

A129531


Triangle read by rows: T(n,k) is the number of 4ary words of length n on {0,1,2,3} having k inversions (n >= 0, k >= 0).


2



1, 4, 10, 6, 20, 20, 20, 4, 35, 45, 65, 60, 35, 15, 1, 56, 84, 144, 180, 200, 152, 120, 60, 24, 4, 84, 140, 266, 386, 526, 584, 590, 524, 424, 290, 164, 86, 26, 6, 120, 216, 440, 700, 1064, 1384, 1720, 1844, 1940, 1820, 1616, 1272, 956, 620, 380, 184, 80, 24, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

Row n has (apparently) ceiling((3n^2+4)/8) terms.
Row sums are equal to 4^n = A000302(n).
Alternating row sums are 4^(ceiling(n/2)).
T(n,0) = (n+1)*(n+2)(n+3)/6 = A000292(n+1).
Sum_{k>=0} k*T(n,k) = 3*n*(n1)*4^(n2) = A129532(n).
This sequence is mentioned in the AndrewsSavageWilf paper.  Omar E. Pol, Jan 30 2012


REFERENCES

G. E. Andrews, The Theory of Partitions, AddisonWesley, 1976.
M. Bona, Combinatorics of Permutations, Chapman & Hall/CRC, Boca Raton, FL, 2004, pp. 5761.


LINKS



FORMULA

Generating polynomial of row n is Sum_{a=0..n} Sum_{b=0..na} Sum_{c=0..nab} binomial[n; a,b,c,nabc], where binomial[n;a,b,c,d] (a+b+c+d=n) is a qmultinomial coefficient.


EXAMPLE

T(2,1)=6 because we have 10, 20, 30, 21, 31 and 32.
Triangle starts:
1;
4;
10, 6;
20, 20, 20, 4;
35, 45, 65, 60, 35, 15, 1;
56, 84, 144, 180, 200, 152, 120, 60, 24, 4;


MAPLE

for n from 0 to 12 do br[n]:=sum(q^i, i=0..n1) od: for n from 0 to 12 do f[n]:=simplify(product(br[j], j=1..n)) od: mbr:=(n, a, b, c, d)>simplify(f[n]/f[a]/f[b]/f[c]/f[d]): for n from 0 to 8 do G[n]:=sort(simplify(sum(sum(sum(mbr(n, a, b, c, nabc), c=0..nab), b=0..na), a=0..n))) od: for n from 0 to 8 do seq(coeff(G[n], q, j), j=0..ceil((3*n^24)/8)) od; # yields sequence in triangular form


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



