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a(n) = lcm(a(n-1), n^2 + n + 1) for n > 1 with a(1) = 1.
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%I #13 Sep 22 2024 14:00:09

%S 1,7,91,273,8463,363909,6914271,504741783,504741783,18675445971,

%T 18675445971,2932045017447,178854746064267,37738351419560337,

%U 9094942692114041217,9094942692114041217,2792147406479010653619,136815222917471522027331

%N a(n) = lcm(a(n-1), n^2 + n + 1) for n > 1 with a(1) = 1.

%C Not the same as if a(n) = lcm(f(1),f(2),...,f(n))/3 with f(x)=x^2+x+1 and a(1)=f(1).

%H Harvey P. Dale, <a href="/A221132/b221132.txt">Table of n, a(n) for n = 1..382</a>

%t f[x_] := x^2 + x + 1; a[1] = 1; a[n_] := LCM[f[n], a[n - 1]]; Array[a, 18]

%t nxt[{n_,a_}]:={n+1,LCM[a,n^2+3n+3]}; NestList[nxt,{1,1},20][[;;,2]] (* _Harvey P. Dale_, Sep 22 2024 *)

%Y Cf. A193181.

%K nonn

%O 1,2

%A _Robert G. Wilson v_, Jan 02 2013