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.

A129722 Number of 0's in even position in all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

0, 0, 1, 1, 5, 6, 19, 25, 65, 90, 210, 300, 654, 954, 1985, 2939, 5911, 8850, 17345, 26195, 50305, 76500, 144516, 221016, 411900, 632916, 1166209, 1799125, 3283145, 5082270, 9197455, 14279725, 25655489, 39935214, 71293590, 111228804, 197452746, 308681550
Offset: 0

Views

Author

Emeric Deutsch, May 13 2007

Keywords

Examples

			a(4)=5 because in 1110', 1111, 1101, 10'10', 10'11, 0110', 0111 and 0101 one has altogether five 0's in even position (marked by ').
		

Crossrefs

Programs

  • Maple
    G:=z^2/(1-z-z^2)^2/(1+z-z^2): Gser:=series(G,z=0,45): seq(coeff(Gser,z,n),n=0..42);
  • Mathematica
    CoefficientList[Series[x^2/((1 + x - x^2)*(1 - x - x^2)^2), {x,0,50}], x] (* G. C. Greubel, Mar 09 2017 *)
    LinearRecurrence[{1,4,-3,-4,1,1},{0,0,1,1,5,6},40] (* Harvey P. Dale, Apr 02 2018 *)
  • PARI
    x='x+O('x^50); concat([0,0], Vec(x^2/((1 + x - x^2)*(1 - x - x^2)^2))) \\ G. C. Greubel, Mar 09 2017

Formula

G.f.: z^2/( (1+z-z^2)*(1-z-z^2)^2 ).
a(2*n+1) = a(2*n) + a(2*n-1) (n>=1).
a(2*n+1) = A001871(n-1) (n>=1).
a(2*n) = A129720(2*n) = A001870(n-1).
a(n) = Sum_{ k=0..floor(n/2)} k*A129721(n,k).
a(n) = F(n)*(n+1)/5 + F(n+1)*(2*n - 5 + 5*(-1)^n)/20 where F = A000045. - Greg Dresden, Jan 01 2021

A129720 Number of 0's in odd position in all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

0, 1, 1, 4, 5, 14, 19, 46, 65, 145, 210, 444, 654, 1331, 1985, 3926, 5911, 11434, 17345, 32960, 50305, 94211, 144516, 267384, 411900, 754309, 1166209, 2116936, 3283145, 5914310, 9197455, 16458034, 25655489, 45638101, 71293590, 126159156
Offset: 0

Views

Author

Emeric Deutsch, May 13 2007

Keywords

Examples

			a(4)=5 because in 1110, 1111, 110'1, 1010, 1011, 0'110, 0'111 and 0'10'1 one has altogether five 0's in odd position (marked by ').
		

Crossrefs

Programs

  • Maple
    g:=z*(1-z^2)/(1-z-z^2)^2/(1+z-z^2): gser:=series(g,z=0,43): seq(coeff(gser,z,n),n=0..40);

Formula

G.f.: z(1-z^2)/((1-z-z^2)^2*(1+z-z^2)).
a(2n) = a(2n-1) + a(2n-2) (n >= 1).
a(2n-1) = A030267(n).
a(2n) = A129722(2n) = A001870(n-1).
a(n) = Sum_{k=0..ceiling(n/2)} k*A129719(n,k).

A129721 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 0's in even positions (0<=k<=floor(n/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 2, 2, 1, 4, 1, 4, 3, 1, 8, 4, 1, 8, 8, 4, 1, 16, 12, 5, 1, 16, 20, 13, 5, 1, 32, 32, 18, 6, 1, 32, 48, 38, 19, 6, 1, 64, 80, 56, 25, 7, 1, 64, 112, 104, 63, 26, 7, 1, 128, 192, 160, 88, 33, 8, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 448, 432, 280, 129, 42, 9, 1, 256, 576, 688, 552
Offset: 0

Views

Author

Emeric Deutsch, May 13 2007

Keywords

Comments

Row n has 1+floor(n/2) terms. Row sums are the Fibonacci numbers (A000045). T(2n+1,k)=T(2n,k)+T(2n-1,k) (n>=1). T(2n,k)=A129719(2n,k). Sum(k*T(n,k), 0<=k<=floor(n/2))=A129722(n).

Examples

			T(6,2)=4 because we have 111010, 101110, 101011 and 011010.
Triangle starts:
1;
2;
2,1;
4,1;
4,3,1;
8,4,1;
8,8,4,1;
		

Crossrefs

Programs

  • Maple
    G:=(1+2*z-t*z^3)/(1-2*z^2-t*z^2+t*z^4): Gser:=simplify(series(G,z=0,21)): for n from 0 to 18 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 18 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

Formula

G.f.=G(t,z)=(1+2z-tz^3)/[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].
Showing 1-3 of 3 results.