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A174973
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Numbers whose divisors increase by a factor of at most 2.
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53
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1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 66, 72, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 228, 234, 240, 252, 256
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OFFSET
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1,2
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COMMENTS
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That is, if the divisors of a number are listed in increasing order, the ratio of adjacent divisors is at most 2. The only odd number in this sequence is 1. Every term appears to be a practical number (A005153). The first practical number not here is 78.
Let p1^e1 * p2^e2 * ... * pr^er be the prime factorization of a number, with primes p1 < p2 < ... < pr and ek > 0. Then the number is in this sequence if and only if pk <= 2*Product_{j < k} p_j^e_j. This condition is similar to, but more restrictive than, the condition for practical numbers.
The polymath8 project led by Terry Tao refers to these numbers as "2-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
Is this the same as numbers k with the property that the symmetric representation of sigma(k) has only one part? If not, where is the first place these sequences differ? (cf. A237593). - Omar E. Pol, Mar 06 2014
Yes, the sequence so defined is the same as this sequence; see proof in the links. - Hartmut F. W. Hoft, Nov 26 2014
Saias (1997) called these terms "2-dense numbers" and proved that if N(x) is the number of terms not exceeding x, then there are two positive constants c_1 and c_2 such that c_1 * x/log_2(x) <= N(x) <= c_2 * x/log_2(x) for all x >= 2. - Amiram Eldar, Jul 23 2020
Weingartner (2015, 2019) showed that N(x) = c*x/log(x) + O(x/(log(x))^2), where c = 1.224830... As a result, a(n) = C*n*log(n*log(n)) + O(n), where C = 1/c = 0.816439... - Andreas Weingartner, Jun 22 2021
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LINKS
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FORMULA
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a(n) = C*n*log(n*log(n)) + O(n), where C = 0.816439... (see comments). - Andreas Weingartner, Jun 23 2021
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EXAMPLE
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The divisors of 12 are 1, 2, 3, 4, 6, 12. The ratios of adjacent divisors is 2, 3/2, 4/3, 3/2, and 2, which are all <= 2. Hence 12 is in this sequence.
The symmetric representation of sigma(6) = 12 in the first quadrant looks like this:
y
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. . . . . |_| . . x
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6 is in the sequence because the symmetric representation of sigma(6) = 12 has only one part. The 6th row of A237593 is [4, 1, 1, 1, 1, 4] and the 5th row of A237593 is [3, 2, 2, 3] therefore between both symmetric Dyck paths there is only one region (or part) of size 12.
70 is not in the sequence because the symmetric representation of sigma(70) = 144 has three parts. The 70th row of A237593 is [36, 12, 6, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 6, 12, 36] and the 69th row of A237593 is [35, 12, 7, 4, 2, 3, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 2, 4, 7, 12, 35] therefore between both symmetric Dyck paths there are three regions (or parts) of size [54, 36, 54]. (End)
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MAPLE
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a:= proc() option remember; local k; for k from 1+`if`(n=1, 0,
a(n-1)) while (l-> ormap(x-> x, [seq(l[i]>l[i-1]*2, i=2..
nops(l))]))(sort([(numtheory[divisors](k))[]])) do od; k
end:
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MATHEMATICA
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OK[n_] := Module[{d=Divisors[n]}, And@@(#<=2& /@ (Rest[d]/Most[d]))]; Select[Range[1000], OK]
dif2Q[n_]:=AllTrue[#[[2]]/#[[1]]&/@Partition[Divisors[n], 2, 1], #<=2&]; Select[Range[300], dif2Q] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 29 2020 *)
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PROG
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(Haskell)
a174973 n = a174973_list !! (n-1)
a174973_list = filter f [1..] where
f n = all (<= 0) $ zipWith (-) (tail divs) (map (* 2) divs)
where divs = a027750_row' n
(PARI) is(n)=my(d=divisors(n)); for(i=2, #d, if(d[i]>2*d[i-1], return(0))); 1 \\ Charles R Greathouse IV, Jul 06 2013
(Magma) [k:k in [1..260]|forall{i:i in [1..#Divisors(k)-1]|d[i+1]/d[i] le 2 where d is Divisors(k)}]; // Marius A. Burtea, Jan 09 2020
(Python)
from sympy import divisors
def ok(n):
d = divisors(n)
return all(d[i]/d[i-1] <= 2 for i in range(1, len(d)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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