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A141665
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A signed half of Pascal's triangle A007318: p(x,n) = (1+I*x)^n; t(n,m) = real part of coefficients(p(x,n)).
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1
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1, 1, 0, 1, 0, -1, 1, 0, -3, 0, 1, 0, -6, 0, 1, 1, 0, -10, 0, 5, 0, 1, 0, -15, 0, 15, 0, -1, 1, 0, -21, 0, 35, 0, -7, 0, 1, 0, -28, 0, 70, 0, -28, 0, 1, 1, 0, -36, 0, 126, 0, -84, 0, 9, 0, 1, 0, -45, 0, 210, 0, -210, 0, 45, 0, -1
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OFFSET
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0,9
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COMMENTS
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Polynomials like these are seen in complex dynamics.
This method symmetrically breaks up Pascal's triangle A007318 into two parts as polynomial coefficient vectors. See the examples for the s(n,m) = imaginary part of coefficients(p(x,n)).
The row sums equal A146559 and the two antidiagonal sums lead to A104862 (minus a(0)) and A110161 (minus a(0)).
The mirror of this triangle (for the absolute values of the coefficients) is A119467. (End)
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LINKS
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FORMULA
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p(x,n) = (1+I*x)^n
t(n,m) = real part of coefficients(p(x,n))
s(n,m) = imaginary part of coefficients(p(x,n))
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EXAMPLE
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s(n,m) = imaginary part of coefficients(p(x,n))
{0},
{0, 1},
{0, 2, 0},
{0, 3, 0, -1},
{0, 4, 0, -4, 0},
{0, 5, 0, -10, 0, 1},
{0, 6, 0, -20, 0, 6, 0},
{0, 7, 0, -35, 0, 21, 0, -1},
{0, 8, 0, -56, 0, 56, 0, -8, 0},
{0, 9, 0, -84, 0, 126, 0, -36, 0, 1},
{0, 10, 0, -120, 0, 252, 0, -120, 0, 10, 0}
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MAPLE
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nmax:=10: for n from 0 to nmax do p(x, n) := (1+I*x)^n: for m from 0 to n do t(n, m) := Re(coeff(p(x, n), x, m)) od: od: seq(seq(t(n, m), m=0..n), n=0..nmax);
nmax:=10: for n from 0 to nmax do for m from 0 to n do A119467(n, m) := binomial(n, m) * (1+(-1)^(n-m))/2: if (m mod 4 = 2) then x(n, m):= -1 else x(n, m):= 1 end if: od: od: for n from 0 to nmax do for m from 0 to n do t(n, m) := A119467(n, n-m)*x(n, m) od: od: seq(seq(t(n, m), m=0..n), n=0..nmax); # (End)
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MATHEMATICA
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p[x_, n_] := If[n == 0, 1, Product[(1 + I*x), {i, 1, n}]]; Table[Expand[p[x, n]], {n, 0, 10}]; Table[Im[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[%] Table[Re[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[%]
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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