A207824 Triangle of coefficients of Chebyshev's S(n,x+5) polynomials (exponents of x in increasing order).

1, 5, 1, 24, 10, 1, 115, 73, 15, 1, 551, 470, 147, 20, 1, 2640, 2828, 1190, 246, 25, 1, 12649, 16310, 8631, 2400, 370, 30, 1, 60605, 91371, 58275, 20385, 4225, 519, 35, 1, 290376, 501150, 374115, 157800, 41140, 6790, 693, 40, 1

Offset: 0

Author

Philippe Deléham, Feb 20 2012

Keywords

• easy
• nonn
• tabl

Comments

Riordan array (1/(1-5*x+x^2), x/(1-5*x+x^2)).

Subtriangle of triangle given by (0, 5, -1/5, 1/5, 0, 0, 0, 0, 0, 0, 0, 0...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Unsigned version of A123967 and A179900.

For 1<=k<=n, T(n,k) equals the number of (n-1)-length words over {0,1,2,3,4,5} containing k-1 letters equal 5 and avoiding 01. - Milan Janjic, Dec 20 2016

Examples

			Triangle begins :
  1
  5, 1
  24, 10, 1
  115, 73, 15, 1
  551, 470, 147, 20, 1
  2640, 2828, 1190, 246, 25, 1
  12649, 16310, 8631, 2400, 370, 30, 1
  ...
Triangle (0, 5, -1/5, 1/5, 0, 0, 0,...) DELTA (1, 0, 0, 0, ...) begins :
  1
  0, 1
  0, 5, 1
  0, 24, 10, 1
  0, 115, 73, 15, 1
  0, 551, 470, 147, 20, 1
  0, 2640, 2828, 1190, 246, 25, 1
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
• Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Crossrefs

Cf. Triangles of coefficients of Chebyshev's S(n,x+k) polynomials : A207824 (k = 5), A207823 (k = 4), A125662 (k = 3), A078812 (k = 2), A101950 (k = 1), A049310 (k = 0), A104562 (k = -1), A053122 (k = -2), A207815 (k = -3), A159764 (k = -4), A123967 (k = -5).

Programs

Mathematica

With[{n = 8}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - 5 x + x^2 - y x), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

PARI

row(n) = Vecrev(polchebyshev(n, 2, (x+5)/2)); \\ Michel Marcus, Apr 26 2018

Formula

Recurrence : T(n,k) = 5*T(n-1,k) + T(n-1,k-1) - T(n-2,k).

G.f.: 1/(1-5*x+x^2-y*x).

Diagonal sums are 5^n = A000351(n).

Row sums are A001109(n+1).

T(n,0) = A004254(n+1), T(n+1,n) = 5n+5 = A008587(n+1).