A172004 Let y = y(u,v) be implicitly defined by g(u,v,y(u,v)) = 0. Read as a triangle by rows k = 1,2,..., the sequence represents the number of terms a(i,k-i) in the expansion of the partial derivatives d^k y/du^i dv^{k-i} in terms of partial derivatives of g.

1, 1, 3, 4, 3, 9, 15, 15, 9, 24, 47, 59, 47, 24, 61, 136, 195, 195, 136, 61, 145, 360, 580, 663, 580, 360, 145, 333, 904, 1586, 2032, 2032, 1586, 904, 333, 732, 2152, 4077, 5684, 6350, 5684, 4077, 2152, 732, 1565, 4927, 9948, 14938, 18123, 18123, 14938, 9948

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),...

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 A003262, which is the corresponding sequence for univariate implicit functions.

Examples

			The formulas dy/du = -g_u/g_y,
d^2y/du^2 = -g_yy g_u^2/g_y^3 + 2g_uy g_u/g_y^2 - g_uu/g_y,
d^2y/dudv = -2g_yy g_u g_v / g_y^3 + g_uy g_v/g_y^2 + g_vy g_u/g_y^3 - g_uv/g_y
imply that a(1,0) = 1, a(2,0) = 3, and a(1,1) = 4.
		

Links

• Wilde, T., Implicit higher derivatives and a formula of Comtet and Fiolet, preprint arXiv:0805.2674 [math.CO], 2008.

Crossrefs

Cf. A003262, which is the univariate variant of this sequence.

Cf. A172003, which is the analogous sequence for implicit divided differences, and A162326 for its univariate variant.

Programs

Sage

# Upon executing the following code in Sage 4.2 (using Singular as a backend), it
# computes the number of terms a(n1,n2) and stores it in the entry A[n1][n2] of the
# double list A.
N = 9
E1 = N
E2 = N
p = [[[0 for i1 in range(E1+1)] for i2 in range(E2+1)] for j in range(E1 + E2)]
q = [[[0 for i1 in range(E1+1)] for i2 in range(E2+1)] for j in range(E1 + E2)]
for m in range(1, E1 + E2):
    for d in range(1, m+1):
        quotient, remainder = divmod(m, d)
        if remainder == 0:
            for i1 in range(quotient + 1 + 1):
                for i2 in range(quotient + 1 - i1 + 1):
                    if d*i1 <= E1 and d*i2 <= E2:
                        q[m][i1*d][i2*d] += 1/d
for i1 in range(E1 + 1):
    for i2 in range(E2 + 1):
        p[0][i1][i2] = 1
for n in range(1, E1 + E2):
    for s in range(n+1):
        for k1 in range(E1+1):
            for k2 in range(E2+1):
                for i1 in range(k1 + 1):
                    for i2 in range(k2 + 1):
                        p[n][k1][k2] += 1/n * s * q[s][k1-i1][k2-i2] * p[n-s][i1][i2]
A = [[ p[n1+n2-1][n1][n2] for n1 in range(E1+1)] for n2 in range(E2+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} in Product_{(r,s,t) in E} (1 - u^r * v^s * y^{r+s+t-1})^{-1}.