cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A003586 3-smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.

Original entry on oeis.org

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
Offset: 1

Views

Author

Paul Zimmermann, Dec 11 1996

Keywords

Comments

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
Where record values greater than 1 occur in A088468: A160519(n) = A088468(a(n)). - Reinhard Zumkeller, May 16 2009
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
A204455(3*a(n)) = 3, and only for these numbers. - Wolfdieter Lang, Feb 04 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
Named 3-friables in French. - Michel Marcus, Jul 17 2013
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
Range of values of A000005(n) (and also A181819(n)) for cubefree numbers n. - Matthew Vandermast, May 14 2014
A036561 is a permutation of this sequence. - L. Edson Jeffery, Sep 22 2014
Also the sorted union of A000244 and A007694. - Lei Zhou, Apr 19 2017
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
In music theory, a comma is a ratio, close to 1 (typically less than 1.04), between two natural numbers divisible by only small primes (typically single digit). In this sequence, a(131) / a(130) = 531441 / 524288 ~ 1.013643 is the Pythagorean comma (A221363), the difference between 12 perfect fifths and 7 octaves. - Hal M. Switkay, Mar 23 2025

References

  • 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.

Crossrefs

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).
Cf. also A000244, A007694. - Lei Zhou, Apr 19 2017
Cf. A191475 (successive values of i), A191476 (successive values of j), A022330 (indices of the pure terms 2^i), A022331 (indices of the pure terms 3^j). - N. J. A. Sloane, May 26 2024
Cf. A221363.

Programs

  • 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)
    -- Reinhard Zumkeller, Dec 16 2010
    
  • Magma
    [n: n in [1..4000] | PrimeDivisors(n) subset [2,3]]; // Bruno Berselli, Sep 24 2012
  • Maple
    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
  • 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}, 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 *)
  • 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
    is_A003586(n)=n<5||vecmax(factor(n,5)[, 1])<5 \\ M. F. Hasler, Jan 16 2015
    
  • 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
    
  • Python
    from itertools import count, takewhile
    def aupto(lim):
        pows2 = list(takewhile(lambda x: xMichael S. Branicky, Jul 08 2022
    
  • Python
    from sympy import integer_log
    def A003586(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024
    
  • Python
    # faster for initial segment of sequence
    import heapq
    from itertools import islice
    def A003586gen(): # generator of terms
        v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3]
        while True:
            v = heapq.heappop(h)
            if v != oldv:
                yield v
                oldv = v
                for p in psmooth_primes:
                    heapq.heappush(h, v*p)
    print(list(islice(A003586gen(), 65))) # Michael S. Branicky, Sep 17 2024
    (C++) // Returns A003586 <= threshold without approximations nor sorting
    #include 
    std::forward_list A003586(const int threshold) {
        std::forward_list sequence;
        auto start_it = sequence.before_begin();
        for (int i = 1; i <= threshold; i *= 2) {
            for (int inc = 1; std::next(start_it) != sequence.end() && inc <= i; inc *= 3)
                ++start_it;
            auto it = start_it;
            for (int j = 1; i * j <= threshold; j *= 3) {
                sequence.emplace_after(it, i * j);
                for (int inc = 1; std::next(it) != sequence.end() && inc <= i; inc *= 2)
                    ++it;
            }
        }
        return sequence;
    } // Eben Gino Lester, Apr 17 2025
    
  • Sage
    def isA003586(n) :
        return not any(d != 2 and d != 3 for d in prime_divisors(n))
    @CachedFunction
    def A003586(n) :
        if n == 1 : return 1
        k = A003586(n-1) + 1
        while not isA003586(k) : k += 1
        return k
    [A003586(n) for n in (1..55)] # Peter Luschny, Jul 20 2012
    

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
A061987(n) = a(n + 1) - a(n), a(A084791(n)) = A084789(n), a(A084791(n) + 1) = A084790(n). - Reinhard Zumkeller, Jun 03 2003
Union of powers of 2 and 3 with n such that psi(n) = 2*n, where psi(n) = n*Product_(1 + 1/p) over all prime factors p of n = A001615(n). - Lekraj Beedassy, Sep 07 2004; corrected by Franklin T. Adams-Watters, Mar 19 2009
a(n) = 2^A022328(n)*3^A022329(n). - N. J. A. Sloane, Mar 19 2009
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
a(n) = A007694(n+1)/2. - Lei Zhou, Apr 19 2017

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

A071521 Number of 3-smooth numbers <= n.

Original entry on oeis.org

1, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18
Offset: 1

Views

Author

Benoit Cloitre, Jun 02 2002

Keywords

Comments

A 3-smooth number is a number of the form 2^x * 3^y where x >= 0 and y >= 0.

References

  • Bruce C. Berndt and Robert A. Rankin, "Ramanujan : letters and commentary", History of Mathematics Volume 9, AMS-LMS, p. 23, p. 35.
  • G. H. Hardy, Ramanujan: Twelve lectures on subjects suggested by his life and work, AMS Chelsea Pub., 1999, pages 67-82.

Crossrefs

Programs

  • Haskell
    a071521 n = length $ takeWhile (<= n) a003586_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Maple
    N:= 10000: # to get a(1) to a(N)
    V:= Vector(N):
    for y from 0 to floor(log[3](N)) do
      for x from 0 to ilog2(N/3^y) do
        V[2^x*3^y]:= 1
    od od:
    convert(map(round,Statistics:-CumulativeSum(V)),list); # Robert Israel, Dec 16 2014
  • Mathematica
    a[n_] := Sum[ MoebiusMu[6k]*Floor[n/k], {k, 1, n}]; Table[a[n], {n, 1, 75}] (* Jean-François Alcover, Oct 11 2011, after Benoit Cloitre *)
    f[n_] := Sum[Floor@Log[3, n/2^i] + 1, {i, 0, Log[2, n]}]; Array[f, 75] (* faster, or *)
    f[n_] := Sum[Floor@Log[2, n/3^i] + 1, {i, 0, Log[3, n]}]; Array[f, 75] (* Robert G. Wilson v, Aug 18 2012 *)
    Accumulate[Table[If[Max[FactorInteger[n][[All,1]]]<4,1,0],{n,80}]] (* Harvey P. Dale, Jan 11 2017 *)
  • PARI
    for(n=1,100,print1(sum(k=1,n,if(sum(i=3,n,if(k%prime(i),0,1)),0,1)),","))
    
  • PARI
    a(n)=sum(k=1,n,moebius(2*3*k)*floor(n/k)) \\ Benoit Cloitre, Jun 14 2007
    
  • PARI
    a(n)=my(t=1/3); sum(k=0,logint(n,3), t*=3; logint(n\t,2)+1) \\ Charles R Greathouse IV, Jan 08 2018
    
  • Python
    from sympy import integer_log
    def A071521(n): return sum((n//3**i).bit_length() for i in range(integer_log(n,3)[0]+1)) # Chai Wah Wu, Sep 15 2024

Formula

a(n) = Card{ k | A003586(k) <= n }. Asymptotically: let a=1/(2*log(2)*log(3)), b=sqrt(6), then from Ramanujan a(n) ~ a*log(2*n)*log(3*n) or equivalently a(n) ~ a*log(b*n)^2.
A022331(n) = a(A000079(n)); A022330(n) = a(A000244(n)). - Reinhard Zumkeller, May 09 2006
a(n) = Sum_{k=1..n} mu(6k)*floor(n/k). - Benoit Cloitre, Jun 14 2007
a(n) = Sum_{k=1..n} (floor(6^k/k)-floor((6^k-1)/k)). - Anthony Browne, May 19 2016
From Ridouane Oudra, Jul 17 2020: (Start)
a(n) = Sum_{i=0..floor(log_2(n))} (floor(log_3(n/2^i)) + 1).
a(n) = Sum_{i=0..floor(log_3(n))} (floor(log_2(n/3^i)) + 1). (End)
A322026(n) = a(A065331(n)). - Antti Karttunen, Sep 08 2024

A022330 Index of 3^n within sequence of numbers of form 2^i*3^j (A003586).

Original entry on oeis.org

1, 3, 7, 12, 19, 27, 37, 49, 62, 77, 93, 111, 131, 152, 175, 199, 225, 252, 281, 312, 344, 378, 413, 450, 489, 529, 571, 614, 659, 705, 753, 803, 854, 907, 961, 1017, 1075, 1134, 1195, 1257, 1321, 1386, 1453, 1522, 1592, 1664, 1737, 1812, 1889, 1967, 2047, 2128
Offset: 0

Views

Author

Keywords

Comments

a(1000)=793775, a(10000)=79261054, a(100000)=7924941755, a(1000000)=792482542841.

Crossrefs

Cf. A022331, A020914 (first differences).

Programs

  • Mathematica
    c[0] = 1; c[n_] := 1 + Sum[Ceiling[j*Log[2, 3]], {j, n}]; Table[c[i], {i, 0, 51}] (* Norman Carey, Jun 13 2012 *)
  • PARI
    listsm(lim)=my(v=List(),N); for(n=0,log(lim)\log(3),N=3^n; while(N<=lim,listput(v,N);N<<=1)); v=Vec(v); vecsort(v)
    list(lim)=my(v=listsm(3^floor(lim)));vector(floor(lim+1),i,setsearch(v,3^(i-1))) \\ Charles R Greathouse IV, Aug 19 2011
    
  • PARI
    a(n)=sum(k=0,n, logint(3^k,2))+n+1 \\ Charles R Greathouse IV, Nov 22 2022
    
  • Python
    def A022330(n): return sum((3**i).bit_length() for i in range(n+1)) # Chai Wah Wu, Sep 16 2024

Formula

a(n) = A071521(A000244(n)); A003586(a(n)) = A000244(n). - Reinhard Zumkeller, May 09 2006
a(n) ~ kn^2 with k = log(3)/log(4) = 0.792.... More exact asymptotics? - Zak Seidov, Dec 22 2011
a(n+1) = a(n) + A020914(n+1). - Ruud H.G. van Tol, Nov 25 2022
kn^2 + kn + 1 <= a(n) <= kn^2 + (k+1)n + 1, so a(n) = kn^2 + O(n) with k = log(3)/log(4). The law of the iterated logarithm suggests that a better error term might be possible. - Charles R Greathouse IV, Nov 28 2022

A020915 Number of digits in base-3 representation of 2^n.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 47
Offset: 0

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Author

Keywords

Comments

For n > 0, first differences of A022331. - Michel Marcus, Oct 03 2013

Crossrefs

Cf. A022924 (first differences).

Programs

Formula

a(n) = 1 + floor(n*log_3(2)) = 1 + floor(n*A102525) = 1 + A136409(n). - R. J. Mathar, May 23 2009
a(n) = A081604(A000079(n)). - Reinhard Zumkeller, Jul 12 2011
a(A020914(n)) = n + 1. - Reinhard Zumkeller, Mar 28 2015

Extensions

More terms from James Sellers

A202821 Position of 6^n among 3-smooth numbers A003586.

Original entry on oeis.org

1, 5, 14, 26, 43, 64, 89, 119, 153, 191, 233, 279, 330, 385, 444, 507, 575, 646, 722, 802, 886, 975, 1067, 1164, 1266, 1371, 1481, 1595, 1713, 1835, 1961, 2092, 2227, 2366, 2509, 2657, 2809, 2965, 3125, 3289, 3458, 3630, 3807, 3989, 4174, 4364, 4558, 4756
Offset: 0

Views

Author

Zak Seidov, Dec 25 2011

Keywords

Examples

			a(0) = 1 because A003586(1) = 6^0 = 1.
a(1) = 5 because A003586(5) = 6^1 = 6.
a(2) = 14 because A003586(14) = 6^2 = 36.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Floor[Log[3, 6^n/2^i]] + 1, {i, 0, Log2[6^n]}]; Array[a, 50, 0] (* Amiram Eldar, Jul 15 2023 *)
  • Python
    # uses imports/function in A372401
    print(list(islice(A372401gen(p=3), 1000))) # Michael S. Branicky, Jun 06 2024
    
  • Python
    from sympy import integer_log
    def A202821(n): return 1+n*(n+1)+sum((m:=3**i).bit_length()+((1<Chai Wah Wu, Oct 22 2024

Formula

A003586(a(n)) = 6^n, for n >= 0.
a(n) ~ (log(6))^2/(log(3)*log(4))*n^2 = 2.1079...*n^2.

A025723 Index of 7^n within sequence of numbers of form 5^i*7^j.

Original entry on oeis.org

1, 3, 6, 10, 15, 22, 30, 39, 49, 60, 73, 87, 102, 118, 135, 154, 174, 195, 217, 240, 265, 291, 318, 346, 376, 407, 439, 472, 506, 542, 579, 617, 656, 696, 738, 781, 825, 870, 916, 964, 1013, 1063, 1114, 1166, 1220, 1275, 1331, 1388, 1447, 1507, 1568, 1630, 1693
Offset: 0

Views

Author

Keywords

Comments

Positions of zeros in A025652. - R. J. Mathar, Jul 06 2025

Crossrefs

Cf. A025708, A022330, A022331, etc.

Programs

  • Maple
    ListTools:-PartialSums([1,seq(ceil(k*log[5](7)),k=1..100)]); # Robert Israel, Nov 16 2016
  • Mathematica
    Table[1 + Sum[Ceiling[k Log[5, 7]], {k, n}], {n, 0, 52}] (* Michael De Vlieger, Nov 16 2016 *)
  • PARI
    a(n)=my(N=1); n+1+sum(i=1, n, logint(N*=7, 5)); \\ Charles R Greathouse IV, Jan 11 2018
    
  • PARI
    first(n)=my(s, N=1/7); vector(n+1, i, s+=logint(N*=7, 5)+1) \\ Charles R Greathouse IV, Jan 11 2018

Formula

a(n) = 1 + Sum_{k=1..n} ceiling(k*log_5(7)). - Robert Israel, Nov 16 2016

Extensions

Offset changed to 0 by Robert Israel, Nov 16 2016

A096075 Least common multiple of first n 3-smooth numbers.

Original entry on oeis.org

1, 2, 6, 12, 12, 24, 72, 72, 144, 144, 144, 432, 864, 864, 864, 864, 1728, 1728, 5184, 5184, 5184, 10368, 10368, 10368, 10368, 10368, 31104, 62208, 62208, 62208, 62208, 62208, 62208, 124416, 124416, 124416, 373248, 373248, 373248, 373248, 746496
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 21 2004

Keywords

Comments

Subsequence of A003586.

Examples

			The first seven 3-smooth numbers are {1, 2, 3, 4, 6, 8, 9} and their lcm is 72. - _David A. Corneth_, Jul 13 2023
		

Crossrefs

Programs

  • Mathematica
    seq[max_] := Module[{sm3 = Sort[Flatten[Table[2^i*3^j, {i, 0, Log2[max]}, {j, 0, Log[3, max/2^i]}]]], e2, e3}, e2 = FoldList[Max, IntegerExponent[sm3, 2]]; e3 = FoldList[Max, IntegerExponent[sm3, 3]]; 2^e2*3^e3]; seq[1000] (* Amiram Eldar, Jul 13 2023 *)

Formula

a(n) > a(n-1) iff A003586(n) is a power of 2 or of 3 (cf. A006899, A022330, A022331).

A374484 Index of A006899(n) in A003586.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 12, 13, 17, 19, 22, 27, 28, 34, 37, 41, 48, 49, 56, 62, 65, 74, 77, 84, 93, 95, 106, 111, 118, 130, 131, 143, 152, 157, 171, 175, 186, 199, 202, 218, 225, 235, 252, 253, 271, 281, 290, 309, 312, 329, 344, 350, 371, 378, 393, 413, 416, 439
Offset: 1

Views

Author

Chai Wah Wu, Sep 16 2024

Keywords

Comments

Index of powers of 2 and 3 in 3-smooth numbers.

Examples

			A006899(10) = 64 which is the 17th term of A003586, therefore a(10) = 17.
		

Crossrefs

Disjoint union of A022330 and A022331.

Programs

  • Mathematica
    seq[lim_] := Position[Times @@ IntegerExponent[#, {2, 3}] & /@ Sort[Flatten[ Table[2^i*3^j, {i, 0, Log2[lim]}, {j, 0, Log[3, lim/2^i]}] ]], 0] // Flatten; seq[10^11] (* Amiram Eldar, Sep 18 2024 *)
  • Python
    from sympy import integer_log
    def A374484(n): return sum(((1<
    				

Formula

A003586(a(n)) = A006899(n).
a(n) ~ c * n^2, where c = log(2)*log(3)/(2*(log(2) + log(3))^2) = 0.118598856384648... - Vaclav Kotesovec and Amiram Eldar, Sep 19 2024

A379258 a(n) is the number of iterations of the Euler phi function needed to reach 1 starting at the n-th 3-smooth number.

Original entry on oeis.org

1, 2, 3, 3, 3, 4, 4, 4, 5, 4, 5, 5, 6, 5, 6, 5, 7, 6, 6, 7, 6, 8, 7, 6, 8, 7, 7, 9, 8, 7, 9, 8, 7, 10, 9, 8, 8, 10, 9, 8, 11, 10, 9, 8, 11, 10, 9, 12, 9, 11, 10, 9, 12, 11, 10, 13, 9, 12, 11, 10, 13, 10, 12, 11, 14, 10, 13, 12, 11, 14, 10, 13, 12, 15, 11, 14, 11
Offset: 1

Views

Author

Amiram Eldar, Dec 19 2024

Keywords

Examples

			a(6) = 4 because the 6th 3-smooth number is A003586(6) = 8, and 4 iterations of phi are needed to reach 1: 8 -> 4 -> 2 -> 1.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3]}, e2 + e3 + 1 + Boole[e2 == 0]]; f[1] = 1; With[{max = 3*10^4}, f /@ Sort[Flatten[Table[2^i*3^j, {i, 0, Log2[max]}, {j, 0, Log[3, max/2^i]}]]]]
  • PARI
    list(lim) = {my(e2, e3); print1(1, ", "); for(k = 2, lim, e2 = valuation(k, 2); e3 = valuation(k, 3); if(k == (1 << e2) * 3^e3, print1(e2 + e3 + 1 + (e2 == 0), ", ")));}

Formula

a(n) = A049108(A003586(n)).
a(n) = valuation(A003586(n), 2) + valuation(A003586(n), 3) + 1 + [valuation(A003586(n), 2) == 0] for n > 1, where [] is the Iverson bracket.
a(n) = A022328(n) + A022329(n) + 1 + [n is in A022330], for n > 1.
a(A022330(n)) = n + 2 for n >= 1.
a(A022331(n)) = n + 1 for n >= 0.
a(A202821(n)) = 2*n + 1, for n >= 0.
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