OFFSET
1,3
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.
LINKS
Wilde, T., Implicit higher derivatives and a formula of Comtet and Fiolet, preprint arXiv:0805.2674 [math.CO], 2008.
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}.
EXAMPLE
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.
PROG
(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)]
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Georg Muntingh, Jan 22 2010
STATUS
approved