A341846 a(1) = 0; thereafter, if a(n) has not appeared before, a(n+1) = number of existing terms which are greater than a(n); otherwise, a(n) = n-m where a(m) is the most recent copy of a(n).

0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 1, 4, 2, 7, 0, 6, 8, 0, 3, 6, 4, 9, 0, 5, 14, 0, 3, 8, 11, 1, 19, 0, 6, 13, 2, 22, 0, 5, 14, 14, 1, 11, 13, 9, 22, 9, 2, 12, 8, 21, 2, 4, 31, 0, 17, 5, 18, 5, 2, 8, 11, 19, 31, 10, 18, 8, 6, 34, 0, 15, 11, 10, 8, 7, 60, 0, 7

Offset: 1

Author

David James Sycamore, Feb 19 2021

Keywords

• nonn

Comments

The sequence is nontrivial only with a(1) = 0 or 1. a(1)=0 here for contrast with the Van Eck sequence A181391, with which data agrees up to a(12). For n > 1, a(n)=0 if and only if a(n-1) is a record novel term, whereas every non-record novel term is followed by a term > 0. Definition implies a(n) < n for all n. Open questions: Is it true that a(n) + a(n+1) < n for all n? (This is true for n <= 65000.) Do records ever arise from rule 1? Does every number appear? (If so, 0 appears infinitely many times.)

Examples

			a(2)=0 because a(1)=0 is a novel term, and there are 0 terms > 0.
a(3)=1 because a(2) has been seen before at a(1), and 2-1=1.
a(13)=1, because a(12)=5, a novel term with 1 earlier term (a(10)=6) greater than it.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10001
• Michael De Vlieger, Scatterplot of a(n) for 1 <= n <= 4096, color coded to show records in red, zeros in blue, terms instigated by novel a(n) in green, and records in the last-mentioned terms in orange. The least unused number is indicated by a yellow step function line.
• Michael De Vlieger, Plot of b(n) = a(n) + a(n+1) for 1 <= n <= 65536 showing a partial bifurcation resulting from conditions involving novel a(n+1) such that, for such conditions, b(n) > n/Pi.
• Michael De Vlieger, Logarithmic scatterplot of a(n) for 1 <= n <= 65536.

Crossrefs

Cf. A181391.

Programs

Mathematica

Block[{nn = 120, a = {0}, c}, Do[If[IntegerQ@ c[#], AppendTo[a, i - c[#] ]; Set[c[#], i], Set[c[#], i]; AppendTo[a, Count[Most@ a, ?(# > a[[-1]] &)]]] &[ a[[-1]] ], {i, nn}]; a] (* _Michael De Vlieger, Feb 21 2021 *)

Python

def aupton(terms):
  alst, an = [], 0
  for n in range(1, terms+1):
    if an not in alst: anp1 = sum(ai > an for ai in alst)
    else: anp1 = alst[::-1].index(an) + 1
    alst, an = alst + [an], anp1
  return alst
print(aupton(79)) # Michael S. Branicky, Feb 21 2021