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A357081 Leader at step n of the THROWBACK procedure (see definition in comments).

%I A357081 #26 Mar 24 2024 11:03:05
%S A357081 3,4,5,6,3,7,4,8,3,5,9,4,3,6,10,5,3,4,7,11,3,6,4,5,3,8,12,4,3,7,5,6,3,
%T A357081 4,9,13,3,5,4,8,3,6,7,4,3,5,10,14,3,4,6,5,3,9,4,7,3,8,5,4,3,6,11,15,3,
%U A357081 4,5,7,3,6,4,10,3,5,8,4,3,9,6,5,3,4,7,12,3,16,4,5,3,6,8,4,3,7,5,11,3,4,6,9
%N A357081 Leader at step n of the THROWBACK procedure (see definition in comments).
%C A357081 The THROWBACK procedure: Start with the infinite sequence of natural numbers beginning with 3, that is 3, 4, 5, 6, 7, 8, ... The first number in the sequence at any step of the procedure is called the "leader". At each step, the leader is moved back in the sequence the number of places equal to its value.
%C A357081 It is conjectured that every number (n >= 3) appears an infinite number of times in this sequence.
%C A357081 The indices of records, ignoring the initial 3, appear to match A155167.
%C A357081 Every fourth term is 3. Values k > 3 occur at nonconstant intervals and the sequence of intervals for each k appears to be cyclic with a period of 3^(k-3). Ignoring the last value, the first 3^(k-3)-1 values of any of these cycles of intervals appear to be a palindrome. E.g., a(n)=5 for n=2,9,15,23,30,37,45,51,58,66,... The intervals between the 5's appear to repeat the pattern 7,6,8,7,7,8,6,7,8 and 7,6,8,7,7,8,6,7 is a palindrome.
%C A357081 Appears to be A087165 with every term incremented by 2. If so, then the recurrence a(n)=3 when n == 0 (mod 4), otherwise a(n) = a(n - ceiling(n/4)) + 1 holds. Also appears to be A087165 with every 1 and every 2 removed. See the comment by _Benoit Cloitre_ on A087165 for another possible way to construct this sequence.
%C A357081 If every 3 is removed then the result appears to be the original sequence with every term incremented by 1.
%C A357081 If the THROWBACK procedure is performed on all natural numbers including 1 and 2, then the sequence of leaders appears to be A001511. Other initial values appear to produce similar patterns to this sequence.
%H A357081 Popular Computing (Calabasas, CA), <a href="/A155167/a155167.pdf">Coding Fun: Rearranging All The Numbers</a>, Annotated and scanned copy of pages PC55-2, PC55-3, and PC55-1(cover) of Vol. 5 (No. 55, Oct 1977).
%H A357081 Anthony M. Kozar Jr., <a href="http://www.anthonykozar.net/posts/2022/09/05/throwback-all-of-your-number-sequences/">THROWBACK All of Your Number Sequences</a>
%F A357081 a(n) = A087165(n+1) + 2 (conjectured).
%e A357081 Before the first step, 3 is the leader, so a(0) = 3. In the first step, 3 is moved back 3 places giving the new sequence 4, 5, 6, 3, 7, 8, ..., so a(1) = 4.
%o A357081 (Python)
%o A357081 from collections import deque
%o A357081 from itertools import count, islice
%o A357081 def tgen(): yield from count(3) # generator of sequence to throwback
%o A357081 def agen(): # generator of terms
%o A357081     g = tgen()
%o A357081     a = deque([next(g)])
%o A357081     while True:
%o A357081         leader = a.popleft()
%o A357081         yield leader
%o A357081         while leader > len(a): a.append(next(g))
%o A357081         a.insert(leader, leader)
%o A357081 print(list(islice(agen(), 100))) # _Michael S. Branicky_, Sep 11 2022
%Y A357081 Cf. A087165, A155167.
%Y A357081 Cf. A354223, A355080. (Other variants of the THROWBACK procedure).
%K A357081 nonn,easy
%O A357081 0,1
%A A357081 _Anthony M. Kozar Jr._, Sep 08 2022