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A365399
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Length of the longest subsequence of 1, ..., n on which the number of divisors function tau A000005 is nondecreasing.
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12
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1, 2, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 15, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 23, 23, 24, 24, 24, 24, 25, 25, 25, 25, 26, 26, 27, 27, 27, 27
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OFFSET
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1,2
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COMMENTS
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The sequence was inspired by A365339.
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LINKS
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FORMULA
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a(n+1) - a(n) <= 1.
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EXAMPLE
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The terms of the subsequences of A000005 are marked by '*'. They start:
1*, 2, 2 , 3, 2, 4, 2, 4, ... -> a(1) = 1
1*, 2*, 2 , 3, 2, 4, 2, 4, ... -> a(2) = 2
1*, 2*, 2*, 3, 2, 4, 2, 4, ... -> a(3) = 3
1*, 2*, 2*, 3*, 2, 4, 2, 4, ... -> a(4) = 4
1*, 2*, 2*, 3*, 2, 4, 2, 4, ... -> a(5) = 4
1*, 2*, 2*, 3*, 2, 4*, 2, 4, ... -> a(6) = 5
1*, 2*, 2*, 3*, 2, 4*, 2, 4, ... -> a(7) = 5
1*, 2*, 2*, 3*, 2, 4*, 2, 4*, ... -> a(8) = 6
Example: a(2000000) = 450033.
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PROG
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(Julia)
# Computes the first N terms of the sequence using function tau from A000005.
function LLS_list(seq, N)
lst = zeros(Int64, N)
dyn = zeros(Int64, N)
for n in 1:N
p = seq(n)
nxt = dyn[p] + 1
while p <= N && dyn[p] < nxt
dyn[p] = nxt
p += 1
end
lst[n] = dyn[n]
end
return lst
end
A365399List(N) = LLS_list(tau, N)
println(A365399List(69))
(Python)
from bisect import bisect
from sympy import divisor_count
plist, qlist, c = tuple(divisor_count(i) for i in range(1, n+1)), [0]*(n+1), 0
for i in range(n):
qlist[a:=bisect(qlist, plist[i], lo=1, hi=c+1, key=lambda x:plist[x])]=i
c = max(c, a)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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