A343332 a(1) = 0; thereafter a(n+1) = floor((a(n)+y)/2), where y is the number of numbers m < n such that a(m) = a(n).

0, 0, 0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 2, 2, 2, 3, 2, 3, 3, 3, 4, 2, 4, 2, 4, 3, 4, 3, 5, 2, 5, 3, 5, 3, 6, 3, 6, 3, 7, 3, 7, 4, 4, 4, 5, 4, 5, 4, 6, 4, 6, 4, 7, 4, 7, 5, 5, 5, 6, 5, 6, 5, 7, 5, 7, 6, 6, 6, 7, 6, 7, 7, 7, 8, 4, 8, 4, 8, 5, 8, 5, 8, 6, 8

Offset: 1

Author

Pontus von Brömssen, Apr 12 2021

Keywords

• nonn
• look

Comments

Variant of A340488, with XOR(a,y) replaced by floor((a+y)/2).

Every number appears, and their first occurrences are in increasing order.

Apparently, a(n) <= A343333(n) for all n.

Links

• Pontus von Brömssen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A340488, A343333, A343334.

Programs

Python

def A343332_list(n_max):
  a=0
  a_list=[0]
  count=[]
  for i in range(n_max-1):
    if a==len(count): count.append(0)
    else: count[a]+=1
    a=(a+count[a])//2
    a_list.append(a)
  return a_list