OFFSET
3,2
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>=3 and k be of the same parity. Consider a set X consisting of (n+k)/2-3 blocks of the size 2 and an additional block of the size 3, then (-1)^((n-k)/2)a(n,k) is the number of n-3-subsets of X intersecting each block of the size 2.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 3..10197 (rows 3 <= n <= 200, flattened).
Milan Janjic, Two enumerative functions.
Milan 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>=3 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-3, i)*binomial(n+k-3-2*i, n-3), i=0..(n+k)/2-3) and a(n,k)=0 if n and k are of different parity.
EXAMPLE
Rows are (1),(-3,2),(3,-7,4),(-1,9,-16,8),(-5,25,-36,16),...
since P_{3,3}=x^3, P_{4,3}=-3x^2+2x^4, P_{5,3}=3x-7x^3+4x^5,...
MAPLE
if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-3, i)*binomial(n+k-3-2*i, n-3), i=0..(n+k)/2-3); end if;
MATHEMATICA
DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2)*Sum[(-1)^i*Binomial[(n + k)/2 - 3, i] Binomial[n + k - 3 - 2 i, n - 3], {i, 0, (n + k)/2 - 3}], {n, 3, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Milan Janjic, Mar 30 2008, revised Apr 05 2008
STATUS
approved