|
Search: id:A127675
|
|
|
| A127675 |
|
Coefficient table for Chebyshev's U(2*n,x) polynomials in decreasing powers of (1-x^2). |
|
+0
3
|
|
| 1, -4, 3, 16, -20, 5, -64, 112, -56, 7, 256, -576, 432, -120, 9, -1024, 2816, -2816, 1232, -220, 11, 4096, -13312, 16640, -9984, 2912, -364, 13, -16384, 61440, -92160, 70400, -28800, 6048, -560, 15, 65536, -278528, 487424, -452608, 239360, -71808, 11424, -816, 17, -262144, 1245184
(list; table; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
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
|
W. Lang, First 15 rows and more.
|
|
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)*diff(U(2*n,x),x$2) - 3*x*diff(U(2*n,x),x) + 4*n*(n+1)*U(2*n,x)=0.)
a(n,m)=0 if n<m else a(n,m)= sum(binomial(m+k,k)*binomial(2*n+1,2*(m+k))*(-1)^(n-m),k=0..n-m) (from de Moivre's formula for sin((2*n+1)*phi) after replacing cos(phi)^2 by 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
|
Row sums (signed) A033999(n)=(-1)^n. Row sums (unsigned) A002315(n).
Cf. A082985 (scaled coefficient table).
Sequence in context: A065679 A062776 A084471 this_sequence A058557 A038233 A046162
Adjacent sequences: A127672 A127673 A127674 this_sequence A127676 A127677 A127678
|
|
KEYWORD
|
sign,tabl,easy
|
|
AUTHOR
|
Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Mar 07 2007
|
|
|
Search completed in 0.002 seconds
|
