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A038547
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Least number with exactly n odd divisors.
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59
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1, 3, 9, 15, 81, 45, 729, 105, 225, 405, 59049, 315, 531441, 3645, 2025, 945, 43046721, 1575, 387420489, 2835, 18225, 295245, 31381059609, 3465, 50625, 2657205, 11025, 25515, 22876792454961, 14175, 205891132094649, 10395, 1476225, 215233605
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OFFSET
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1,2
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COMMENTS
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Also least odd number with exactly n divisors. - Lekraj Beedassy, Aug 30 2006
a(2n-1) = {1, 9, 81, 729, 225, 59049, ...} are the squares. A122842(n) = sqrt(a(2n-1)) = {1, 3, 9, 27, 15, 243, 729, 45, 6561, 19683, 135, 177147, 225, 105, 4782969, 14348907, 1215, ...}. - Alexander Adamchuk, Sep 13 2006
Also the least number k such that there are n partitions of k whose elements are consecutive integers. I.e., 1=1, 3=1+2=3, 9=2+3+4=4+5=9, 15=1+2+3+4+5=4+5+6=7+8=15, etc. - Robert G. Wilson v, Jun 02 2007
The politeness of an integer, A069283(n), is defined to be the number of its nontrivial runsum representations, and the sequence 3, 9, 15, 81, 45, 729, 105, ... represents the least integers to have a politeness of 1, 2, 3, 4, ... This is also the sequence of smallest integers with n+1 odd divisors and so apart from the leading 1, is precisely this sequence. - Ant King, Sep 23 2009
a(n) is also the least number k with the property that the symmetric representation of sigma(k) has n subparts. - Omar E. Pol, Dec 31 2016
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LINKS
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FORMULA
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a(p) = 3^(p-1) for primes p. - Zak Seidov, Apr 18 2006
It was suggested by Alexander Adamchuk that for all n >= 1, we have a(3^(n-1)) = (p(n)#/2)^2 = (A002110(n)/2)^2 = A070826(n)^2. But this is false! E.g., (p(n)#/2)^2 = 3^2 * 5^2 * 7^2 * ... * 23^2 * 29^2 does indeed have 3^9 odd factors, but it is greater than 3^8 * 5^2 * 7^2 * ... * 23^2 which has 9*3*3*3*3*3*3*3 = 9*3^7 = 3^9 odd factors. - Richard Sabey, Oct 06 2007
a(p^k) = Product_{i=1..k} prime(i+1)^(p-1), p prime and k >= 0, only when p_(k+1) < 3^p. - Hartmut F. W. Hoft, Nov 03 2022
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EXAMPLE
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a(2^3) = 105 = 3*5 while a(2^4) = 945 = 3^3 * 5 * 7. There are 5 partition lists for the exponents of numbers with 16 odd divisors; they are {1, 1, 1, 1}, {3, 1, 1}, {3, 3}, {7, 1}, and {15} that result in the 5 numbers 1155, 945, 3375, 10935, and 14348907. Number a(3^8) = a(6561) = 3^2 * 5^2 * ... * 19^2 * 23^2 = 12442607161209225 while a(3^9) = a(19683) = 3^8 * 5^2 * ... * 19^2 * 23^2 = 9070660620521525025. The numbers a(5^52) = 3^4 * 5^4 * 7^4 * ... and a(5^53) = 3^24 * 5^4 * 7^4 * ... have 393 and 402 digits, respectively. - Hartmut F. W. Hoft, Nov 03 2022
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MATHEMATICA
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Table[Select[Range[1, 532000, 2], DivisorSigma[0, #]==k+1 &, 1], {k, 0, 15}]//Flatten (* Ant King, Nov 28 2010 *)
2#-1&/@With[{ds=DivisorSigma[0, Range[1, 600000, 2]]}, Table[Position[ds, n, 1, 1], {n, 16}]]//Flatten (* The program is not suitable for generating terms beyond a(16) *) (* Harvey P. Dale, Jun 06 2017 *)
(* direct computation of A038547(n) *)
mp[1, m_] := {{}}; mp[n_, 1] := {{}}; mp[n_?PrimeQ, m_] := If[m<n, {}, {{n}}];
mp[n_, m_] := Join@@Table[Map[Prepend[#, d]&, mp[n/d, d]], {d, Select[Rest[Divisors[n]], #<=m&]}];
mp[n_]:=mp[n, n];
a038547[bound_] := Module[{mulpar}, Table[mulpar=mp[n]-1; Min[Table[Product[Prime[s+1]^mulpar[[j, s]], {s, 1, Length[mulpar[[j]]]}], {j, 1, Length[mulpar]}]], {n, 1, bound}]]
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PROG
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(PARI) for(nd=1, 15, forstep(k=1, 10^66, 2, if(nd==numdiv(k), print1(k, ", "); break())))
(Haskell)
import Data.List (find)
import Data.Maybe (fromJust)
a038547 n = fromJust $ find ((== n) . length . divisors) [1, 3..]
where divisors m = filter ((== 0) . mod m) [1..m]
(Python)
from math import prod
from sympy import isprime, divisors, prime
def mult_factors(n):
if isprime(n):
return [(n, )]
c = []
for d in divisors(n, generator=True):
if 1<d<n:
for a in mult_factors(n//d):
c.append(tuple(sorted((d, )+a)))
return list(set(c))
return min((prod(prime(i)**(j-1) for i, j in enumerate(reversed(d), 2)) for d in mult_factors(n)), default=1) # Chai Wah Wu, Aug 17 2024
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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