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a(1) = 4, a(n) = a(n-1)*(a(n-1) - 1)/2.
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%I #55 Feb 08 2023 09:39:57

%S 4,6,15,105,5460,14903070,111050740260915,

%T 6166133456248548335768188155,

%U 19010600900133834176644234577571914951562754277857057935

%N a(1) = 4, a(n) = a(n-1)*(a(n-1) - 1)/2.

%C The next two terms, a(10) and a(11), have 111 and 221 digits. - _Harvey P. Dale_, Jun 10 2011

%C Interpretation through plane geometry: Start with the a(n)-sided regular polygon, connect all the vertices to create a figure having a(n+1)=A000217(a(n)-1) edges. Repeat to obtain this sequence. - _T. D. Noe_, May 13 2016

%C Let y(1) = x1+x2+x3+x4, and define y(n+1) as the plethysm e2[y(n)], where e2 represents the second elementary symmetric function. Then a(n) is y(n) evaluated at x1=x2=x3=x4=1. - _Per W. Alexandersson_, Jun 06 2020

%C Each term is the number of coordinate planes in Euclidean space of the dimensionality of the previous term. - _Shel Kaphan_, Feb 06 2023

%H Vincenzo Librandi, <a href="/A086714/b086714.txt">Table of n, a(n) for n = 1..13</a>

%F Limit_{n->oo} a(n)^(1/2^n) = 1.280497808541657066685323460209089278782... (see A251794). - _Vaclav Kotesovec_, Feb 15 2014, updated Dec 09 2014

%F a(n) ~ 2 * A251794^(2^n). - _Vaclav Kotesovec_, Dec 09 2014

%F a(n+1) = binomial(a(n), 2). - _Shel Kaphan_, Feb 06 2023

%e a(2) = a(1)*(a(1)-1)/2 = 4*3/2 = 6.

%t RecurrenceTable[{a[1]==4,a[n]==(a[n-1](a[n-1]-1))/2},a[n],{n,10}] (* _Harvey P. Dale_, Jun 10 2011 *)

%o (PARI) v=vector(10,i,(i==1)*4); for(i=2,10,v[i]=v[i-1]*(v[i-1]-1)/2); v

%o (PARI) a086714(upto)={my(a217(n)=n*(n+1)/2,a=4);for(k=1,upto,print1(a,", ");a=a217(a-1))};

%o a086714(9) \\ _Hugo Pfoertner_, Sep 18 2021

%Y Cf. A251702, A251794, A006893.

%Y Cf. A000217, A007501, A013589, A050542, A050548, A050536, A050909.

%K nonn

%O 1,1

%A _Jon Perry_, Jul 29 2003