A181878 Coefficient array for square of Chebyshev S-polynomials.

1, 1, 1, -2, 1, 4, -4, 1, 1, -6, 11, -6, 1, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1, 36, -420, 1897, -4352, 5776, -4744, 2486, -832, 172, -20, 1, 1, -42, 581, -3108, 8518, -13672, 13820, -9142, 4013, -1158, 211, -22, 1

Offset: 0

Author

Wolfdieter Lang, Dec 22 2010

Keywords

• sign
• easy
• tabf

Comments

For the coefficients of Chebyshev polynomials S(n,x) see A049310.

The row length sequence for this array is A109613 = {1,1,3,3,5,5,...}.

The row polynomials (in x^2) for even row numbers are

S(2*k,x)^2 = Sum_{m=0..2*k} a(2*k,m)*x^(2*m), k >= 0.

For odd row numbers the row polynomials (in x^2) are

(S(2*k+1,x)^2)/x^2 = Sum_{m=0..2*k} a(2*k+1,m)*x^(2*m), k >= 0.

The o.g.f. for the polynomials S(n,x)^2 is

S(x,z):=((1+z)/(1-z))/(1 + (2-x^2)z +z^2). See the link for a proof. Therefore the coefficients constitute the Riordan array (1/(1-x^2),x/(1+x)^2) found as A158454.

The o.g.f. for S(2*k,sqrt(x))^2 is

(1-2*(1-x)*z+z^2)/((1-z)*(1 - (2-4*x+x^2)*z + z^2)).

The o.g.f. for (S(2*k+1,sqrt(x))^2)/x is

((1+z)/(1-z))/(1 - (2-4*x+x^2)*z + z^2).

The row sums A011655(n+1) are the same as those for the triangle A158454.

The alternating row sums for even numbered rows (-1)^n*A007598(n+1) coincide with those of triangle A158454. For odd row numbers n=2k+1 these sums are A049684(k+1), k >= 0 (squares of even-indexed Fibonacci numbers).

Examples

			The irregular triangle a(n,m) begins:
  n\m  0    1    2      3     4      5     6    7   8   9  10 ...
  0:   1
  1:   1
  2:   1   -2    1
  3:   4   -4    1
  4:   1   -6   11     -6     1
  5:   9  -24   22     -8     1
  6:   1  -12   46    -62    37    -10     1
  7:  16  -80  148   -128    56    -12     1
  8:   1  -20  130   -314   367   -230    79  -14   1
  9:  25 -200  610   -920   771   -376   106  -16   1
  10:  1  -30  295  -1106  2083  -2232  1444 -574 137 -18   1
  ... Reformatted and extended by _Wolfdieter Lang_, Nov 24 2012
		

Links

• Wolfdieter Lang, First ten rows with more details and proofs.

Crossrefs

Cf. A158454, A129818.

Programs

Mathematica

Join[{{1}, {1}}, CoefficientList[Table[ChebyshevU[n, Sqrt[x]/2]^2, {n, 2, 10}], x]] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
Join[{{1}, {1}}, CoefficientList[ChebyshevU[Range[2, 10], Sqrt[x]/2]^2, x]]  // Flatten (* Eric W. Weisstein, Apr 04 2018 *)

Formula

a(2*k,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+m-1-2*j, 2*m-1), k >= 0.

a(2*k+1,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+1+m-2*j, 2*m+1), k >= 0.

This derives from the formula for the entries of the Riordan array A158454.

For the o.g.f.s see the comment.

Extensions

Corrected by Wolfdieter Lang, Jan 21 2011