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

1, 1, 4, 36, 1764, 3261636, 10650053687364, 113423713055411194304049636, 12864938683278671740537145884937248491231415124195364

Offset: 1

Author

Giovanni Teofilatto, Mar 31 2010

Keywords

• easy
• nonn

Comments

a(n) divides a(n+1) with result a square.

Except for first two terms, partial sum k of a(n) is divisible by 6.

These numbers are divisible by their digital roots, which makes the sequence a subsequence of A064807. - Ivan N. Ianakiev, Oct 09 2013

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..13
• Index entries for sequences of form a(n+1)=a(n)^2 + ...

Crossrefs

Cf. A000058, A007018.

Programs

Mathematica

t = {1}; Do[AppendTo[t, Total[t]^2], {n, 9}]; t (* Vladimir Joseph Stephan Orlovsky, Feb 24 2012 *)
Join[{1},FoldList[(#+Sqrt[#])^2&,1,Range[7]]] (* Ivan N. Ianakiev, May 08 2015 *)

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

Formula

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

a(n) = A000058(n-1) - A000058(n-2), n>=2. - Ivan N. Ianakiev, Oct 09 2013

a(n+2) + 1 = ( A000058(n+1)^2+1 ) / ( A000058(n)^2+1 ). - Bill Gosper, Hugo Pfoertner, May 09 2021