OFFSET
0,4
COMMENTS
Row n has 1+ceiling(n/2) terms.
T(n,0) = Fibonacci(n+1) = A000045(n+1).
T(n,1) = A029907(n).
Sum_{k>=0} k*T(n,k) = A059570(n).
Triangle, with zeros included, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,-1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 07 2011
T(n,k) is the number of compositions of n+1 that have exactly k even parts. - Geoffrey Critzer, Mar 03 2012
REFERENCES
I. Goulden and D. Jackson, Combinatorial Enumeration, John Wiley and Sons, 1983, page 54.
LINKS
Alois P. Heinz, Rows n = 0..200, flattened
Rigoberto Flórez, Javier González, Mateo Matijasevick, Cristhian Pardo, José Luis Ramírez, Lina Simbaqueba, and Fabio Velandia, Lattice paths in corridors and cyclic corridors, Contrib. Disc. Math. (2024) Vol. 19. No. 2, 36-55. See p. 17.
Ralph Grimaldi and Silvia Heubach, Binary strings without odd runs of zeros, Ars Combinatoria 75 (2005), 241-255.
FORMULA
G.f.: (1+t*z)/(1-z-z^2-t*z^2).
G.f. of column k (k>=1): z^(2*k-1)*(1-z^2)/(1-z-z^2)^(k+1).
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Dec 07 2011
Sum_{k=0..n} T(n,k)*x^k = A000045(n+1), A000079(n), A105476(n+1), A159612(n+1), A189732(n+1) for x = 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Dec 07 2011
G.f.: (1+x*y)*T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
EXAMPLE
T(5,2)=8 because we have 00010, 01000, 01011, 01101, 01110, 10101, 10110 and 11010.
T(5,2)=8 because there are 8 compositions of 6 that have 2 even parts: 4+2, 2+4, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - Geoffrey Critzer, Mar 03 2012
Triangle starts:
1;
1, 1;
2, 2;
3, 4, 1;
5, 8, 3;
8, 15, 8, 1;
From Philippe Deléham, Dec 07 2011: (Start)
Triangle (1,1,-1,0,0,0...) DELTA (1,-1,0,0,0,...) begins:
1;
1, 1;
2, 2, 0;
3, 4, 1, 0;
5, 8, 3, 0, 0;
8, 15, 8, 1, 0, 0;
13, 28, 19, 4, 0, 0, 0;
21, 51, 42, 13, 1, 0, 0, 0;
34, 92, 89, 36, 5, 0, 0, 0, 0; ... (End)
MAPLE
G:=(1+t*z)/(1-z-z^2-t*z^2): Gser:=simplify(series(G, z=0, 18)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser, z^n)) od: for n from 0 to 14 do seq(coeff(P[n], t, j), j=0..ceil(n/2)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; local j; if n=0 then 1
else []; for j to n do zip((x, y)->x+y, %,
[`if`(irem(j, 2)=0, 0, NULL), b(n-j)], 0) od; %[] fi
end:
T:= n-> b(n+1):
seq(T(n), n=0..14); # Alois P. Heinz, May 23 2013
MATHEMATICA
f[list_] := Select[list, # > 0 &]; nn = 15; a = (x + y x^2)/(1 - x^2); Map[f, Drop[CoefficientList[Series[1/(1 - a), {x, 0, nn}], {x, y}], 1]] // Flatten (* Geoffrey Critzer, Mar 03 2012 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, May 22 2006
STATUS
approved