%I
%S 1,2,2,2,3,3,3,3,4,5,6,5,5,5,5,6,7,6,7,7,8,9,10,8,9,10,11,11,12,12,11,
%T 10,11,12,13,13,14,14,14,14,15,13,14,14,15,16,17,15,16,17,18,19,20,19,
%U 20,19,19,20,21,19,20,20,20,21,22,23,24,24,25,26,27
%N Number of disjoint trees that appear while iterating the sum of divisors function up to n.
%C A tree like (2, 3, 4, 7, 8) contains all numbers below 8 such that iterating the sum of divisors function to any of them, while staying below 8, will lead to 8.
%C Inspired by the article in link, where a (p1, p2, p3)tree is defined with p1 the smallest number in the tree, and p2, p3, such that all sequences {sigma^i(n)} (iterations of sigma), with p1<=n<=p2 and sigma^i(n)<p3 have nonempty intersection with {sigma^i(p1)}. For instance, 21 (p1, 200, 10^10)trees and 64 (p1, 1000, 10^100)trees were found.
%H G. L. Cohen and H. J. J. te Riele, <a href="http://projecteuclid.org/euclid.em/1047565640">Iterating the sumofdivisors function</a>, Experimental Mathematics, 5 (1996), pp. 93100.
%F For n>1 a(n) = a(n1) + 1  A054973(n), a(1) = 1.  _Michel Marcus_, Oct 22 2013
%e For n=23, there are 10 disjoint trees: (1), (2, 3, 4, 7, 8, 15), (5, 6, 11, 12), (9, 13, 14), (10, 17, 18), (16), (19, 20), (21), (22), (23). With the arrival of 24, 3 trees are united, that is those that contain 15, 14 and 23, so that there are now 8 trees. Here are some further values : a(100)=33, a(500)=167, a(1000)=333.
%o (PARI) lista(n) = {vecb = readvec("b000203.log"); tree = vector(n, x, []); for (i=1, n, tree[i] = concat(tree[i], i); for (j=1, i, for (k=1, length(tree[j]), if (vecb[tree[j][k]] == i, tree[j] = concat(tree[j], i)););); myset = Set(); for (j=1, i, lenj = length(tree[j]); if (lenj > 0, myset = setunion (myset, Set(tree[j][lenj])));); print1(length(myset), ", "););} \\ with b000203.log being the second column of b000203.txt. _Michel Marcus_, Mar 12 2013
%Y Cf. A000203.
%K nonn
%O 1,2
%A _Michel Marcus_, Mar 12 2013
