OFFSET
1,3
COMMENTS
a(n) < n (with exception a(1) = 1). Proof: Suppose a(s) = s+x, x >= 0, is the first occurrence of a(n) >= n. From here we branch into two possibilities: Possibility #1: a(s-2) < a(s-1), from which it follows that a(s) = a(s-1-a(s-1))+1 and therefore a(s-1-a(s-1)) = s-1+x is an earlier example of a(n) >= n. Possibility #2: a(s-2) >= a(s-1) and the terms can be expressed as a(s-1) = (s+x)/2 and a(s-2) = (s+x)/2+y with y >= 0. From this it follows that a(s-2-((s+x)/2+y))+1 = (s+x)/2, which when simplified reveals that a((s+x)/2-2-x-y) = (s+x)/2-1 is an earlier example of a(n) >= n. Both possibilities lead to a contradiction of the first statement, therefore we conclude that there is no occurrence of a(n) >= n (with exception a(1) = 1).
Some numbers never seem to appear in the sequence; the smallest of these are 328, 329, 331, 332, 333, 445, 667, 668, 669, ...
EXAMPLE
a(3) = 2*a(2) = 2 because a(2) <= a(1).
a(4) = a(3-a(3))+1 = 2 because a(3) > a(2).
MATHEMATICA
Nest[Append[#, If[Less @@ Take[#, -2], #[[Length@ # - #[[-1]] ]] + 1, 2 #[[-1]] ]] &, {1, 1}, 73] (* Michael De Vlieger, Dec 23 2019 *)
PROG
(Python)
a = [1, 1]
for n in range(1, 1000):
if(a[n] > a[n-1]):
a.append(a[n-a[n]]+1)
else:
a.append(2*a[n])
CROSSREFS
KEYWORD
nonn
AUTHOR
Rok Cestnik, Dec 21 2019
STATUS
approved