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A047836
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"Nullwertzahlen" (or "inverse prime numbers"): n=p1*p2*p3*p4*p5*...*pk, where pi are primes with p1 <= p2 <= p3 <= p4 ...; then p1 = 2 and p1*p2*...*pi >= p(i+1) for all i < k.
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11
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2, 4, 8, 12, 16, 24, 32, 36, 40, 48, 56, 60, 64, 72, 80, 84, 96, 108, 112, 120, 128, 132, 144, 160, 168, 176, 180, 192, 200, 208, 216, 224, 240, 252, 256, 264, 280, 288, 300, 312, 320, 324, 336, 352, 360, 384, 392, 396, 400, 408, 416, 420, 432, 440, 448
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OFFSET
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1,1
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COMMENTS
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Start with n and reach 2 by repeatedly either dividing by d where d <= the square root or by adding or subtracting 1. The division steps are free, but adding or subtracting 1 costs 1 point. The "value" of n (A047988) is the smallest cost to reach 2. Sequence gives numbers with value 0.
a(n) is also the length of the largest Dyck path of the symmetric representation of sigma of the n-th number whose symmetric representation of sigma has only one part. For an illustration see A317305. (Cf. A237593.) - Omar E. Pol, Aug 25 2018
This sequence can be defined equivalently as the increasing terms of the set containing 2 and all the integers such that if n is in the set, then all m * n are in the set for all m <= n. - Giuseppe Melfi, Oct 21 2019
The subsequence giving the largest term with k prime factors (k >= 1) starts 2, 4, 12, 132, 17292, 298995972, ... . - Peter Munn, Jun 04 2020
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LINKS
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FORMULA
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The number of terms <= x is c*x/log(x) + O(x/(log(x))^2), where c = 0.612415..., and a(n) = C*n*log(n*log(n)) + O(n), where C = 1/c = 1.63287... This follows from the formula just above. - Andreas Weingartner, Jun 30 2021
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EXAMPLE
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Starting at 24 we divide by 3, 2, then 2, reaching 2.
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MATHEMATICA
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PROG
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(Haskell)
import Data.List.Ordered (union)
a047836 n = a047836_list !! (n-1)
a047836_list = f [2] where
f (x:xs) = x : f (xs `union` map (x *) [2..x])
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Thomas Kantke (bytes.more(AT)ibm.net)
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EXTENSIONS
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STATUS
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approved
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