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A068914
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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.
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7
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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
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OFFSET
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0,13
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COMMENTS
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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
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LINKS
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FORMULA
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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
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EXAMPLE
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Rows start:
1,0,0,0,0,...;
1,1,1,1,1,...;
1,1,2,2,4,...;
1,1,2,3,5,...;
etc.
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MAPLE
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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
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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