A172003 Let y = y(u,v) be implicitly defined by g(u,v,y(u,v)) = 0. Read as a triangle by rows, the sequence represents the number of terms a(i,k-i) in the expansion of the bivariate divided difference [u_0,...,u_i; v_0,...,v_{k-i}]y in terms of trivariate divided differences of g.

1, 1, 3, 5, 3, 13, 33, 33, 13, 71, 245, 351, 245, 71, 441, 1921, 3597, 3597, 1921, 441, 2955, 15525, 35931, 46709, 35931, 15525, 2955, 20805, 127905, 352665, 563821, 563821, 352665, 127905, 20805, 151695, 1067925, 3417975, 6483285, 7963151

Offset: 1

Author

Georg Muntingh, Jan 22 2010

Keywords

• nonn
• tabl

Comments

The sequence starts with a(1,0),a(0,1),a(2,0),a(1,1),a(0,2),a(3,0),...

Examples

			The subsequences a(1,0),a(2,0),a(3,0),... and a(0,1),a(0,2),a(0,3),... coincide with the sequence A162326.
For (m,n) = (1,1), one expresses [u_0,u_1;v_0,v_1]y as a sum of 5 terms,
[01;01]y =
- [0;0;(0,0),(1,0),(1,1)]g * [01;0;(1,0)]g * [1;01;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;0;(0,0),(1,0)]g * [1;0;(1,0),(1,1)]g )
+ [01;0;(1,0),(1,1)]g * [1;01;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [1;0;(1,0),(1,1)]g )
- [01;01;(1,1)]g / [0;0;(0,0),(1,1)]g
- [0;0;(0,0),(0,1),(1,1)]g * [0;01;(0,1)]g * [01;1;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;0;(0,0),(0,1)]g * [0;1;(0,1),(1,1)]g )
+ [0;01;(0,1),(1,1)]g * [01;1;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;1;(0,1),(1,1)]g ),
where the numbers refer to the indices of the corresponding variable, e.g.
[1;01;(1,1)]g = [u_1;v_0,v_1;y(u_1,v_1)]g.
		

Crossrefs

Cf. A162326, which is the univariate variant of this sequence.

Cf. A172004, which is the analogous sequence for implicit derivatives, and A003262 for its univariate variant.

Programs

Sage

R. = PolynomialRing(ZZ,3)
def P(n1,n2,q):
    E = cartesian_product([list(range(n1+1)), list(range(n2+1)), list(range(n1+n2+1))])
    E = [(i1,i2,j) for (i1,i2,j) in E if (i1,i2,j) != (0,0,0) and
         (i1,i2,j) != (0,0,1) and i1 + i2 + j <= n1 + n2 and
         2*(i1 + i2) + j - 1 <= 2*(n1+n2) - q]
    return R.sum(X1^s1 * X2^s2 * Y^(s1+s2+t-1) for s1,s2,t in E)
n1, n2 = 4, 4
L = [[0 for _ in range(n1 + 1)]] * (n2 + 1)
h = 1 + sum(((P(n1,n2,q))^q)/q for q in range(1,2*(n1+n2)))
for k1 in range(n1+1):
    for k2 in range(k1+1):
        if (k1, k2) != (0, 0):
            print(k1, k2, h.coefficient({X1:k1, X2:k2, Y:k1+k2-1}))

Formula

Let E = N^3 \ {(0,0,0), (0,0,1)} be a set of triples of natural numbers. The number of terms a(m,n) is the coefficient of u^m * v^n * y^{m+n-1} of the generating function

- log(1 - Sum_{(r,s,t) in E} u^r * v^s * y^{r+s+t-1})

= Sum_{q >= 1} (Sum_{(r,s,t) in E} u^r * v^s * y^{r+s+t-1})^q / q.