A274763 Number of linear extensions of the one-level grid poset G[(1^n), (1^(n-1)), (0^(n-1))].

1, 10, 215, 7200, 328090, 18914190, 1318595475, 107813147200, 10112867995550, 1070215246700100, 126122386636230950, 16378717184245443000, 2323753119238888045500, 357594668486650175355750, 59323244552378848484536875, 10553747415214416889115286000, 2004246729406751177924041663750, 404685181230584369889138573637500, 86569650968075614116679243211951250, 19558042902565983702641321883519060000

Offset: 1

Author

Ran Pan, Jul 05 2016

Keywords

• nonn

Comments

The definition of a one-level grid poset can be found in the Pan links. The number of linear extensions of the one-level grid poset G[(0^n), (0^(n-1)), (0^(n-1))] is given by Catalan number A000108(n) and the number of linear extensions of the one-level grid poset G[(1^n), (0^(n-1)), (0^(n-1))] is given by A274644(n).

Links

• Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
• Ran Pan, Problem 1, Project P.
• Ran Pan, Algorithmic Solution to Problem 1 (and linear extensions of general one-level grid-like posets), Project P.

Crossrefs

Cf. A000108, A274644.

Programs

Maple

N := 100;
ff[1] := y-x;
for n from 1 to N-1 do
   ff[n+1] := simplify((y-x)*int(int((x-u)*subs(x=u,y=v,ff[n]),v=u..y),u=0..x));
end:
for n from 1 to N do
   a[n] := factorial(4*n-1)*int(int(ff[n],x=0..y),y=0..1);
end:
seq(a[n],n=1..10);
# Michael Wallner, Feb 14 2024

Formula

a(n) = (4*n-1)!*Integral_{y=0..1} Integral_{x=0..y} f_{n}(x,y) dx dy where f_{n+1}(x,y) = (y-x)*Integral_{u=0..x} Integral_{v=u..y} (x-u)*f_{n}(u,v) dv du for n>=1 and f_{1}(x,y) = y-x (Derived using the density method; see [Banderier, Wallner 2021]). - Michael Wallner, Feb 14 2024

Extensions

Corrected and extended by Michael Wallner, Feb 14 2024