A244420 Numerators of coefficient triangle for expansion of x^n in terms of polynomials Todd(k, x) = T(2*k+1, sqrt(x))/sqrt(x) (A084930), with the Chebyshev polynomials of the first kind (type T).

1, 3, 1, 5, 5, 1, 35, 21, 7, 1, 63, 21, 9, 9, 1, 231, 165, 165, 55, 11, 1, 429, 1287, 715, 143, 39, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 12155, 2431, 1547, 1547, 595, 85, 17, 17, 1, 46189, 37791, 12597, 6783, 2907, 969, 969, 171, 19, 1, 88179, 146965, 101745, 14535, 6783, 20349, 5985, 665, 105, 21, 1

Offset: 0

Author

Wolfdieter Lang, Aug 04 2014

Keywords

• nonn
• easy
• tabl
• frac

Comments

This expansion is due to the Riordan property of the triangle A084930. The inverse of the lower triangular matrix built by A084930 is therefore also a (rational) Riordan triangle, namely ((2 - c(z/4))/(1-z), -1 + c(z/4)) in the standard notation, where c is the o.g.f. of A000108 (Catalan).

For the denominators of this triangle see A244421.

The expansion is x^n = sum(R(n,m)*Todd(m, x), m=0..n), n >= 0, with the rational triangle with entries R(n,m) = a(n, m)/b(n, m) with b(n, m) = A244421(n, m).

If one uses instead the expansion of (4*x)^n one finds the integer triangle A111418: (4*x)^n = sum(A111418(n,k) * Todd(k, x), k=0..n).

The row sums of the rational triangle R(n,m) are identically 1. The alternating row sums have o.g.f. 1/sqrt(1-x) which generates A001790(n)/A046161(n) (see a Michael Somos comment on A046161), namely 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, ...

From Wolfdieter Lang, Jun 13 2016: (Start)

R(n,m) = a(n, m)/ A244421(n, m) is also the rational triangle for the expansion cos^{2*n+1}(x) = Sum_{m=0..n} R(n, m)*cos((2*m+1)*x), n >= 0, m = 0..n. Compare with the odd numbered rows of A273496. In terms of Chebyshev T-polynomials (A053120) this is the identity x^(2*n+1) = Sum_{m=0..n} R(n, m)*T(2*m+1, x).

S(n,m) = (-1)^m*a(n, m)/ A244421(n, m) is the rational triangle for the expansion sin^{2*n+1}(x) = Sum_{m=0..n} S(n, m)*sin((2*m+1)*x), n >= 0, m = 0..n. In terms of Chebyshev S-polynomials (A049310) this is equivalent to the identity (4 - x^2)*n = Sum_{m=0..n} (-1)^m * binomial(n, n-m)*S(2*m,x), n >= 0.

(End)

Examples

			The numerator triangle a(n,m) begins:
  n\m      0      1      2     3    4     5    6   7   8   9
  0:       1
  1:       3      1
  2:       5      5      1
  3:      35     21      7     1
  4:      63     21      9     9    1
  5:     231    165    165    55   11     1
  6:     429   1287    715   143   39    13    1
  7:    6435   5005   3003  1365  455   105   15   1
  8:   12155   2431   1547  1547  595    85   17  17   1
  9:   46189  37791  12597  6783 2907   969  969 171  19   1
  ...
The rational triangle R(n,m) begins:
  n\m       0        1         2       3        4        5
  0:        1
  1:      3/4      1/4
  2:      5/8     5/16      1/16
  3:    35/64    21/64      7/64    1/64
  4:   63/128    21/64      9/64   9/256    1/256
  5:  231/512  165/512  165/1024 55/1024  11/1024   1/1024
  ...
The next rows are:
  n=6: 429/1024, 1287/4096, 715/4096, 143/2048, 39/2048, 13/4096, 1/4096,
  n=7: 6435/16384, 5005/16384, 3003/16384, 1365/16384, 455/16384, 105/16384, 15/16384, 1/16384,
  n=8: 12155/32768, 2431/8192, 1547/8192, 1547/16384, 595/16384, 85/8192, 17/8192, 17/65536, 1/65536,
  n=9: 46189/131072, 37791/131072, 12597/65536, 6783/65536, 2907/65536, 969/65536, 969/262144, 171/262144, 19/262144, 1/262144,
  ...
Expansions:
x^2 = 5/8 * Todd(0,x) + 5/16 * Todd(1,x) + 1/16 * Todd(2,x) = 5/8 + (5/16)*(-3 + 4*x) +(1/16)*(5 -20*x + 16*x^2).
x^3 = (35*Todd(0, x) + 21*Todd(1, x) + 7*Todd(2, x) + 1*Todd(3, x))/64 = (35 + 21*(-3+4*x) + 7*( 5-20*x+16*x^2) + (-7+56*x-112*x^2+64*x^3))/64.
For the Todd polynomials see the coefficient table A084930.
		

Links

• Wolfdieter Lang, First rows of the triangles.

Crossrefs

Cf. A084930, A244421, A000108, A001790, A046161, A111418, A273496.

Formula

a(n, m) = numerator(R(n, m)) with the rationals Riordan matrix elements R(n, m)= [x^m]R(n, x), with the row polynomials R(n, x) generated by ((2 - c(z/4))/(1-z))/(1 - x*(-1 + c(z/4))) = 2*((1+x)*(z-1) + (1-x)*sqrt(1-z))/((1-z)*((1+x)^2*z - 4*x)), where c(x) is the o.g.f. of the Catalan numbers A000108.

The rationals R(n, m) = binomial(2*n+1, m)/2^(2*n). - Wolfdieter Lang, Jun 12 2016