A052287 -id:A052287 - cp's OEIS frontend

A047836 "Nullwertzahlen" (or "inverse prime numbers"): n=p1p2p3p4p5...pk, where pi are primes with p1 <= p2 <= p3 <= p4 ...; then p1 = 2 and p1p2...*pi >= p(i+1) for all i < k.

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

Offset: 1

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

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

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

T. D. Noe, Table of n, a(n) for n = 1..1000
Thomas Kantke, Das Spiel Minimum und die Zerlegung natürlicher Zahlen, Spektrum der Wissenschaft, No. 4, 1993, pp. 11-13.
Andreas Weingartner, Uniform distribution of alpha*n modulo one for a family of integer sequences, arXiv:2303.16819 [math.NT], 2023.

Cf. A047984, A047985, A047986, A047987, A047988, A052287, A237593, A317305.

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])
-- Reinhard Zumkeller, Jun 25 2015, Sep 28 2011

nMax = 100; A174973 = Select[Range[10*nMax], AllTrue[Rest[dd = Divisors[#]] / Most[dd], Function[r, r <= 2]]&]; a[n_] := 2*A174973[[n]]; Array[a, nMax] (* Jean-François Alcover, Nov 10 2016, after Reinhard Zumkeller *)

a(n) = 2 * A174973(n). - Reinhard Zumkeller, Sep 28 2011

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

More terms from David W. Wilson

A196149 Numbers whose divisors increase by a factor of 3 or less.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 24, 27, 28, 30, 32, 36, 40, 42, 44, 45, 48, 50, 54, 56, 60, 63, 64, 66, 70, 72, 75, 78, 80, 81, 84, 88, 90, 96, 99, 100, 102, 104, 105, 108, 110, 112, 117, 120, 126, 128, 130, 132, 135, 136, 140, 144, 147, 150

Offset: 1

Reinhard Zumkeller, Sep 28 2011

nonn

The polymath8 project led by Terry Tao refers to these numbers as "3-densely divisible". In general they say that n is y-densely divisible if its divisors increase by a factor of y or less, or equivalently, if for every R with 1 <= R <= n, there is a divisor in the interval [R/y,R]. They use this as a weakening of the condition that n be y-smooth. - David S. Metzler, Jul 02 2013

Let D(x) denote the number of such integers up to x. D(x) has order of magnitude x/log(x) (See Saias 1997). Moreover, we have D(x) = c*x/log(x) + O(x/(log(x))^2), where c = 2.05544... (See Weingartner 2015, 2019). As a result, a(n) = C*n*log(n*log(n)) + O(n), where C = 1/c = 0.486513... - Andreas Weingartner, Jun 25 2021

			14 is not a term because its divisors are 1,2,7,14, and the gap from 2 to 7 is more than a factor of 3. - _N. J. A. Sloane_, Aug 03 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Saias, Entiers à diviseurs denses 1, Journal of Number Theory, Vol. 62, No. 1 (1997), pp. 163-191.
Terence Tao, A Truncated Elementary Selberg Sieve of Pintz. (blog entry defining y-densely divisible)
Terence Tao et al., Polymath8 home page.
Andreas Weingartner, Practical numbers and the distribution of divisors, The Quarterly Journal of Mathematics, Vol. 66, No. 2 (2015), pp. 743-758; arXiv preprint, arXiv:1405.2585 [math.NT], 2014-2015.
Andreas Weingartner, On the constant factor in several related asymptotic estimates, Math. Comp., Vol. 88, No. 318 (2019), pp. 1883-1902; arXiv preprint, arXiv:1705.06349 [math.NT], 2017-2018.

A174973 is a subsequence.

Cf. A027750.

a196149 n = a196149_list !! (n-1)
a196149_list = filter f [1..] where
   f n = all (<= 0) $ zipWith (-) (tail divs) (map (* 3) divs)
                      where divs = a027750_row' n
-- Reinhard Zumkeller, Jun 25 2015, Sep 28 2011

dif3[n_]:=Max[#[[2]]/#[[1]]&/@Partition[Divisors[n],2,1]]<=3; Select[ Range[ 200],dif3] (* Harvey P. Dale, Jun 08 2015 *)

is(n)=my(d=divisors(n));for(i=2,#d,if(d[i]>3*d[i-1],return(0)));1 \\ Charles R Greathouse IV, Jul 06 2013

from sympy import divisors
def ok(n):
    d = divisors(n)
    return all(d[i]/d[i-1] <= 3 for i in range(1, len(d)))
print(list(filter(ok, range(1, 151)))) # Michael S. Branicky, Jun 25 2021

a(n) = A052287(n) / 3.

a(n) = C*n*log(n*log(n)) + O(n), where C = 0.486513… (See comments). - Andreas Weingartner, Jun 25 2021

cp's OEIS Frontend

A047836 "Nullwertzahlen" (or "inverse prime numbers"): n=p1p2p3p4p5...pk, where pi are primes with p1 <= p2 <= p3 <= p4 ...; then p1 = 2 and p1p2...*pi >= p(i+1) for all i < k.

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A196149 Numbers whose divisors increase by a factor of 3 or less.

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cp's OEIS Frontend

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.

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Haskell

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Extensions

A196149 Numbers whose divisors increase by a factor of 3 or less.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Python

Formula

A047836 "Nullwertzahlen" (or "inverse prime numbers"): n=p1p2p3p4p5...pk, where pi are primes with p1 <= p2 <= p3 <= p4 ...; then p1 = 2 and p1p2...*pi >= p(i+1) for all i < k.