OFFSET
0,13
COMMENTS
The (n,k)-entry of the square array is p(n,k) in the R. Kemp reference (see Table 1 on p. 160 and Theorem 2 on p. 159). - Emeric Deutsch, Jun 16 2011
LINKS
Stefano Spezia, First 151 antidiagonals of the array, flattened
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015. See formula 0.2, p. 2.
Nancy S. S. Gu, Helmut Prodinger, Combinatorics on lattice paths in strips, arXiv:2004.00684 [math.CO], 2020. See p. 2.
R. Kemp, On the average depth of a prefix of the Dycklanguage D_1, Discrete Math., 36, 1981, 155-170.
FORMULA
An explicit expression for the (n,k)-entry of the square array can be found in the R. Kemp reference (Theorem 2 on p. 159). - Emeric Deutsch, Jun 16 2011
The g.f. of column k is (1 + v^2)*(1 - v^(k+1))/((1 - v)*(1 + v^(k+2))), where v = (1 - sqrt(1-4*z^2))/(2*z) (see p. 159 of the R. Kemp reference). - Emeric Deutsch, Jun 16 2011
EXAMPLE
Rows start:
1,0,0,0,0,...;
1,1,1,1,1,...;
1,1,2,2,4,...;
1,1,2,3,5,...;
etc.
MAPLE
v := ((1-sqrt(1-4*z^2))*1/2)/z: G := proc (k) options operator, arrow: (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) end proc: a := proc (n, k) options operator, arrow: coeff(series(G(k), z = 0, 80), z, n) end proc: for n from 0 to 15 do seq(a(n, k), k = 0 .. 15) end do; # yields the first 16 entries of the first 16 rows of the square array
v := ((1-sqrt(1-4*z^2))*1/2)/z: G := proc (k) options operator, arrow: (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) end proc: a := proc (n, k) options operator, arrow: coeff(series(G(k), z = 0, 80), z, n) end proc: for n from 0 to 13 do seq(a(n-i, i), i = 0 .. n) end do; # yields the first 14 antidiagonals of the square array in triangular form
MATHEMATICA
v = (1-Sqrt[1-4z^2])/(2z); f[k_] = (1+v^2)*(1-v^(k+1))/((1-v)*(1+v^(k+2))) ; m = 14; a = Table[ PadRight[ CoefficientList[ Series[f[k], {z, 0, m}], z], m], {k, 0, m}]; Flatten[Table[a[[n+1-k, k]], {n, m}, {k, n, 1, -1}]][[;; 95]] (* Jean-François Alcover, Jul 13 2011, after Emeric Deutsch *)
PROG
(PARI) T(n, k) = sum(j=floor(-n/(k+2)), ceil(n/(k+2)), (-1)^j*binomial(n, floor((n+(k+2)*j)/2))); \\ Stefano Spezia, May 08 2020
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Henry Bottomley, Mar 06 2002
STATUS
approved