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.

A129719 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 0's in odd positions (0 <= k <= ceiling(n/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 1, 4, 3, 1, 4, 5, 3, 1, 8, 8, 4, 1, 8, 12, 9, 4, 1, 16, 20, 13, 5, 1, 16, 28, 25, 14, 5, 1, 32, 48, 38, 19, 6, 1, 32, 64, 66, 44, 20, 6, 1, 64, 112, 104, 63, 26, 7, 1, 64, 144, 168, 129, 70, 27, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 128, 320, 416, 360, 225, 104, 35
Offset: 0

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Author

Emeric Deutsch, May 13 2007

Keywords

Comments

Row n has 1+ceiling(n/2) terms.

Examples

			T(6,2)=4 because we have 110101, 011101, 010110 and 010111.
Triangle starts:
  1;
  1, 1;
  2, 1;
  2, 2, 1;
  4, 3, 1;
  4, 5, 3, 1;
  8, 8, 4, 1;
		

Crossrefs

Programs

  • Maple
    G:=(1+z)*(1+t*z-t*z^2)/(1-(2+t)*z^2+t*z^4): Gser:=simplify(series(G,z=0,20)): for n from 0 to 17 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 17 do seq(coeff(P[n],t,j),j=0..ceil(n/2)) od; # yields sequence in triangular form

Formula

G.f.: G(t,z) = (1+z)(1+tz-tz^2)/(1-(2+t)z^2+tz^4). The trivariate generating function H(t,s,z), where t marks number of 0's in odd position and s marks number of 0's in even position, is given by H(t,s,z) = (1+(1+t)z-tsz^3)/(1-(1+t+s)z^2+tsz^4).
Row sums are the Fibonacci numbers (A000045).
T(2n,k) = T(2n-1,k) + T(2n-2,k) (n >= 1).
T(2n,k) = A129721(2n,k).
Sum_{k=0..ceiling(n/2)} k*T(n,k) = A129720(n).