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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)).
7
2, 48, 15360, 65601536, 3737426853888, 2839095978202497024, 28748176693620694822420480, 3879520049632381491007256002560000, 6976271067658190025590579601863413334016000, 167148731069381900203656839566190759098759848866611200
OFFSET
1,1
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.
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
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 *)
CROSSREFS
Sequence in context: A196448 A053290 A056989 * A230886 A368132 A090770
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Mar 31 2022
STATUS
approved