A121462 Triangle read by rows: T(n,k) is the number of nondecreasing Dyck paths of semilength n, having pyramid weight k (1 <= k <= n).
1, 0, 2, 0, 1, 4, 0, 1, 4, 8, 0, 1, 5, 12, 16, 0, 1, 6, 18, 32, 32, 0, 1, 7, 25, 56, 80, 64, 0, 1, 8, 33, 88, 160, 192, 128, 0, 1, 9, 42, 129, 280, 432, 448, 256, 0, 1, 10, 52, 180, 450, 832, 1120, 1024, 512, 0, 1, 11, 63, 242, 681, 1452, 2352, 2816, 2304, 1024, 0, 1, 12, 75, 316
Offset: 1
Examples
T(4,3)=4 because we have (UD)U(UD)(UD)D, U(UD)(UD)(UD)D, U(UD)(UUDD)D and U(UUDD)(UD)D, where U=(1,1) and D=(1,-1) (the maximal pyramids are shown between parentheses). Triangle starts: 1; 0, 2; 0, 1, 4; 0, 1, 4, 8; 0, 1, 5, 12, 16; 0, 1, 6, 18, 32, 32;
Links
- Elena Barcucci, Alberto del Lungo, S. Fezzi and Renzo Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
- Nicolas Bělohoubek and Antonín Slavík, L-Tetromino Tilings and Two-Color Integer Compositions, Univ. Karlova (Czechia, 2025). See p. 10.
- Alain Denise and Rodica Simion, Two combinatorial statistics on Dyck paths, Discrete Math., 137, 1995, 155-176.
- Emeric Deutsch and Helmut Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.
Programs
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Maple
T:=proc(n,k) if n=1 and k=1 then 1 elif k=1 then 0 elif k<=n then sum(binomial(k-1,j)*binomial(n-k-1+j,j-1),j=0..k-1) else 0 fi end: for n from 1 to 13 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
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Mathematica
u[1, x_] := 1; v[1, x_] := 1; z = 16; u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x]; v[n_, x_] := x*u[n - 1, x] + (x + 1) v[n - 1, x]; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n, 1, z}]; TableForm[cu] Flatten[%] (* A121462 *) Table[Expand[v[n, x]], {n, 1, z}] cv = Table[CoefficientList[v[n, x], x], {n, 1, z}]; TableForm[cv] Flatten[%] (* A208341 *) (* Clark Kimberling, Mar 11 2012 *)
Formula
T(n,k) = Sum_{j=0..k-1} binomial(k-1,j)*binomial(n-k-1+j,j-1) for 2 <= k <= n; T(1,1)=1; T(n,1)=0 for n >= 2.
G.f.: G = G(t,z) = tz(1-z)/(1-2tz-z+tz^2).
T(n+1,k+1) = A062110(n,k)*2^(2*k-n). - Philippe Deléham, Aug 01 2006
Comments