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.

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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

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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

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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].

A177352 The triangle t(n,k) of the binomial sum as in A177351 in the column index range -floor(n/2)-1 <=k <= floor(n/2)-1.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 2, 5, 5, 5, 4, 1, 8, 8, 8, 7, 3, 13, 13, 13, 13, 12, 7, 1, 21, 21, 21, 21, 20, 14, 4, 34, 34, 34, 34, 34, 33, 26, 11, 1, 55, 55, 55, 55, 55, 54, 46, 25, 5, 89, 89, 89, 89, 89, 89, 88, 79, 51, 16, 1
Offset: 0

Views

Author

Roger L. Bagula, Dec 10 2010

Keywords

Comments

Row sums are 1, 1, 5, 8, 20, 34, 72, 122, 241, 405, 769, 1284, 2375, 3947, 7165,
11866, 21238, 35078, 62094, 102340, 179561,.... which apparently is (n+1)*Fibonacci(n+1)- A129722(n) for even n, and n*Fibonacci(n+1)-A129722(n) for odd n.
The first column is A000045 by construction. The change in the column index range adds the Fibonacci numbers as a first column and removes the trailing zero in the rows compared to A177351.
Comment R. J. Mathar, Dec 20 2010 (Start):
If we construct the complements of each row's entries with respect to the Fibonacci number of that row, an array
1; # complement to 2
1,4; # complement to 4,1
1,5 # complement to 7,3
1,6,12 # complement to 12,7,1
1,7,17 # complement to 20,14,4
1,8,23,33 # complement to 33,26,11,1
emerges which appears to be related to A038791. (End).

Examples

			1
1;
2, 2, 1;
3, 3, 2;
5, 5, 5, 4, 1;
8, 8, 8, 7, 3;
13, 13, 13, 13, 12, 7, 1;
21, 21, 21, 21, 20, 14, 4;
34, 34, 34, 34, 34, 33, 26, 11, 1;
55, 55, 55, 55, 55, 54, 46, 25, 5;
89, 89, 89, 89, 89, 89, 88, 79, 51, 16, 1;
		

Crossrefs

Programs

  • Mathematica
    w[n_, m_, k_] = Binomial[n - (m + k), m + k];
    t[n_, k_] := Sum[w[n, m, k], {m, 1, Floor[n/2 - k]}];
    Table[Table[t[n, k], {k, -Floor[n/2 + 1], Floor[n/2 + 1] - 2}], {n, 0,
       10}]
    Flatten[%]
Showing 1-3 of 3 results.