A019555 -id:A019555 - cp's OEIS frontend

A015050 Let m = A013929(n); then a(n) = smallest k such that m divides k^3.

2, 2, 3, 6, 4, 6, 10, 6, 5, 3, 14, 4, 6, 10, 22, 15, 12, 7, 10, 26, 6, 14, 30, 21, 4, 34, 6, 15, 38, 20, 9, 42, 22, 30, 46, 12, 14, 33, 10, 26, 6, 28, 58, 39, 30, 11, 62, 5, 42, 8, 66, 15, 34, 70, 12, 21, 74, 30, 38, 51, 78, 20, 18, 82, 42, 13, 57, 86

Offset: 1

R. Muller

nonn

R. J. Mathar, Table of n, a(n) for n = 1..10491
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.

Cf. A013929, A015049, A015051, A019555.

isA013929 := proc(n)
    not numtheory[issqrfree](n) ;
end proc:
A013929 := proc(n)
    option remember;
    local a;
    if n = 1 then
        4;
    else
        for a from procname(n-1)+1 do
            if isA013929(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
A015050 := proc(n)
    local m ;
    m := A013929(n) ;
    for k from 1 do
        if modp(k^3,m) = 0 then
            return k;
        end if;
    end do:
end proc:

f[p_, e_] := p^Ceiling[e/3]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; s /@ Select[Range[200], !SquareFreeQ[#] &] (* Amiram Eldar, Feb 09 2021 *)

lista(kmax) = {my(f); for(k = 2, kmax, f = factor(k); if(!issquarefree(f), print1(prod(i = 1, #f~, f[i,1]^ceil(f[i,2]/3)), ", ")));} \\ Amiram Eldar, Jan 06 2024

a(n) = A019555(A013929(n)).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2) * (zeta(2) * zeta(5) * Product_{p prime} (1-1/p^2+1/p^3-1/p^4) - 1)/(zeta(2)-1)^2 = 0.6611256641303... . - Amiram Eldar, Jan 06 2024

Description corrected by Diego Torres (torresvillarroel(AT)hotmail.com), Jun 23 2002

A056551 Smallest cube divisible by n divided by largest cube which divides n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 1, 27, 1000, 1331, 216, 2197, 2744, 3375, 8, 4913, 216, 6859, 1000, 9261, 10648, 12167, 27, 125, 17576, 1, 2744, 24389, 27000, 29791, 8, 35937, 39304, 42875, 216, 50653, 54872, 59319, 125, 68921, 74088, 79507, 10648

Offset: 1

Henry Bottomley, Jun 25 2000

			a(16) = 8 since smallest cube divisible by 16 is 64 and smallest cube which divides 16 is 8 and 64/8 = 8.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000189, A000578, A008834, A019555, A048798, A050985, A053149, A053150, A056552.

Cf. A002117, A013670, A330596.

f[p_, e_] := p^If[Divisible[e, 3], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^3; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%3, f[i,1], 1))^3; } \\ Amiram Eldar, Oct 28 2022

a(n) = A053149(n)/A008834(n) = A048798(n)*A050985(n) = A056552(n)^3.
From Amiram Eldar, Oct 28 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 3, and a(p^e) = p^3 otherwise.
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(12)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013670 * A330596 / (4*A002117) = 0.1557163105... . (End)
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(2*s-3)). - Amiram Eldar, Sep 16 2023

A254733 a(n) is the least k > n such that n divides k^3.

Original entry on oeis.org

2, 4, 6, 6, 10, 12, 14, 10, 12, 20, 22, 18, 26, 28, 30, 20, 34, 24, 38, 30, 42, 44, 46, 30, 30, 52, 30, 42, 58, 60, 62, 36, 66, 68, 70, 42, 74, 76, 78, 50, 82, 84, 86, 66, 60, 92, 94, 60, 56, 60, 102, 78, 106, 60, 110, 70, 114, 116, 118, 90, 122, 124, 84

Offset: 1

Peter Kagey, Feb 06 2015

A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^3.

			a(8) = 10 because 8 divides 10^3, but 8 does not divide 9^3.

Peter Kagey, Table of n, a(n) for n = 1..5000

Cf. A073353 (similar, with k^n).
Cf. A254732 (similar, with k^2), A254734 (similar, with k^4).
Cf. A019555 (similar without the restriction that a(n) > n).

lkn[n_]:=Module[{k=n+1},While[PowerMod[k,3,n]!=0,k++];k]; Array[lkn,70] (* Harvey P. Dale, Nov 23 2024 *)

a(n)=for(k=n+1,2*n,if(k^3%n==0,return(k)))
vector(100,n,a(n)) \\ Derek Orr, Feb 07 2015

def a(n)
  (n+1..2*n).find { |k| k**3 % n == 0 }
end

a(n) = n + A019555(n).

A076116 Start of the smallest string of n consecutive positive numbers with a cube sum, or 0 if no such number exists.

Original entry on oeis.org

1, 13, 8, 0, 23, 2, 46, 0, 20, 8, 116, 0, 163, 18, 218, 6, 281, 32, 352, 0, 431, 50, 518, 0, 28, 72, 14, 0, 827, 98, 946, 0, 1073, 128, 1208, 0, 1351, 162, 1502, 0, 1661, 200, 1828, 0, 53, 242, 2186, 98, 32, 43, 2576, 0, 2783, 36, 2998, 0, 3221, 392, 3452, 0, 3691, 450

Offset: 1

Amarnath Murthy, Oct 09 2002

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007814, A019555, A076117, A076114.

f:= proc(n) local y,F,t,k,v;
      if n::odd then
         F:= ifactors(n)[2];
         y:= mul(t[1]^ceil(t[2]/3),t=F);
         k:= 1+floor((n*(n-1)/2)^(1/3)/y);
         (k*y)^3/n - (n-1)/2;
      else
         v:= padic:-ordp(n,2);
         if v mod 3 <> 1 then return 0 fi;
         F:= ifactors(n/2^v)[2];
         y:= mul(t[1]^ceil(t[2]/3),t=F)*2^((v-1)/3);
         k:= 1 + floor((n*(n-1)/2)^(1/3)/y);
         if k::even then k:= k+1 fi;
         (k*y)^3/n - (n-1)/2;
      fi
end proc:
map(f, [$1..100]); # Robert Israel, Nov 15 2023

f[n_] := Module[{y, F, t, k, v},
If[OddQ[n],
   F = FactorInteger[n];
   y = Product[t[[1]]^Ceiling[t[[2]]/3], {t, F}];
   k = 1 + Floor[(n*(n-1)/2)^(1/3)/y];
   (k*y)^3/n - (n-1)/2
,
   v = IntegerExponent[n, 2];
   If[Mod[v, 3] != 1, Return[0]];
   F = FactorInteger[n/2^v];
   y = Product[t[[1]]^Ceiling[t[[2]]/3], {t, F}]*2^((v-1)/3);
   k = 1 + Floor[(n*(n-1)/2)^(1/3)/y];
   If[EvenQ[k], k = k+1];
   (k*y)^3/n - (n-1)/2]];
Map[f, Range[100]] (* Jean-François Alcover, Jul 09 2024, after Robert Israel *)

From Robert Israel, Nov 15 2023: (Start)
If n is odd, then a(n) is the least positive integer of the form (k*A019555(n))^3/n - (n-1)/2 where k is an integer.
If n is even, then let v = A007814(n). If v == 1 (mod 3) then a(n) is the least positive integer of the form (k*A019555(n/2))^3/n - (n-1)/2 where k an odd integer; otherwise, a(n) = 0. (End)

More terms from David Wasserman, Apr 02 2005

A365489 The number of divisors of the smallest cube divisible by n.

Original entry on oeis.org

1, 4, 4, 4, 4, 16, 4, 4, 4, 16, 4, 16, 4, 16, 16, 7, 4, 16, 4, 16, 16, 16, 4, 16, 4, 16, 4, 16, 4, 64, 4, 7, 16, 16, 16, 16, 4, 16, 16, 16, 4, 64, 4, 16, 16, 16, 4, 28, 4, 16, 16, 16, 4, 16, 16, 16, 16, 16, 4, 64, 4, 16, 16, 7, 16, 64, 4, 16, 16, 64, 4, 16, 4

Offset: 1

Amiram Eldar, Sep 05 2023

The number of divisors of the cube root of the smallest cube divisible by n, A019555(n), is A365488(n).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A019555, A053149, A365345, A365488.

f[p_, e_] := 3*Ceiling[e/3] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecprod(apply(x -> 3*((x-1)\3) + 4, factor(n)[, 2]));

a(n) = A000005(A053149(n)).
Multiplicative with a(p^e) = 3*ceiling(e/3) + 1.
Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 3/p^s - 1/p^(3*s)).

cp's OEIS Frontend

A015050 Let m = A013929(n); then a(n) = smallest k such that m divides k^3.

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A056551 Smallest cube divisible by n divided by largest cube which divides n.

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A254733 a(n) is the least k > n such that n divides k^3.

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A076116 Start of the smallest string of n consecutive positive numbers with a cube sum, or 0 if no such number exists.

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A365489 The number of divisors of the smallest cube divisible by n.

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