A136390 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,4}(x) with 0 omitted (exponents in increasing order).

1, -4, 2, 6, -9, 4, -4, 16, -20, 8, 1, -14, 41, -44, 16, 6, -44, 102, -96, 32, -1, 26, -129, 248, -208, 64, -8, 96, -360, 592, -448, 128, 1, -42, 321, -968, 1392, -960, 256, 10, -180, 1002, -2528, 3232, -2048, 512, -1, 62, -681, 2972, -6448, 7424, -4352, 1024

Offset

4,2

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>=4 and k be of the same parity. Consider a set X consisting of (n+k)/2-4 blocks of the size 2 and an additional block of the size 4, then (-1)^((n-k)/2)a(n,k) is the number of n-4-subsets of X intersecting each block of the size 2.

Links

Michael De Vlieger, Table of n, a(n) for n = 4..10194 (rows 4 <= 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>=4 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-4, i)*binomial(n+k-4-2*i, n-4), i=0..(n+k)/2-4) and a(n,k)=0 if n and k are of different parity.

Example

Rows are (1),(-4,2),(6,-9,4),(-4,16,-20,8),... since P_{4,4}=x^4, P_{5,4}=-4x^3+2x^5, P_{6,4}=6x^2-9x^4+4x^6,...

Maple

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-4, i)*binomial(n+k-4-2*i, n-4), i=0..(n+k)/2-4); end if;

Mathematica

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 4, i] Binomial[n + k - 4 - 2 i, n - 4], {i, 0, (n + k)/2 - 4}], {n, 4, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

Crossrefs

Cf. A008310, A053117.

Sequence in context: A302659 A363705 A134239 * A019610 A344792 A058613

Adjacent sequences: A136387 A136388 A136389 * A136391 A136392 A136393

Keyword

sign,tabf

Author

Milan Janjic, Mar 30 2008, revised Apr 05 2008

Status

approved