A008834 -id:A008834 - cp's OEIS frontend

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

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

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Antti Karttunen, Nov 28 2017

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A000188, A003557, A004526, A020639, A028234, A067029, A295658, A295659.
Cf. A004709 (positions of ones), A082020, A212793.
Cf. also A008834, A053150, A061704.

Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; p > 0 :> p^Floor[(e - 1)/2]] &, 105] (* Michael De Vlieger, Nov 28 2017 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^floor((f[i,2]-1)/2));} \\ Amiram Eldar, Nov 30 2022

a(1) = 1; for n > 1, a(n) = A020639(n)^A004526(A067029(n)-1) * a(A028234(n)).
a(n) = A000188(A003557(n)).
a(n) = 1 iff A212793(n) = 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 15/Pi^2 = 1.519817... (A082020). - Amiram Eldar, Nov 30 2022

A368170 The largest cubefull exponentially odd divisor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 27, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, Dec 14 2023

First differs from A008834 at n = 32, and from A366906 at n = 64.

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

Cf. A004709, A335988, A350390, A356192, A368167, A368171.

Cf. A008834, A366906.

f[p_, e_] := If[e <= 2, 1, If[EvenQ[e], p^(e-1), p^e]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f=factor(n)); prod(i=1, #f~, if(f[i,2] <= 2, 1, if(!(f[i,2]%2), f[i,1]^(f[i, 2]-1), f[i, 1]^f[i, 2])))};

Multiplicative with a(p^e) = 1 if e <= 2, a(p^e) = p^e if e is odd and e > 1, and p^(e-1) otherwise.
a(n) = n/A368171(n).
a(n) >= 1, with equality if and only if n is cubefree (A004709).
a(n) <= n, with equality if and only if n is in A335988.

A277780 a(n) is the least k > n such that n*k^2 is a cube.

Original entry on oeis.org

8, 16, 24, 32, 40, 48, 56, 27, 72, 80, 88, 96, 104, 112, 120, 54, 136, 144, 152, 160, 168, 176, 184, 81, 200, 208, 64, 224, 232, 240, 248, 108, 264, 272, 280, 288, 296, 304, 312, 135, 328, 336, 344, 352, 360, 368, 376, 162, 392, 400, 408, 416, 424, 128, 440

Offset: 1

Peter Kagey, Oct 30 2016

a(n) is bounded above by 8*n (A008590) because n*(8*n)^2 = (4*n)^3.
If and only if n is cubefree, a(n) = 8n. - David A. Corneth, Nov 01 2016
Theorem: If n = q*m^3 with q cubefree then k = q*(m+1)^3. - Hartmut F. W. Hoft, Nov 02 2016
Proof: let q have u distinct prime divisors p_i. Then q = Product_{i=1..u}(p_i^e_i) where e_i > 0 since p_i|q and e_i < 3 since q is cubefree. Therefore, e_i = 1 or e_i = 2. This yields q|k, i.e., q*t = k. Now for n*k^2 = q*m^3*q^2*t^2 = (q*m)^3 * t^2 to be a cube, t must be a cube. Now, k > n, so q*t/(q*m^3) = t/m^3. The least cube > m^3 is (m+1)^3 so k = q*(m+1)^3 which completes the proof. - David A. Corneth, Nov 03 2016

			a(24) = 81  because 24 *  81^2 =  54^3;
a(25) = 200 because 25 * 200^2 = 100^3;
a(26) = 208 because 26 * 208^2 = 104^3;
a(27) = 64  because 27 *  64^2 =  48^3.
The cubefree part of 144 is 18. The cubefull part of 144 is 8 = 2^3. Therefore, a(144) = 18 * 3^3 = 486. - _David A. Corneth_, Nov 01 2016

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000578, A008590, A008834, A048766, A050985, A072905, A254767, A277781.

Cf. A002117, A013662, A013663, A013664.

Table[k = n + 1; While[! IntegerQ[(n k^2)^(1/3)], k++]; k, {n, 55}] (* Michael De Vlieger, Nov 04 2016 *)

a(n) = {my(k = n+1); while (!ispower(n*k^2, 3), k++); k;} \\ Michel Marcus, Oct 31 2016

a(n) = {my(f = factor(n)); f[, 2] = f[, 2]%3; f=factorback(f); n = sqrtnint(n/f,3); (n+1)^3 * f} \\ David A. Corneth, Nov 01 2016

a(n) = A050985(n) * A000578(1+A048766(A008834(n))). [Formula given in comments expressed with A-numbers] - Antti Karttunen, Nov 02 2016.

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + (3*zeta(4) + 3*zeta(5) + zeta(6))/zeta(3) = 7.13539675963975495073... . - Amiram Eldar, Feb 17 2024

A365487 The number of divisors of the largest cube dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, Sep 05 2023

The number of divisors of the cube root of the largest cube dividing n, A053150(n), is A061704(n).

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

Cf. A000005, A000578, A004709, A008834, A053150, A061704.

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

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

a(n) = A000005(A008834(n)).
Multiplicative with a(p^e) = 3*floor(e/3) + 1.
a(n) = 1 if and only if n is cubefree (A004709).
a(n) <= A000005(n) with equality if and only if n is a cube (A000578).
Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 2/p^(3*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(3) * Product_{p prime} (1 + 2/p^3) = 1.6552343865608... .

A366123 The number of prime factors of the cube root of the largest cube dividing n, counted with multiplicity.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0

Offset: 1

Amiram Eldar, Sep 30 2023

First differs from A295659 at n = 64.

The number of distinct prime factors of the cube root of the largest cube dividing n is A295659(n).

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

Cf. A000578, A001222, A002264, A008834, A004709, A053150, A286229, A295659.

Cf. A061704 (number of divisors), A333843 (sum of divisors).

f[p_, e_] := Floor[e/3]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecsum(apply(x -> x\3, factor(n)[, 2]));

a(n) = A001222(A053150(n)).
a(n) = A001222(A008834(n))/3.
Additive with a(p^e) = floor(e/3) = A002264(e).
a(n) >= 0, with equality if and only if n is cubefree (A004709).
a(n) <= A001222(n)/3, with equality if and only if n is a positive cube (A000578 \ {0}).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p^3-1) = 0.194118... (A286229).

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A368170 The largest cubefull exponentially odd divisor of n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A277780 a(n) is the least k > n such that n*k^2 is a cube.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A365487 The number of divisors of the largest cube dividing n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A366123 The number of prime factors of the cube root of the largest cube dividing n, counted with multiplicity.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula