A136398 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,6}(x) with 0 omitted (exponents in increasing order).

1, -6, 2, 15, -13, 4, -20, 36, -28, 8, 15, -55, 85, -60, 16, -6, 50, -146, 198, -128, 32, 1, -27, 155, -377, 456, -272, 64, 8, -104, 456, -952, 1040, -576, 128, -1, 43, -363, 1289, -2360, 2352, -1216, 256, -10, 190, -1182, 3530, -5760, 5280, -2560, 512

Offset

6,2

Comments

If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).

Let n>=6 and k be of the same parity. Consider a set X consisting of (n+k)/2-6 blocks of the size 2 and an additional block of the size 6, then (-1)^((n-k)/2)a(n,k) is the number of n-6-subsets of X intersecting each block of the size 2.

Links

Michael De Vlieger, Table of n, a(n) for n = 6..10185 (rows 6 <= n <= 200, flattened).

Milan Janjic, Two enumerative functions.

M. Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2.

Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Formula

If n>=6 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-6, i)*binomial(n+k-6-2*i, n-6), i=0..(n+k)/2-6) and a(n,k)=0 if n and k are of different parity.

Example

Rows are (1),(-6,2),(15,-13,4),... since P_{6,6}=x^6, P_{7,6}=-6x^5+2x^7, P_{8,6}=15x^4-13x^6+4x^8,...

Maple

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-6, i)*binomial(n+k-6-2*i, n-6), i=0..(n+k)/2-6); end if;

Mathematica

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 6, i] Binomial[n + k - 6 - 2 i, n - 6], {i, 0, (n + k)/2 - 6}], {n, 6, 15}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

Crossrefs

Cf. A008310, A053117.

Sequence in context: A348060 A082155 A128225 * A376729 A228692 A368521

Adjacent sequences: A136395 A136396 A136397 * A136399 A136400 A136401

Keyword

sign,tabf

Author

Milan Janjic, Mar 30 2008, revised Apr 05 2008

Status

approved