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A139815
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A triangle of coefficients of a Moebius-transformed Pascal triangle as a sum: b(x,y,n)=Sum[Binomial[n,i]*x^i*y^(n-i),{i,0,n}]; transforms: x'->(a1*x + b1)/(c1*x + d1); y'->(a2*y + b2)/(c2*y + d2); b1(x,y,n)=(c1*x + b1)^(k)*(c2*y + d2)^(k)*b(x',y',n); f(x,y,z,n)=b1(x,y,n)+b1(y,z,n)+b1(z,x,n).
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0
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3, -16, 4, 88, -48, 8, -496, 432, -144, 16, 2848, -3456, 1728, -384, 32, -16576, 25920, -17280, 5760, -960, 64, 97408, -186624, 155520, -69120, 17280, -2304, 128, -576256, 1306368, -1306368, 725760, -241920, 48384, -5376, 256, 3424768, -8957952, 10450944, -6967296, 2903040, -774144, 129024
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OFFSET
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1,1
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COMMENTS
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Row sums are:
{3, -12, 48, -192, 768, -3072, 12288, -49152, 196608, -786432, 3145728}.
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REFERENCES
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P. J. Olver, Classical Invariant Theory, Cambridge Univ. Press, p. 242.
McKean and Moll, Elliptic Curves, Function Theory,Geometry, Arithmetic, Cambridge University Press, New York, 199, page 172
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LINKS
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FORMULA
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b(x,y,n)=Sum[Binomial[n,i]*x^i*y^(n-i),{i,0,n}]; transforms: x'->(a1*x + b1)/(c1*x + d1); y'->(a2*y + b2)/(c2*y + d2); b1(x,y,n)=(c1*x + b1)^(k)*(c2*y + d2)^(k)*b(x',y',n); f(x,y,z,n)=b1(x,y,n)+b1(y,z,n)+b1(z,x,n); Out_n,m=Coefficients(f(x,y,z,n)).
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EXAMPLE
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{3},
{-16,4},
{88, -48, 8},
{-496, 432, -144, 16},
{2848, -3456, 1728, -384, 32},
{-16576, 25920, -17280, 5760, -960,64},
{97408, -186624, 155520, -69120, 17280, -2304, 128},
{-576256, 1306368, -1306368, 725760, -241920, 48384, -5376, 256},
{3424768, -8957952,10450944, -6967296, 2903040, -774144, 129024, -12288, 512}, {-20417536, 60466176, -80621568, 62705664, -31352832, 10450944, -2322432, 331776, -27648, 1024},
{121980928, -403107840, 604661760, -537477120, 313528320, -125411328, 34836480, -6635520, 829440, -61440, 2048}
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MATHEMATICA
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a1 = 1; b1 = -2; c1 = 0; d1 = 1; a2 = 0; b2 = 1; c2 = 1; d2 = -2; p[x_, y_, k_] = (c1*x + b1)^(k)*(c2*y + d2)^(k)*Sum[Binomial[k, i]*((a1*x + b1)/(c1*x + d1))^i*((a2*y + b2)/(c2*y + d2))^(k - i), {i, 0, k}]; f[x_, y_, z_, k_] = p[x, y, k] + p[y, z, k] + p[z, x, k]; Table[ExpandAll[f[x, y, z, k]], {k, 0, 10}]; a = Table[CoefficientList[f[x, y, z, k] /. y -> 1 /. z -> 1, x], {k, 0, 10}]; Flatten[a] Table[Apply[Plus, CoefficientList[f[x, y, z, k] /. y -> 1 /. z -> 1, x]], {k, 0, 10}]
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CROSSREFS
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KEYWORD
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uned,tabf,sign
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AUTHOR
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STATUS
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approved
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