

A174864


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


5




OFFSET

1,3


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 n2. 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 + ...


FORMULA

a(n+1) = [Sum_{i=1..n}{a(i)}]^2, with a(1)=1.  Paolo P. Lava, Apr 23 2010
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(n1)  A000058(n2), n>=2.  Ivan N. Ianakiev, Oct 09 2013


MAPLE

P:=proc(i) local a, s, n; a:=1; print(1); s:=1; for n from 0 by 1 to i do a:=s^2; print(a); s:=s+a; od; end: P(10); # Paolo P. Lava, Apr 23 2010


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 *)


PROG

(PARI) a=vector(10); a[1]=a[2]=1; for(n=3, #a, a[n]=a[n1]*(sqrtint(a[n1])+1)^2); a


CROSSREFS

Cf. A000058, A007018.
Sequence in context: A143764 A152287 A086857 * A175493 A179870 A001152
Adjacent sequences: A174861 A174862 A174863 * A174865 A174866 A174867


KEYWORD

easy,nonn


AUTHOR

Giovanni Teofilatto, Mar 31 2010


STATUS

approved



