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A357081
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Leader at step n of the THROWBACK procedure (see definition in comments).
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0
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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, 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, 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
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OFFSET
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0,1
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COMMENTS
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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.
It is conjectured that every number (n >= 3) appears an infinite number of times in this sequence.
The indices of records, ignoring the initial 3, appear to match A155167.
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.
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.
If every 3 is removed then the result appears to be the original sequence with every term incremented by 1.
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.
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LINKS
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FORMULA
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a(n) = A087165(n+1) + 2 (conjectured).
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EXAMPLE
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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.
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PROG
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(Python)
from collections import deque
from itertools import count, islice
def tgen(): yield from count(3) # generator of sequence to throwback
def agen(): # generator of terms
g = tgen()
a = deque([next(g)])
while True:
leader = a.popleft()
yield leader
while leader > len(a): a.append(next(g))
a.insert(leader, leader)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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