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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. 3
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 (list; table; graph; refs; listen; history; text; internal format)
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
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.
Sequence in context: A158134 A184853 A174280 * A051508 A281230 A323977
KEYWORD
nonn,tabl
AUTHOR
Georg Muntingh, Jan 22 2010
STATUS
approved

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Last modified May 22 10:24 EDT 2024. Contains 372745 sequences. (Running on oeis4.)