A379250 a(1)=1; thereafter, a(n) is the number of coincidences between the sequence thus far and its terms rearranged in descending order.

1, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 4, 5, 6, 7, 8, 7, 8, 7, 8, 7, 8, 7, 6, 5, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 6, 6, 8, 8, 10, 10, 11, 12, 12, 13, 12, 12, 11, 12, 12, 11, 12, 12, 12, 12, 12, 12, 11, 11, 10, 11, 9, 9, 7

Offset: 1

Author

Neal Gersh Tolunsky, Dec 17 2024

Keywords

• nonn

Comments

Equivalently, this is the number of coincidences between the reverse of the sequence and its terms rearranged in ascending order.

Examples

			To find a(8), we compare the first 7 terms of the sequence with the same terms arranged in descending order:
  1, 1, 2, 1, 2, 1, 2
  2, 2, 2, 1, 1, 1, 1
        ^  ^     ^
We find three coincidences, so a(8) = 3.
		

Links

• Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
• Pontus von Brömssen, Plot of n, a(n) for n = 1..100000.

Crossrefs

Cf. A276638, A379265, A379266.

Programs

Mathematica

Nest[Append[#,Count[#-Reverse[Sort[#]],0]]&,{1},79] (* James C. McMahon, Dec 21 2024 *)

Python

from bisect import insort
from itertools import islice
def agen(): # generator of terms
    a, d, an = [], [], 1
    while True:
        a.append(an)
        insort(d, an, key=lambda x: -x)
        yield an
        an = sum(1 for x, y in zip(a, d) if x == y)
print(list(islice(agen(), 80))) # Michael S. Branicky, Dec 21 2024