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A172004
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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.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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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}.
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EXAMPLE
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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.
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PROG
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(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)]
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CROSSREFS
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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.
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KEYWORD
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AUTHOR
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STATUS
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approved
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