A129529 Triangle read by rows: T(n,k) is the number of ternary words of length n on {0,1,2} that have k inversions (n >= 0, k >= 0).
1, 3, 6, 3, 10, 8, 8, 1, 15, 15, 21, 18, 9, 3, 21, 24, 39, 45, 48, 30, 24, 9, 3, 28, 35, 62, 82, 107, 108, 101, 81, 62, 37, 17, 8, 1, 36, 48, 90, 129, 186, 222, 264, 252, 255, 219, 183, 126, 90, 48, 27, 9, 3, 45, 63, 123, 186, 285, 372, 492, 561, 624, 648, 651, 597, 537, 435, 336, 249, 165, 99, 54, 27, 9, 3
Offset: 0
Examples
T(3,2) = 8 because we have 100, 110, 120, 200, 201, 211, 220 and 221. Triangle starts: 1; 3; 6, 3; 10, 8, 8, 1; 15, 15, 21, 18, 9, 3; 21, 24, 39, 45, 48, 30, 24, 9, 3; ...
References
- M. Bona, Combinatorics of Permutations, Chapman & Hall/CRC, Boca Raton, FL, 2004, pp. 57-61.
- G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
Links
- Alois P. Heinz, Rows n = 0..50, flattened
- 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.
Programs
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Maple
for n from 0 to 40 do br[n]:=sum(q^i,i=0..n-1) od: for n from 0 to 40 do f[n]:=simplify(product(br[j],j=1..n)) od: mbr:=(n,a,b,c)->simplify(f[n]/f[a]/f[b]/f[c]): for n from 0 to 9 do G[n]:=sort(simplify(sum(sum(mbr(n,a,b,n-a-b),b=0..n-a),a=0..n))) od: for n from 0 to 9 do seq(coeff(G[n],q,j),j=0..floor(n^2/3)) od; # yields sequence in triangular form # second Maple program: b:= proc(n, l) option remember; `if`(n=0, 1, add(expand(b(n-1, `if`(j<3, subsop(j=l[j]+1, l), l)))*x^([0, l[1], l[1]+l[2]][j]), j=1..3)) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, [0$2])): seq(T(n), n=0..10); # Alois P. Heinz, Feb 12 2025
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Mathematica
b[n_, l_] := b[n, l] = If[n == 0, 1, Sum[Expand[b[n-1, If[j < 3, ReplacePart[l, j -> l[[j]]+1], l]]]*x^({0, l[[1]], l[[1]]+l[[2]]}[[j]]), {j, 1, 3}]]; T[n_] := With[{p = b[n, {0, 0}]}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 13 2025, after Alois P. Heinz *)
Formula
T(n,0) = (n+1)*(n+2)/2 = A000217(n+1).
Sum_{k>=0} k*T(n,k) = 3^(n-1)*n*(n-1)/2 = A129530(n).
Generating polynomial of row n is Sum_{i=0..n} Sum_{j=0..n-i} binomial[n; i,j,n-i-j], where binomial[n;a,b,c] (a+b+c=n) is a q-multinomial coefficient.
Sum_{k=0..floor(n^2/3)} (-1)^k * T(n,k) = A056449(n). - Alois P. Heinz, Feb 12 2025
Comments