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*(n-1)*4^(n-2) = A129532(n).
This sequence is mentioned in the Andrews-Savage-Wilf paper. - Omar E. Pol, Jan 30 2012
REFERENCES
G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
M. Bona, Combinatorics of Permutations, Chapman & Hall/CRC, Boca Raton, FL, 2004, pp. 57-61.
LINKS
G. E. Andrews, C. D. Savage and H. S. Wilf, Hypergeometric identities associated with statistics on words
Mark A. Shattuck and Carl G. Wagner, Parity Theorems for Statistics on Lattice Paths and Laguerre Configurations, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.1.
FORMULA
Generating polynomial of row n is Sum_{a=0..n} Sum_{b=0..n-a} Sum_{c=0..n-a-b} binomial[n; a,b,c,n-a-b-c], where binomial[n;a,b,c,d] (a+b+c+d=n) is a q-multinomial 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..n-1) 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, n-a-b-c), c=0..n-a-b), b=0..n-a), a=0..n))) od: for n from 0 to 8 do seq(coeff(G[n], q, j), j=0..ceil((3*n^2-4)/8)) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Apr 22 2007
STATUS
approved