%I #22 Jun 09 2023 11:04:00
%S 1,20,34220,180318314012420,
%T 26383476911029432816173777932463879690054620
%N Let D(n) = n*(9*n^2-9*n+2)/2 then a(k+1) = D(a(k)) and a(0) = 1.
%H Michel Marcus, <a href="/A099187/b099187.txt">Table of n, a(n) for n = 0..6</a>
%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2003), 65-75.
%F Let D(n) = n*(9*n^2-9*n+2)/2 then a(k+1) = D(a(k)) and a(0) = 1.
%t Dod[n_]:= n*(9*n^2-9*n+2)/2;
%t a[n_]:= If[n==0, Dod[1], If[n==1, Dod[2], Dod[a[n-1]]]];
%t Table[a[n], {n, 0, 4}] (* _G. C. Greubel_, Mar 22 2019 *)
%o (PARI) dod(n) = n*(9*n^2-9*n+2)/2;
%o a(n) = if (n==0, 1, if (n==1, dod(2), dod(a(n-1)))); \\ _Michel Marcus_, Dec 14 2015
%Y Cf. A007501, A006566.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Nov 15 2004