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

%I A379250 #26 Dec 22 2024 09:29:22
%S A379250 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,
%T A379250 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,
%U A379250 12,12,11,12,12,12,12,12,12,11,11,10,11,9,9,7
%N A379250 a(1)=1; thereafter, a(n) is the number of coincidences between the sequence thus far and its terms rearranged in descending order.
%C A379250 Equivalently, this is the number of coincidences between the reverse of the sequence and its terms rearranged in ascending order.
%H A379250 Neal Gersh Tolunsky, <a href="/A379250/b379250.txt">Table of n, a(n) for n = 1..10000</a>
%H A379250 Pontus von Brömssen, <a href="/A379250/a379250.png">Plot of n, a(n) for n = 1..100000</a>.
%e A379250 To find a(8), we compare the first 7 terms of the sequence with the same terms arranged in descending order:
%e A379250   1, 1, 2, 1, 2, 1, 2
%e A379250   2, 2, 2, 1, 1, 1, 1
%e A379250         ^  ^     ^
%e A379250 We find three coincidences, so a(8) = 3.
%t A379250 Nest[Append[#,Count[#-Reverse[Sort[#]],0]]&,{1},79] (* _James C. McMahon_, Dec 21 2024 *)
%o A379250 (Python)
%o A379250 from bisect import insort
%o A379250 from itertools import islice
%o A379250 def agen(): # generator of terms
%o A379250     a, d, an = [], [], 1
%o A379250     while True:
%o A379250         a.append(an)
%o A379250         insort(d, an, key=lambda x: -x)
%o A379250         yield an
%o A379250         an = sum(1 for x, y in zip(a, d) if x == y)
%o A379250 print(list(islice(agen(), 80))) # _Michael S. Branicky_, Dec 21 2024
%Y A379250 Cf. A276638, A379265, A379266.
%K A379250 nonn
%O A379250 1,3
%A A379250 _Neal Gersh Tolunsky_, Dec 17 2024