A381003 Lexicographically earliest sequence with a(0) = 0 and a(n) = a(n + a(n)) - a(n - a(n)) > 0.

0, 1, 1, 2, 1, 3, 1, 4, 4, 1, 5, 6, 5, 1, 6, 8, 1, 9, 1, 10, 10, 1, 11, 12, 1, 13, 13, 1, 14, 11, 15, 1, 16, 17, 1, 18, 5, 7, 18, 14, 12, 6, 20, 3, 22, 23, 15, 24, 17, 1, 18, 3, 26, 27, 20, 2, 28, 29, 1, 30, 4, 16, 31, 1, 32, 18, 33, 10, 34, 1, 35, 36, 36, 2, 21, 38

Offset: 0

Author

Thomas Scheuerle, Feb 11 2025

Keywords

• nonn

Comments

A trivial but not lexicographically earliest solution for this problem is given by a(m) = ceiling(m/2) (A110654).

We assume a(<0) = 0 in this definition but conjecture that a(n) < n for n > 1.

If we iterate recursively k_{n+1} = a(k_{n})+k_{n} starting with k{0}=2, then we will observe a monotone increasing trajectory. An interesting consequence of the existence of this sequence is that if we start a trajectory at some k{m} > 2, such a trajectory would in the majority of cases ( or in all such cases? ) merge into the trajectory which is starting at 2.

Examples

			a(10) = 5 = a(10 + 5) - a(10 - 5) = 8 - 3.
		

Links

• Thomas Scheuerle, Table of n, a(n) for n = 0..10000
• Thomas Scheuerle, Graph of Sum_{k=0..n} a(k)/n over n, it looks almost like a straight line with slope 0.14.. . Is there any relation to a well known constant?
• Thomas Scheuerle, Scatter plot of the first 15000 values, colored by first root ancestor the colormap has period 500.
• Thomas Scheuerle, Scatter plot of the first 15000 values, colored by second root ancestor the colormap has period 500.
• Thomas Scheuerle, Scatter plot of the first 15000 values, colored by length of the ancestor chain for the first ancestors the colormap has period 15.
• Thomas Scheuerle, Scatter plot of the first 15000 values, colored by length of the ancestor chain for the second ancestors the colormap has period 5.

Crossrefs

Cf. A110654.

Formula

a(a(n) + n) >= a(n).