|
Search: id:A136390
|
|
|
| A136390 |
|
Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,4}(x) with 0 omitted (exponents in increasing order). |
|
+0
1
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
4,2
|
|
|
COMMENT
|
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
|
Milan Janjic, Two enumerative functions.
|
|
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;
|
|
CROSSREFS
|
Cf. A008310, A053117.
Adjacent sequences: A136387 A136388 A136389 this_sequence A136391 A136392 A136393
Sequence in context: A077157 A114478 A134239 this_sequence A019610 A058613 A053227
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
Milan R. Janjic (agnus(AT)blic.net), Mar 30 2008, revised Apr 05 2008
|
|
|
Search completed in 0.002 seconds
|
