A068914 Square array read by antidiagonals of number of k step walks (each step +-1 starting from 0) which are never more than n or less than 0.
1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 4, 3, 2, 1, 1, 0, 1, 4, 5, 3, 2, 1, 1, 0, 1, 8, 8, 6, 3, 2, 1, 1, 0, 1, 8, 13, 9, 6, 3, 2, 1, 1, 0, 1, 16, 21, 18, 10, 6, 3, 2, 1, 1, 0, 1, 16, 34, 27, 19, 10, 6, 3, 2, 1, 1, 0, 1, 32, 55, 54, 33, 20, 10, 6, 3, 2, 1, 1, 0, 1, 32, 89
Offset: 0
Examples
Rows start: 1,0,0,0,0,...; 1,1,1,1,1,...; 1,1,2,2,4,...; 1,1,2,3,5,...; etc.
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.
Crossrefs
Programs
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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
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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 *) -
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
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
Comments