A323206 A(n, k) = hypergeometric([-k, k+1], [-k-1], n), square array read by ascending antidiagonals for n,k >= 0.
1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 13, 14, 1, 1, 5, 25, 67, 42, 1, 1, 6, 41, 190, 381, 132, 1, 1, 7, 61, 413, 1606, 2307, 429, 1, 1, 8, 85, 766, 4641, 14506, 14589, 1430, 1, 1, 9, 113, 1279, 10746, 55797, 137089, 95235, 4862, 1
Offset: 0
Examples
Array starts: [n\k 0 1 2 3 4 5 6 7 ...] [0] 1, 1, 1, 1, 1, 1, 1, 1, ... A000012 [1] 1, 2, 5, 14, 42, 132, 429, 1430, ... A000108 [2] 1, 3, 13, 67, 381, 2307, 14589, 95235, ... A064062 [3] 1, 4, 25, 190, 1606, 14506, 137089, 1338790, ... A064063 [4] 1, 5, 41, 413, 4641, 55797, 702297, 9137549, ... A064087 [5] 1, 6, 61, 766, 10746, 161376, 2537781, 41260086, ... A064088 [6] 1, 7, 85, 1279, 21517, 387607, 7312789, 142648495, ... A064089 [7] 1, 8, 113, 1982, 38886, 817062, 17981769, 409186310, ... A064090 [8] 1, 9, 145, 2905, 65121, 1563561, 39322929, 1022586105, ... A064091 A001844 A064096 A064302 A064303 A064304 A064305 diag: A323209 . Seen as a triangle (by reading ascending antidiagonals): 1 1, 1 1, 2, 1 1, 3, 5, 1 1, 4, 13, 14, 1 1, 5, 25, 67, 42, 1 1, 6, 41, 190, 381, 132, 1
Links
- J. Abate, W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011).
- B. Derrida, E. Domany and D. Mukamel, An exact solution of a one-dimensional asymmetric exclusion model with open boundaries, J. Stat. Phys. 69, 1992, 667-687.
Crossrefs
Programs
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Maple
# The function ballot is defined in A238762. A := (n, k) -> add(ballot(2*j, 2*k)*n^j, j=0..k): for n from 0 to 6 do seq(A(n, k), k=0..9) od; # Or by recurrence: A := proc(n, k) option remember; if n = 1 then return `if`(k = 0, 1, (4*k + 2)*A(1, k-1)/(k + 2)) fi: if k < 2 then return [1, n+1][k+1] fi; n*(4*k - 2); ((%*(n - 1) - k - 1)*A(n, k-1) + %*A(n, k-2))/((n - 1)*(k + 1)) end: for n from 0 to 6 do seq(A(n, k), k=0..9) od; # Alternative: Arow := proc(n, len) # Function REVERT is in Sloane's 'Transforms'. [seq(1 + n*k, k=0..len-1)]; REVERT(%); seq((-1)^k*%[k+1], k=0..len-1) end: for n from 0 to 8 do Arow(n, 8) od;
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Mathematica
A[n_, k_] := Hypergeometric2F1[-k, k + 1, -k - 1, n]; Table[A[n, k], {n, 0, 8}, {k, 0, 8}] (* Alternative: *) prev[f_, n_] := InverseSeries[Series[-x f, {x, 0, n}]]/(-x); f[n_, x_] := (1 + (n - 1) x)/((1 - x)^2); For[n = 0, n < 9, n++, Print[CoefficientList[prev[f[n, x], 8], x]]] (* Continued fraction: *) num[k_, n_] := If[k < 2, 1, If[k == 2, -x, -n x]]; cf[n_, len_] := ContinuedFractionK[num[k, n], 1, {k, len + 2}]; Arow[n_, len_] := Rest[CoefficientList[Series[cf[n, len], {x, 0, len}], x]]; For[n = 0, n < 9, n++, Print[Arow[n, 8]]]
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PARI
{A(n,k) = polcoeff((1/x)*serreverse(x*((1+(n-1)*(-x))/((1-(-x))^2)+x*O(x^k))), k)} for(n=0, 8, for(k=0, 8, print1(A(n, k), ", ")); print())
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Sage
# Valid for n > 0. def genCatalan(n): return SR(1/(x- x^2*(1 - sqrt(1 - 4*x*n))/(2*x*n))) for n in (1..8): print(genCatalan(n).series(x).list()) # Alternative: def pseudo_reversion(g, invsign=false): if invsign: g = g.subs(x=-x) g = g.shift(1) g = g.reverse() g = g.shift(-1) return g R.
= PowerSeriesRing(ZZ) for n in (0..6): f = (1+(n-1)*x)/((1-x)^2) s = pseudo_reversion(f, true) print(s.list())
Formula
A(n, k) = [x^k] 1/(x - x^2*C(n*x)) if n > 0 and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function of the Catalan numbers A000108.
A(n, k) = Sum_{j=0..k} (binomial(2*k-j, k) - binomial(2*k-j, k+1))*n^(k-j).
A(n, k) = Sum_{j=0..k} binomial(k + j, k)*(1 - j/(k + 1))*n^j (cf. A009766).
A(n, k) = 1 + Sum_{j=0..k-1} ((1+j)*binomial(2*k-j, k+1)/(k-j))*n^(k-j).
A(n, k) = (1/(2*Pi))*Integral_{x=0..4*n} (sqrt(x*(4*n-x))*x^k)/(1+(n-1)*x), n>0.
A(n, k) ~ ((4*n)^k/(Pi^(1/2)*k^(3/2)))*(1+1/(2*n-1))^2.
If we shift the series f with constant term 1 to the right, invert it with respect to composition and shift the result back to the left then we call this the 'pseudo reversion' of f, prev(f). Row n of the array gives the coefficients of the pseudo reversion of f = (1 + (n - 1)*x)/((1 - x)^2) with an additional inversion of sign. Note that f is not revertible. See also the Sage implementation below.
A(n, k) = [x^k] prev((1 + (n - 1)*(-x))/(1 - (-x))^2).
A(n, k) = [x^(k+1)] cf(n, x) where cf(n, x) = K_{i>=1} c(i)/b(i) in the notation of Gauß with b(i) = 1, c(1) = 1, c(2) = -x and c(i) = -n*x for i > 2.
For a recurrence see the Maple section.
Comments