A352207 a(1) = 2, a(2) = 48; for n >= 1, a(n+2) = c(n)*a(n+1)^2/a(n) where c(n) = 256*(2*n+3)^2*(4*n+1)*(4*n+3)^2*(4*n+5) / (27*(3*n+1)*(3*n+2)^2*(3*n+4)^2*(3*n+5)).

2, 48, 15360, 65601536, 3737426853888, 2839095978202497024, 28748176693620694822420480, 3879520049632381491007256002560000, 6976271067658190025590579601863413334016000, 167148731069381900203656839566190759098759848866611200

Offset: 1

Author

N. J. A. Sloane, Mar 31 2022

Keywords

• nonn

Comments

See Problem 8 of the Propp article. Propp conjectured that a(n) is 2-adically continuous as a function of n. The formula is due to David desJardins. It is only a conjecture that the formula is the solution to the counting problem.

Links

• James Propp, Tiling Problems, Old and New, Rutgers University Math Colloquium, March 30, 2022
• James Propp, Trimer covers in the triangular grid: twenty mostly open problems, arXiv:2206.06472 [math.CO], 2022.

Programs

Maple

c := n -> 256*(2*n+3)^2*(4*n+1)*(4*n+3)^2*(4*n+5) / (27*(3*n+1)*(3*n+2)^2*(3*n+4)^2*(3*n+5));
a:=proc(n) option remember;
if n=1 then 2 elif n=2 then 48
else c(n-2)*a(n-1)^2/a(n-2); fi; end;
[seq(a(n),n=1..10)];

Mathematica

RecurrenceTable[{a[1] == 2, a[2] == 48, a[n+2] == a[n+1]^2 / a[n] * 256*(2*n + 3)^2*(4*n + 1)*(4*n + 3)^2*(4*n + 5)/(27*(3*n + 1)*(3*n + 2)^2*(3*n + 4)^2*(3*n + 5))}, a[n], {n, 1, 10}] (* Vaclav Kotesovec, Mar 31 2022 *)

Formula

a(n) ~ exp(1/24) * 2^(9*n^2 - n + 1/6) / (sqrt(A) * n^(1/24) * 3^(9*n^2/2 - 1/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Mar 31 2022