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A322286
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Lexicographically earliest sequence of positive integers without 4 terms in a weakly increasing arithmetic progression.
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1
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1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 2, 2, 3, 1, 1, 1, 2, 1, 2, 2, 2, 3, 3, 3, 1, 1, 3, 1, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 3, 3, 2, 3, 2, 3, 3, 5, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 2, 3, 2, 2, 2, 3, 3, 1, 3, 3, 3, 5, 5, 4, 1, 1, 1, 3, 1, 2, 3, 1, 5, 3, 2, 6, 1, 3, 2, 1, 3, 2, 1, 1, 3, 3, 1, 1, 1
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OFFSET
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1,4
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COMMENTS
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This is a variation of A248641 (where we only exclude weakly increasing arithmetic progressions): they differ from the 101st term.
It is also a variation of A309890 where 3-term is replaced by 4-term.
The numbers n for which the n-th term is 1 are given by A005837.
There is no upper bound, because if there were an upper bound r then there must be s <= r such that the set of numbers n for which the n-th term is s has positive density and this contradicts Szemerédi's theorem.
Assuming Erdős's conjecture on arithmetic progressions, for a fixed positive integer r, the sum of the reciprocals of the numbers n for which the n-th term is r converges.
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LINKS
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PROG
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(SageMath)
cpdef FourFree(int n):
cdef int i, r, k, s, L1, L2, L3
cdef list L, Lb
cdef set b
L=[1, 1, 1]
for k in range(3, n):
b=set()
for i in range(k):
if 3*((k-i)/3)==k-i:
r=(k-i)/3
L1, L2, L3=L[i], L[i+r], L[i+2*r]
s=3*(L2-L1)+L1
if s>0 and L3==2*(L2-L1)+L1:
if L1<=L2:
b.add(s)
if 1 not in b:
L.append(1)
else:
Lb=list(b)
Lb.sort()
for t in Lb:
if t+1 not in b:
L.append(t+1)
break
return L
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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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