

A003586


3smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.


322



1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2187, 2304, 2592, 2916, 3072, 3456, 3888
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OFFSET

1,2


COMMENTS

These numbers were once called "harmonic numbers", see Lenstra links.  N. J. A. Sloane, Jul 03 2015
Successive numbers k such that phi(6k) = 2k.  Artur Jasinski, Nov 05 2008
Also numbers that are divisible by neither 6k  1 nor 6k + 1, for all k > 0.  Robert G. Wilson v, Oct 26 2010
Also numbers m such that the rooted tree with MatulaGoebel number m has m antichains. The MatulaGoebel number of a rooted tree can be defined in the following recursive manner: to the onevertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the tth prime number, where t is the MatulaGoebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the MatulaGoebel numbers of the m branches of T. The vertices of a rooted tree can be regarded as a partially ordered set, where u<=v holds for two vertices u and v if and only if u lies on the unique path between v and the root. An antichain is a nonempty set of mutually incomparable vertices. Example: m=4 is in the sequence because the corresponding rooted tree is \/=ARB (R is the root) having 4 antichains (A, R, B, AB).  Emeric Deutsch, Jan 30 2012
The number of terms less than or equal to n is Sum_{i=0..floor(log_2(n))} floor(log_3(n/2^i) + 1), or Sum_{i=0..floor(log_3(n))} floor(log_2(n/3^i) + 1), which requires fewer terms to compute.  Robert G. Wilson v, Aug 17 2012
In the 14th century Levi Ben Gerson proved that the only pairs of terms which differ by 1 are (1,2), (2,3), (3,4), and (8,9); see A235365, A235366, A236210.  Jonathan Sondow, Jan 20 2014
The sum of the reciprocals of the 3smooth numbers is equal to 3. Brief proof: 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + ... = (Sum_{k>=0} 1/2^k) * (Sum_{m>=0} 1/3^m) = (1/(11/2)) * (1/(11/3)) = (2/(21)) * (3/(31)) = 3.  Bernard Schott, Feb 19 2019
Also those integers k for which, for every prime p > 3, p^(2k)  1 == 0 (mod 24k).  Federico Provvedi, May 23 2022
For n>1, the exponents’ parity {parity(i), parity(j)} of one out of four consecutive terms is {odd, odd}. Therefore, for n>1, at least one out of every four consecutive terms is a Zumkeller number (A083207). If for the term whose parity is {even, odd}, even also means nonzero, then this term is also a Zumkeller number (as is the case with the last of the four consecutive terms 1296, 1458, 1536, 1728).  Ivan N. Ianakiev, Jul 10 2022
Except the initial terms 2, 3, 4, 8, 9 and 16, these are numbers k such that k^6 divides 6^k. Except the initial terms 2, 3, 4, 6, 8, 9, 16, 18 and 27, these are numbers k such that k^12 divides 12^k.  Mohammed Yaseen, Jul 21 2022


REFERENCES

J.M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 654 pp. 85, 2878, Ellipses Paris 2004.
S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962, p. xxiv.
R. Tijdeman, Some applications of Diophantine approximation, pp. 261284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.


LINKS



FORMULA

An asymptotic formula for a(n) is roughly a(n) ~ 1/sqrt(6)*exp(sqrt(2*log(2)*log(3)*n)).  Benoit Cloitre, Nov 20 2001
The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} moebius(6*n)*x^n/(1  x^n).  Paul D. Hanna, Sep 18 2011


MAPLE

A003586 := proc(n) option remember; if n = 1 then 1; else for a from procname(n1)+1 do numtheory[factorset](a) minus {2, 3} ; if % = {} then return a; end if; end do: end if; end proc: # R. J. Mathar, Feb 28 2011
with(numtheory): for i from 1 to 23328 do if(i/phi(i)=3)then print(i/6) fi od; # Gary Detlefs, Jun 28 2011


MATHEMATICA

a[1] = 1; j = 1; k = 1; n = 100; For[k = 2, k <= n, k++, If[2*a[k  j] < 3^j, a[k] = 2*a[k  j], {a[k] = 3^j, j++}]]; Table[a[i], {i, 1, n}] (* Hai He (hai(AT)mathteach.net) and Gilbert Traub, Dec 28 2004 *)
aa = {}; Do[If[EulerPhi[6 n] == 2 n, AppendTo[aa, n]], {n, 1, 1000}]; aa (* Artur Jasinski, Nov 05 2008 *)
fQ[n_] := Union[ MemberQ[{1, 5}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 6]] == {False}; fQ[1] = True; Select[ Range@ 4000, fQ] (* Robert G. Wilson v, Oct 26 2010 *)
powerOfTwo = 12; Select[Nest[Union@Join[#, 2*#, 3*#] &, {1}, powerOfTwo1], # < 2^powerOfTwo &] (* Robert G. Wilson v and T. D. Noe, Mar 03 2011 *)
fQ[n_] := n == 3 EulerPhi@ n; Select[6 Range@ 4000, fQ]/6 (* Robert G. Wilson v, Jul 08 2011 *)
mx = 4000; Sort@ Flatten@ Table[2^i*3^j, {i, 0, Log[2, mx]}, {j, 0, Log[3, mx/2^i]}] (* Robert G. Wilson v, Aug 17 2012 *)
f[n_] := Block[{p2, p3 = 3^Range[0, Floor@ Log[3, n] + 1]}, p2 = 2^Floor[Log[2, n/p3] + 1]; Min[ Select[ p2*p3, IntegerQ]]]; NestList[f, 1, 54] (* Robert G. Wilson v, Aug 22 2012 *)
Select[Range@4000, Last@Map[First, FactorInteger@#] <= 3 &] (* Vincenzo Librandi, Aug 25 2016 *)
Select[Range[4000], Max[FactorInteger[#][[All, 1]]]<4&] (* Harvey P. Dale, Jan 11 2017 *)


PROG

(PARI) test(n)=for(p=2, 3, while(n%p==0, n/=p)); n==1;
for(n=1, 4000, if(test(n), print1(n", ")))
(PARI) list(lim)=my(v=List(), N); for(n=0, log(lim\1+.5)\log(3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
(PARI) list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018
(Haskell)
import Data.Set (Set, singleton, insert, deleteFindMin)
smooth :: Set Integer > [Integer]
smooth s = x : smooth (insert (3*x) $ insert (2*x) s')
where (x, s') = deleteFindMin s
a003586_list = smooth (singleton 1)
a003586 n = a003586_list !! (n1)
(Sage)
def isA003586(n) :
return not any(d != 2 and d != 3 for d in prime_divisors(n))
@CachedFunction
if n == 1 : return 1
while not isA003586(k) : k += 1
return k
(Python)
from itertools import count, takewhile
def aupto(lim):
pows2 = list(takewhile(lambda x: x<lim, (2**i for i in count(0))))
pows3 = list(takewhile(lambda x: x<lim, (3**i for i in count(0))))
return sorted(c*d for c in pows2 for d in pows3 if c*d <= lim)
(Magma) [n: n in [1..4000]  PrimeDivisors(n) subset [2, 3]]; // Bruno Berselli, Sep 24 2012


CROSSREFS

Cf. A051037, A002473, A051038, A080197, A080681, A080682, A117221, A105420, A062051, A117222, A117220, A090184, A131096, A131097, A186711, A186712, A186771, A088468, A061987, A080683 (psmooth numbers with other values of p), A025613 (a subsequence).


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS

Deleted claim that this sequence is union of 2^n (A000079) and 3^n (A000244) sequences  this does not include the terms which are not pure powers.  Walter Roscello (wroscello(AT)comcast.net), Nov 16 2008


STATUS

approved



