A141665 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)).
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
Offset: 0
Examples
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}
Links
- G. C. Greubel, Rows n=0..100 of triangle, flattened
- Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4, section 5.
Programs
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Maple
From Johannes W. Meijer, Mar 10 2012: (Start) 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
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[%]
Formula
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))
Extensions
Edited and information added by Johannes W. Meijer, Mar 10 2012
Comments