A001696 a(n) = a(n-1)*(1 + a(n-1) - a(n-2)), a(0) = 0, a(1) = 1.

0, 1, 2, 4, 12, 108, 10476, 108625644, 11798392680793836, 139202068568601568785946949658348, 19377215893777651167043206536157529523359277782016064519251404524

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Also, numbers remaining after the following sieving process: In step 1, keep all numbers of the set N={0,1,2,...}. In step 2, keep only every second number after a(2)=2: N'={0,1,2,4,6,8,10,...}. In step 3, keep every 4th of the numbers following a(3)=4, N"={0,1,2,4,12,20,28,...}. In step 4, keep every 12th of the numbers beyond a(4)=12: {0,1,2,4,12,108,204,...}. In step 5, keep every 108th of the numbers beyond a(5)=108: {0,1,2,4,12,108,10476,...}, and so on. The next "gap" a(n+1)-a(n) is always a(n) times the former gap, i.e., a(n+1)-a(n) = a(n)*(a(n)-a(n-1)). - M. F. Hasler, Oct 28 2010

Number of plane trees where the root has fewer than n children and the i-th child of any node has fewer than i children. - David Eppstein, Dec 18 2021

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• John Cerkan, Table of n, a(n) for n = 0..13
• A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
• A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
• Index entries for sequences of form a(n+1)=a(n)^2 + ...

Crossrefs

a(n)=A039941(2*n); first difference sequence of this sequence is A001697. - Michael Somos, May 19 2000

Programs

Haskell

a001696 n = a001696_list !! n
a001696_list = 0 : 1 : zipWith (-)
   (zipWith (+) a001696_list' $ map (^ 2) a001696_list')
   (zipWith (*) a001696_list a001696_list')
   where a001696_list' = tail a001696_list
-- Reinhard Zumkeller, Apr 29 2013

Mathematica

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n-1]*(1 + a[n-1] - a[n-2]); Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jul 02 2013 *)

PARI

a(n)=if(n<2,n>0,a(n-1)*(1+a(n-1)-a(n-2)))

Formula

a(n) ~ c^(2^n), where c = 1.15552822483840350150537253088299651035583896919522349372370013726451673646... . - Vaclav Kotesovec, May 21 2015