 23 2018
    

Formula

a(n+3) = (A185048(n+3)=2,2,4,2,... ) + 1.
a(n+2) - a(n) = 0, 2, 2, 2. (Period 4).
a(n) = 2*a(n-4) - a(n-8).
a(2*n) = A109613(n).
a(n+1) - a(n) = 2* (-1)^n * A059169(n).
G.f. : ( 1+x+2*x^3+x^4+x^5 ) / ( (x^2+1)*(x-1)^2*(1+x)^2 ). - Jean-François Alcover, Aug 14 2012