A136389 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,3}(x) with 0 omitted (exponents in increasing order).
1, -3, 2, 3, -7, 4, -1, 9, -16, 8, -5, 25, -36, 16, 1, -19, 66, -80, 32, 7, -63, 168, -176, 64, -1, 33, -192, 416, -384, 128, -9, 129, -552, 1008, -832, 256, 1, -51, 450, -1520, 2400, -1792, 512, 11, -231, 1452, -4048, 5632, -3840, 1024, -1, 73, -912, 4424, -10496, 13056, -8192, 2048
Offset: 3
Examples
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,...
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.
Programs
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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;
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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 *)
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.
Comments