A136397 Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,5}(x) with 0 omitted (exponents in increasing order).

1, -5, 2, 10, -11, 4, -10, 25, -24, 8, 5, -30, 61, -52, 16, -1, 20, -85, 146, -112, 32, -7, 70, -231, 344, -240, 64, 1, -34, 225, -608, 800, -512, 128, 9, -138, 681, -1560, 1840, -1088, 256, -1, 52, -501, 1970, -3920, 4192, -2304, 512

Offset

5,2

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>=5 and k be of the same parity. Consider a set X consisting of (n+k)/2-5 blocks of the size 2 and an additional block of the size 5, then (-1)^((n-k)/2)a(n,k) is the number of n-5-subsets of X intersecting each block of the size 2.

Links

Michael De Vlieger, Table of n, a(n) for n = 5..10190 (rows 5 <= 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>=5 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5) and a(n,k)=0 if n and k are of different parity.

Example

Rows are (1),(-5,2),(10,-11,4),... since P_{5,5}=x^5, P_{6,5}=-5x^4+2x^6, P_{7,5}=10x^3-11x^5+4x^7,...

Maple

if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5); end if;

Mathematica

DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 5, i] Binomial[n + k - 5 - 2 i, n - 5], {i, 0, (n + k)/2 - 5}], {n, 5, 14}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)

Crossrefs

Cf. A008310, A053117.

Sequence in context: A249368 A055682 A187875 * A065292 A090812 A319408

Adjacent sequences: A136394 A136395 A136396 * A136398 A136399 A136400

Keyword

sign,tabf

Author

Milan Janjic, Mar 30 2008, revised Apr 05 2008

Status

approved