|
Search: id:A130138
|
|
|
| A130138 |
|
Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 1011's (n>=0, 0<=k<=floor((n-1)/3)). A Fibonacci binary word is a binary word having no 00 subword. |
|
+0 1
|
|
| 1, 2, 3, 5, 7, 1, 9, 4, 11, 10, 13, 20, 1, 15, 35, 5, 17, 56, 16, 19, 84, 40, 1, 21, 120, 86, 6, 23, 165, 166, 23, 25, 220, 296, 68, 1, 27, 286, 496, 171, 7, 29, 364, 791, 382, 31, 31, 455, 1211, 781, 105, 1, 33, 560, 1792, 1488, 300, 8, 35, 680, 2576, 2678, 756, 40, 37
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Row n has 1+floor((n-1)/3) terms. Row sums are the Fibonacci numbers (A000045). T(n,0)=A004280(n). Sum(k*T(n,k), k>=0)=A004798(n-3) (n>=4).
|
|
FORMULA
|
G.f.=G(t,z)=(1+z)(1+z^3-tz^3)/[1-z-z^2+z^3-tz^3].
|
|
EXAMPLE
|
T(7,2)=1 because we have 1011011.
Triangle starts:
1;
2;
3;
5;
7,1;
9,4;
11,10;
13,20,1;
15,35,5;
|
|
MAPLE
|
G:=(1+z)*(1+z^3-t*z^3)/(1-z-z^2+z^3-t*z^3): Gser:=simplify(series(G, z=0, 24)): for n from 0 to 21 do P[n]:=sort(coeff(Gser, z, n)) od: 1; for n from 1 to 21 do seq(coeff(P[n], t, j), j=0..floor((n-1)/3)) od; # yields sequence in triangular form
|
|
CROSSREFS
|
Cf. A000045, A004280, A004798.
Sequence in context: A053666 A101987 A126052 this_sequence A130136 A032759 A093338
Adjacent sequences: A130135 A130136 A130137 this_sequence A130139 A130140 A130141
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), May 13 2007
|
|
|
Search completed in 0.002 seconds
|