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A174864 a(1) = 1, a(n) = square of the sum of previous terms.

%I A174864 #47 Mar 17 2024 02:15:32
%S A174864 1,1,4,36,1764,3261636,10650053687364,113423713055411194304049636,
%T A174864 12864938683278671740537145884937248491231415124195364
%N A174864 a(1) = 1, a(n) = square of the sum of previous terms.
%C A174864 a(n) divides a(n+1) with result a square.
%C A174864 Except for first two terms, partial sum k of a(n) is divisible by 6.
%C A174864 These numbers are divisible by their digital roots, which makes the sequence a subsequence of A064807. - _Ivan N. Ianakiev_, Oct 09 2013
%C A174864 a(n) is the number of binary trees in which the nodes are labeled by nonnegative integer heights, the left and right children of each node (if present) must have smaller height, and the root has height n-2. For instance, there are four trees with root height 1: the left and right children of the root may or may not be present, and must each be at height 0 if present. - _David Eppstein_, Oct 25 2018
%H A174864 Vincenzo Librandi, <a href="/A174864/b174864.txt">Table of n, a(n) for n = 1..13</a>
%H A174864 <a href="/index/Aa#AHSL">Index entries for sequences of form a(n+1)=a(n)^2 + ...</a>
%F A174864 a(n+1) = (a(n) + sqrt(a(n)))^2 = a(n) * (sqrt(a(n)) + 1)^2 for n > 1. - _Charles R Greathouse IV_, Jun 30 2011
%F A174864 a(n) = A000058(n-1) - A000058(n-2), n>=2. - _Ivan N. Ianakiev_, Oct 09 2013
%F A174864 a(n+2) + 1 = ( A000058(n+1)^2+1 ) / ( A000058(n)^2+1 ). - _Bill Gosper_, _Hugo Pfoertner_, May 09 2021
%t A174864 t = {1}; Do[AppendTo[t, Total[t]^2], {n, 9}]; t (* _Vladimir Joseph Stephan Orlovsky_, Feb 24 2012 *)
%t A174864 Join[{1},FoldList[(#+Sqrt[#])^2&,1,Range[7]]] (* _Ivan N. Ianakiev_, May 08 2015 *)
%o A174864 (PARI) a=vector(10);a[1]=a[2]=1;for(n=3,#a,a[n]=a[n-1]*(sqrtint(a[n-1])+1)^2);a
%Y A174864 Cf. A000058, A007018.
%K A174864 easy,nonn
%O A174864 1,3
%A A174864 _Giovanni Teofilatto_, Mar 31 2010