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a(0) = 0; a(n) = (1+a(n-1))^3 for n > 0.
3

%I #29 Dec 21 2023 18:51:38

%S 0,1,8,729,389017000,58871587162270593034051001,

%T 204040901322752673844230437877671861543858084850895762746141813554591014612008

%N a(0) = 0; a(n) = (1+a(n-1))^3 for n > 0.

%C Take the rooted ternary tree of depth n, with (3^(n+1) - 1) / 2 labeled nodes. Let the number of rooted subtrees be a(n). For example, for n = 1 the a(2) = 8 subtrees are:

%C R...R...R...R......R.......R...R.......R

%C .../....|....\..../.\...../|...|\...../|\

%C ..o.....o.....o..o...o...o.o...o.o...o.o.o

%C Then a(n+1) = (1+a(n))^3.

%H Geir Agnarsson, Elie Alhajjar, and Aleyah Dawkins, <a href="https://arxiv.org/abs/2312.11379">On locally finite ordered rooted trees and their rooted subtrees</a>, arXiv:2312.11379 [math.CO], 2023.

%H A. V. Aho and N. J. A. Sloane, <a href="https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, <a href="http://neilsloane.com/doc/doubly.html">alternative link</a>.

%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>

%F As for A004019, it follows from Aho and Sloane that there is a constant c such that a(n) is the nearest integer to c^(3^n). In fact a(n) = nearest integer to b^(3^n) - 1 where b = 2.0804006677503193521177452323719035237099784935372250879749088464344434056773788...

%t {0}~Join~RecurrenceTable[{a[n]==(a[n-1]+1)^3, a[0]==1},a,{n,0,8}] (* _Vaclav Kotesovec_, May 21 2015 *)

%Y Cf. A000578, A004019, A368326, A368327.

%K easy,nonn

%O 0,3

%A _Paolo Bonzini_, Mar 15 2006; corrected Apr 06 2006 and Jan 19 2007

%E Name edited by _Michael De Vlieger_, Dec 21 2023