A115590 a(0) = 0; a(n) = (1+a(n-1))^3 for n > 0.

0, 1, 8, 729, 389017000, 58871587162270593034051001, 204040901322752673844230437877671861543858084850895762746141813554591014612008

Offset

0,3

Comments

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:

R...R...R...R......R.......R...R.......R

.../....|....\..../.\...../|...|\...../|\

..o.....o.....o..o...o...o.o...o.o...o.o.o

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

Links

Table of n, a(n) for n=0..6.

Geir Agnarsson, Elie Alhajjar, and Aleyah Dawkins, On locally finite ordered rooted trees and their rooted subtrees, arXiv:2312.11379 [math.CO], 2023.

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, alternative link.

Index entries for sequences related to rooted trees

Formula

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...

Mathematica

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

Crossrefs

Cf. A000578, A004019, A368326, A368327.

Sequence in context: A268106 A023813 A068895 * A134923 A144230 A309377

Adjacent sequences: A115587 A115588 A115589 * A115591 A115592 A115593

Keyword

easy,nonn

Author

Paolo Bonzini, Mar 15 2006; corrected Apr 06 2006 and Jan 19 2007

Extensions

Name edited by Michael De Vlieger, Dec 21 2023

Status

approved