OFFSET
0,2
COMMENTS
This table gives therefore sin((2*n+1)*phi) in terms of falling odd powers of sin(phi).
The unsigned triangle with reversed rows is A084930 (the signs differ).
LINKS
FORMULA
a(n,m)=0 if n < m else a(n,m) = ((-4)^(n-m))*binomial(2n-m,m)*(2*n+1)/(2*(n-m)+1), n >= m >= 0. (Proof from the differential eq. for U(2*n,x): (1-x^2)*(d^2/dx^2)U(2*n,x) - 3*x*(d/dx)U(2*n,x) + 4*n*(n+1)*U(2*n,x) = 0.)
a(n,m)=0 if n < m else a(n,m) = Sum_{k=0..n-m} (binomial(m+k,k)*binomial(2*n+1,2*(m+k))*(-1)^(n-m)) (from de Moivre's formula for sin((2*n+1)*phi) after replacing cos(phi)^2 with 1-sin(phi)^2).
EXAMPLE
[1];[ -4,3];[16,-20,5];[ -64,112,-56,7];[256,-576,432,-120,9]; ...
Row n=3: -64*(1-x^2)^3+ 112*(1-x^2)^2 -56*(1-x^2)^1 + 7 = 64*x^6 - 80*x^4 + 24* x^2 -1 =U(6,x).
Row n=3: sin(7*phi)=-64*sin(phi)^7 + 112*sin(phi)^5 - 56*sin(phi)^3 + 7*sin(phi).
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Mar 07 2007
STATUS
approved