OFFSET
0,5
COMMENTS
Conjecture: A(n, k) is odd if and only if n is even or (n is odd and k + 2 = 2^j for some j > 0).
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.
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).
For a recurrence see the Maple section.
EXAMPLE
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
.
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
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;
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]]]
PROG
(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.<x> = PowerSeriesRing(ZZ)
for n in (0..6):
f = (1+(n-1)*x)/((1-x)^2)
s = pseudo_reversion(f, true)
print(s.list())
CROSSREFS
Sums of antidiagonals: A323207
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Feb 21 2019
STATUS
editing