

A047836


"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.


11



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

1,1


COMMENTS

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 nth 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


LINKS



FORMULA

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


EXAMPLE

Starting at 24 we divide by 3, 2, then 2, reaching 2.


MATHEMATICA



PROG

(Haskell)
import Data.List.Ordered (union)
a047836 n = a047836_list !! (n1)
a047836_list = f [2] where
f (x:xs) = x : f (xs `union` map (x *) [2..x])


CROSSREFS



KEYWORD

nonn,nice,easy


AUTHOR

Thomas Kantke (bytes.more(AT)ibm.net)


EXTENSIONS



STATUS

approved



