cp's OEIS Frontend

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)).

%I A352207 #36 Aug 24 2023 11:53:10
%S A352207 2,48,15360,65601536,3737426853888,2839095978202497024,
%T A352207 28748176693620694822420480,3879520049632381491007256002560000,
%U A352207 6976271067658190025590579601863413334016000,167148731069381900203656839566190759098759848866611200
%N 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)).
%C A352207 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.
%H A352207 James Propp, <a href="https://faculty.uml.edu//jpropp/rutgers22.pdf">Tiling Problems, Old and New</a>, Rutgers University Math Colloquium, March 30, 2022
%H A352207 James Propp, <a href="https://arxiv.org/abs/2206.06472">Trimer covers in the triangular grid: twenty mostly open problems</a>, arXiv:2206.06472 [math.CO], 2022.
%F A352207 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
%p A352207 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));
%p A352207 a:=proc(n) option remember;
%p A352207 if n=1 then 2 elif n=2 then 48
%p A352207 else c(n-2)*a(n-1)^2/a(n-2); fi; end;
%p A352207 [seq(a(n),n=1..10)];
%t A352207 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 *)
%K A352207 nonn
%O A352207 1,1
%A A352207 _N. J. A. Sloane_, Mar 31 2022