Search: a160519 -id:a160519
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A003586
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3-smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.
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+10
330
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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
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1,2
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COMMENTS
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This sequence is easily confused with A033845, which gives numbers of the form 2^i*3^j with i, j >= 1. Don't simply say "numbers of the form 2^i*3^j", but specify which sequence you mean. - N. J. A. Sloane, May 26 2024
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 Matula-Goebel number m has m antichains. The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel 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 Matula-Goebel 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 3-smooth 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/(1-1/2)) * (1/(1-1/3)) = (2/(2-1)) * (3/(3-1)) = 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
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REFERENCES
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J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 654 pp. 85, 287-8, 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. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.
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LINKS
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FORMULA
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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
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MAPLE
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A003586 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+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
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MATHEMATICA
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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}, powerOfTwo-1], # < 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 *)
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PROG
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(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 !! (n-1)
(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
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CROSSREFS
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Cf. A051037, A002473, A051038, A080197, A080681, A080682, A117221, A105420, A062051, A117222, A117220, A090184, A131096, A131097, A186711, A186712, A186771, A088468, A061987, A080683 (p-smooth numbers with other values of p), A025613 (a subsequence).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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A061987
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Number of times n-th distinct value is repeated in sequence b(k) = 1 + b(floor(k/2)) + b(floor(k/3)) with b(0) = 0, i.e., in A061984; also number of times n-th distinct row is repeated in square array T(n,k) = T(n-1,k) + T(n-1,floor(k/2)) + T(n-1,floor(k/3)) with T(0,0) = 1, i.e., in A061980.
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+10
16
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1, 1, 1, 1, 2, 2, 1, 3, 4, 2, 6, 3, 5, 4, 12, 6, 10, 8, 9, 15, 12, 20, 16, 18, 30, 24, 27, 13, 32, 36, 60, 48, 54, 26, 64, 72, 81, 39, 96, 108, 52, 128, 144, 162, 78, 192, 216, 104, 139, 117, 288, 324, 156, 384, 432, 208, 278, 234, 576, 648, 312, 417, 351, 864, 416, 556
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OFFSET
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0,5
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COMMENTS
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LINKS
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FORMULA
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PROG
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(Haskell)
import Data.List (group)
a061987 n = a061987_list !! n
a061987_list = map length $ group a061984_list
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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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A088468
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a(0) = 1, a(n) = a(floor(n/2)) + a(floor(n/3)) for n > 0.
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+10
9
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1, 2, 3, 4, 5, 5, 7, 7, 8, 9, 9, 9, 12, 12, 12, 12, 13, 13, 16, 16, 16, 16, 16, 16, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 33, 33, 33, 33, 33, 33, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 38, 38, 38, 38, 38, 38, 38, 38, 48
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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Limit_{n->oo} a(n)/n = 0, as proved in Michael Penn's Youtube video (see Links). Michael Penn states in the video that this is a simplification of a problem of Paul Erdős, where the original problem is to show that limit_{n->oo} b(n)/n = 12/log(432) for b(0) = 1, b(n) = b(floor(n/2)) + b(floor(n/3)) + b(floor(n/6)) for n > 0 ({b(n)} is the sequence A007731). - Jianing Song, Sep 27 2023
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MATHEMATICA
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a[0]=1; a[n_]:=a[n]=a[Floor[n/2]]+a[Floor[n/3]]; Array[a, 75, 0] (* Harvey P. Dale, Aug 23 2020 *)
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PROG
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(PARI) a(n)=if(n<1, n==0, a(n\2)+a(n\3))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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