

A136397


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


0



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
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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)/25 blocks of the size 2 and an additional block of the size 5, then (1)^((nk)/2)a(n,k) is the number of n5subsets of X intersecting each block of the size 2.


LINKS

Table of n, a(n) for n=5..53.
Milan Janjic, Two enumerative functions.


FORMULA

If n>=5 and k are of the same parity then a(n,k)= (1)^((nk)/2)*sum((1)^i*binomial((n+k)/25, i)*binomial(n+k52*i, n5), i=0..(n+k)/25) 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^311x^5+4x^7,...


MAPLE

if modp(nk, 2)=0 then a[n, k]:=(1)^((nk)/2)*sum((1)^i*binomial((n+k)/25, i)*binomial(n+k52*i, n5), i=0..(n+k)/25); end if;


CROSSREFS

Cf. A008310, A053117.
Sequence in context: A249368 A055682 A187875 * A065292 A090812 A050004
Adjacent sequences: A136394 A136395 A136396 * A136398 A136399 A136400


KEYWORD

sign,tabf


AUTHOR

Milan Janjic, Mar 30 2008, revised Apr 05 2008


STATUS

approved



