A000189 -id:A000189 - cp's OEIS frontend

A003557 n divided by largest squarefree divisor of n; if n = Product p(k)^e(k) then a(n) = Product p(k)^(e(k)-1), with a(1) = 1.

1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 8, 1, 3, 1, 2, 1, 1, 1, 4, 5, 1, 9, 2, 1, 1, 1, 16, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 8, 7, 5, 1, 2, 1, 9, 1, 4, 1, 1, 1, 2, 1, 1, 3, 32, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 5, 2, 1, 1, 1, 8, 27, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 16, 1, 7

Offset: 1

Marc LeBrun

a(n) is the size of the Frattini subgroup of the cyclic group C_n - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 07 2001.
Also of the Frattini subgroup of the dihedral group with 2*n elements. - Sharon Sela (sharonsela(AT)hotmail.com), Jan 01 2002
Number of solutions to x^m==0 (mod n) provided that n < 2^(m+1), i.e. the sequence of sequences A000188, A000189, A000190, etc. converges to this sequence. - Henry Bottomley, Sep 18 2001
a(n) is the number of nilpotent elements in the ring Z/nZ. - Laszlo Toth, May 22 2009
The sequence of partial products of a(n) is A085056(n). - Peter Luschny, Jun 29 2009
The first occurrence of n in this sequence is at A064549(n). - Franklin T. Adams-Watters, Jul 25 2014
From Hal M. Switkay, Jul 03 2025: (Start)
For n > 1, a(n) is a proper divisor of n. Thus the sequence n, a(n), a(a(n)), ... eventually becomes 1. This yields a minimal factorization of n as a product of squarefree numbers (A005117), each factor dividing all larger factors, in a factorization that is conjugate to the minimal factorization of n as a product of prime powers (A000961), as follows.
Let f(n,0) = n, and let f(n,k) = a(f(n,k-1)) for k > 0. A051903(n) is the minimal value of k such that f(n,k) = 1. A051903(n) <= log(n)/log(2). Since n/a(n) = A007947(n) is always squarefree by definition, n is a product of squarefree factors in the form Product_{i=1..A051903(n)} [f(n,i-1)/f(n,i)].
The two factorizations correspond to conjugate partitions of bigomega(n) = A001222(n). (End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Steven R. Finch, Idempotents and Nilpotents Modulo n, arXiv:math/0605019 [math.NT], 2006-2017.

Cf. A007947, A062378, A062379, A064549, A300717 (Möbius transform), A326306 (inv. Möbius transf.), A328572.
Sequences that are multiples of this sequence (the other factor of a pointwise product is given in parentheses): A000010 (A173557), A000027 (A007947), A001615 (A048250), A003415 (A342001), A007434 (A345052), A057521 (A071773).
Cf. A082695 (Dgf at s=2), A065487 (Dgf at s=3).
Cf. A000961, A001222, A005117, A051903.

a003557 n = product $ zipWith (^)
                      (a027748_row n) (map (subtract 1) $ a124010_row n)
-- Reinhard Zumkeller, Dec 20 2013

using Nemo
function A003557(n)
    n < 4 && return 1
    q = prod([p for (p, e) ∈ Nemo.factor(fmpz(n))])
    return n == q ? 1 : div(n, q)
end
[A003557(n) for n in 1:90] |> println  # Peter Luschny, Feb 07 2021

[(&+[(Floor(k^n/n)-Floor((k^n-1)/n)): k in [1..n]]): n in [1..100]]; // G. C. Greubel, Nov 02 2018

A003557 := n -> n/ilcm(op(numtheory[factorset](n))):
seq(A003557(n), n=1..98); # Peter Luschny, Mar 23 2011
seq(n / NumberTheory:-Radical(n), n = 1..98); # Peter Luschny, Jul 20 2021

Prepend[ Array[ #/Times@@(First[ Transpose[ FactorInteger[ # ] ] ])&, 100, 2 ], 1 ] (* Olivier Gérard, Apr 10 1997 *)

a(n)=n/factorback(factor(n)[,1]) \\ Charles R Greathouse IV, Nov 17 2014

for(n=1, 100, print1(direuler(p=2, n, (1 - p*X + X)/(1 - p*X))[n], ", ")) \\ Vaclav Kotesovec, Jun 20 2020

from sympy.ntheory.factor_ import core
from sympy import divisors
def a(n): return n / max(i for i in divisors(n) if core(i) == i)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Apr 16 2017

from math import prod
from sympy import primefactors
def A003557(n): return n//prod(primefactors(n)) # Chai Wah Wu, Nov 04 2022

def A003557(n) : return n*mul(1/p for p in prime_divisors(n))
[A003557(n) for n in (1..98)] # Peter Luschny, Jun 10 2012

Multiplicative with a(p^e) = p^(e-1). - Vladeta Jovovic, Jul 23 2001
a(n) = n/rad(n) = n/A007947(n) = sqrt(J_2(n)/J_2(rad(n))), where J_2(n) is A007434. - Enrique Pérez Herrero, Aug 31 2010
a(n) = (J_k(n)/J_k(rad(n)))^(1/k), where J_k is the k-th Jordan Totient Function: (J_2 is A007434 and J_3 A059376). - Enrique Pérez Herrero, Sep 03 2010
Dirichlet convolution of A000027 and A097945. - R. J. Mathar, Dec 20 2011
a(n) = A000010(n)/|A023900(n)|. - Eric Desbiaux, Nov 15 2013
a(n) = Product_{k = 1..A001221(n)} (A027748(n,k)^(A124010(n,k)-1)). - Reinhard Zumkeller, Dec 20 2013
a(n) = Sum_{k=1..n}(floor(k^n/n)-floor((k^n-1)/n)). - Anthony Browne, May 11 2016
a(n) = e^[Sum_{k=2..n} (floor(n/k)-floor((n-1)/k))*(1-A010051(k))*Mangoldt(k)] where Mangoldt is the Mangoldt function. - Anthony Browne, Jun 16 2016
a(n) = Sum_{d|n} mu(d) * phi(d) * (n/d), where mu(d) is the Moebius function and phi(d) is the Euler totient function (rephrases formula of Dec 2011). - Daniel Suteu, Jun 19 2018
G.f.: Sum_{k>=1} mu(k)*phi(k)*x^k/(1 - x^k)^2. - Ilya Gutkovskiy, Nov 02 2018
Dirichlet g.f.: Product_{primes p} (1 + 1/(p^s - p)). - Vaclav Kotesovec, Jun 24 2020
From Richard L. Ollerton, May 07 2021: (Start)
a(n) = Sum_{k=1..n} mu(n/gcd(n,k))*gcd(n,k).
a(n) = Sum_{k=1..n} mu(gcd(n,k))*(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). (End)
a(n) = A001615(n)/A048250(n) = A003415/A342001(n) = A057521(n)/A071773(n). - Antti Karttunen, Jun 08 2021

Secondary definition added to the name by Antti Karttunen, Jun 08 2021

A000188 (1) Number of solutions to x^2 == 0 (mod n). (2) Also square root of largest square dividing n. (3) Also max_{ d divides n } gcd(d, n/d).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 2, 5, 1, 3, 2, 1, 1, 1, 4, 1, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 4, 7, 5, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 3, 8, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 5, 2, 1, 1, 1, 4, 9, 1, 1, 2, 1, 1, 1, 2, 1, 3

Offset: 1

N. J. A. Sloane

Shadow transform of the squares A000290. - Vladeta Jovovic, Aug 02 2002
Labos Elemer and Henry Bottomley independently proved that (2) and (3) define the same sequence. Bottomley also showed that (1) and (2) define the same sequence.
Proof that (2) = (3): Let max{gcd(d, n/d)} = K, then d = Kx, n/d = Ky so n = KKxy where xy is the squarefree part of n, otherwise K is not maximal. Observe also that g = gcd(K, xy) is not necessarily 1. Thus K is also the "maximal square-root factor" of n. - Labos Elemer, Jul 2000
We can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n) and b*c = A019554(n) = "outer square root" of n.

			a(8) = 2 because the largest square dividing 8 is 4, the square root of which is 2.
a(9) = 3 because 9 is a perfect square and its square root is 3.
a(10) = 1 because 10 is squarefree.

T. D. Noe, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, preprint.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
Gerry Myerson, Trifectas in Geometric Progression, Australian Mathematical Society Gazette 35(3) (2008), 189-194
Andrew Reiter, On (mod n) spirals (2014), and posting to Number Theory Mailing List, Mar 23 2014.
N. J. A. Sloane, Transforms.
Florentin Smarandache, Collected Papers, Vol. II, Tempus Publ. Hse, Bucharest, 1996.
László Tóth, Counting solutions of quadratic congruences in several variables revisited, arXiv preprint arXiv:1404.4214 [math.NT], 2014.
László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), # 14.11.6.

Cf. A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).
Cf. A000189, A000190, A007947, A008833, A027748, A046951, A055210, A007913, A117811, A124010. For partial sums see A120486.
Cf. A240976 (Dgf at s=2).

a000188 n = product $ zipWith (^)
                      (a027748_row n) $ map (`div` 2) (a124010_row n)
-- Reinhard Zumkeller, Apr 22 2012

with(numtheory):A000188 := proc(n) local i: RETURN(op(mul(i,i=map(x->x[1]^floor(x[2]/2),ifactors(n)[2])))); end;

Array[Function[n, Count[Array[PowerMod[#, 2, n ] &, n, 0 ], 0 ] ], 100]
(* Second program: *)
nMax = 90; sList = Range[Floor[Sqrt[nMax]]]^2; Sqrt[#] &/@ Table[ Last[ Select[ sList, Divisible[n, #] &]], {n, nMax}] (* Harvey P. Dale, May 11 2011 *)
a[n_] := With[{d = Divisors[n]}, Max[GCD[d, Reverse[d]]]] (* Mamuka Jibladze, Feb 15 2015 *)
f[p_, e_] := p^Floor[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

PARI
```
a(n)=if(n<1,0,sum(i=1,n,i*i%n==0))
```

a(n)=sqrtint(n/core(n)) \\ Zak Seidov, Apr 07 2009

a(n)=core(n, 1)[2] \\ Michel Marcus, Feb 27 2013

from sympy.ntheory.factor_ import core
from sympy import integer_nthroot
def A000188(n): return integer_nthroot(n//core(n),2)[0] # Chai Wah Wu, Jun 14 2021

a(n) = n/A019554(n) = sqrt(A008833(n)).
a(n) = Sum_{d^2|n} phi(d), where phi is the Euler totient function A000010.
Multiplicative with a(p^e) = p^floor(e/2). - David W. Wilson, Aug 01 2001
Dirichlet series: Sum_{n >= 1} a(n)/n^s = zeta(2*s - 1)*zeta(s)/zeta(2*s), (Re(s) > 1).
Dirichlet convolution of A037213 and A008966. - R. J. Mathar, Feb 27 2011
Finch & Sebah show that the average order of a(n) is 3 log n/Pi^2. - Charles R Greathouse IV, Jan 03 2013
a(n) = sqrt(n/A007913(n)). - M. F. Hasler, May 08 2014
Sum_{n>=1} lambda(n)*a(n)*x^n/(1-x^n) = Sum_{n>=1} n*x^(n^2), where lambda() is the Liouville function A008836 (cf. A205801). - Mamuka Jibladze, Feb 15 2015
a(2*n) = a(n)*(A096268(n-1) + 1). - observed by Velin Yanev, Jul 14 2017, The formula says that a(2n) = 2*a(n) only when 2-adic valuation of n (A007814(n)) is odd, otherwise a(2n) = a(n). This follows easily from the definition (2). - Antti Karttunen, Nov 28 2017
Sum_{k=1..n} a(k) ~ 3*n*((log(n) + 3*gamma - 1)/Pi^2 - 12*zeta'(2)/Pi^4), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Dec 01 2020
Conjecture: a(n) = Sum_{k=1..n} A010052(n*k). - Velin Yanev, Jul 04 2021
G.f.: Sum_{k>=1} phi(k) * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Aug 20 2021

Edited by M. F. Hasler, May 08 2014

A053150 Cube root of largest cube dividing n.

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, 2, 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, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Henry Bottomley, Feb 28 2000

This can be thought as a "lower 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "upper 3rd root" of positive integers is given by A019555. - Petros Hadjicostas, Sep 15 2019

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Vaclav Kotesovec, Graph - the asymptotic ratio.

Cf. A000188 (inner square root), A019554 (outer square root), A019555 (outer third root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A000190, A008834, A008835, A015051, A061704.

f[list_] := list[[1]]^Quotient[list[[2]], 3]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)
Table[SelectFirst[Reverse@ Divisors@ n, IntegerQ[#^(1/3)] &]^(1/3), {n, 105}] (* Michael De Vlieger, Jul 28 2017 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

A053150(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= (f[k, 1]^(f[k, 2]\3)); ); m; } \\ Antti Karttunen, Jul 28 2017

a(n) = my(f = factor(n)); for (k=1, #f~, f[k,2] = f[k,2]\3); factorback(f); \\ Michel Marcus, Jul 28 2017

from math import prod
from sympy import factorint
def A053150(n): return prod(p**(q//3) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

Multiplicative with a(p^e) = p^[e/3]. - Mitch Harris, Apr 19 2005
a(n) = A008834(n)^(1/3) = sqrt(A000189(n)/A000188(A050985(n))).
Dirichlet g.f.: zeta(3s-1)*zeta(s)/zeta(3s). - R. J. Mathar, Apr 09 2011
Sum_{k=1..n} a(k) ~ Pi^2 * n / (6*zeta(3)) + 3*zeta(2/3) * n^(2/3) / Pi^2. - Vaclav Kotesovec, Jan 31 2019
a(n) = Sum_{d^3|n} phi(d). - Ridouane Oudra, Dec 30 2020
G.f.: Sum_{k>=1} phi(k) * x^(k^3) / (1 - x^(k^3)). - Ilya Gutkovskiy, Aug 20 2021

More terms from Antti Karttunen, Jul 28 2017

A019555 Smallest number whose cube is divisible by n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

R. Muller

This can be thought as an "upper 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "lower 3rd root" of positive integers is given by A053150. - Petros Hadjicostas, Sep 15 2019

Peter Kagey, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Vaclav Kotesovec, Graph - the asymptotic ratio (1000000 terms).
Ana Rechtman, Mai 2021, 4e défi, Images des Mathématiques, CNRS, 2021 (in French).
Florentin Smarandache, Collected Papers, Vol. II, Tempus Publ. Hse, Bucharest, 1996.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A015050.

f:= n -> mul(t[1]^ceil(t[2]/3), t = ifactors(n)[2]):
map(f, [$1..100]); # Robert Israel, Sep 22 2015

cubes=Range[85]^3; Table[Position[Divisible[cubes,i],True,1,1][[1,1]],{i,85}] (* Harvey P. Dale, Jan 12 2011 *)
f[p_, e_] := p^Ceiling[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]  (* Amiram Eldar, Jan 06 2024 *)

a(n)=my(r=1);while(r^3%n!=0,r++);r \\ Anders Hellström, Sep 22 2015

for(n=1, 100, print1(direuler(p=2, n, (1 + p*X + p*X^2)/(1 - p*X^3))[n], ", ")) \\ Vaclav Kotesovec, Aug 30 2021

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^ceil(f[i,2]/3));} \\ Amiram Eldar, Jan 06 2024

from math import prod
from sympy import factorint
def A019555(n): return prod(p**((q%3 != 0)+(q//3)) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

[prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1,79)] # Danny Rorabaugh, Sep 22 2015

Replace any cubic factors in n by their cube roots.
a(n) = n/A000189(n).
Multiplicative with a(p^e) = p^ceiling(e/3). - R. J. Mathar, May 29 2011
From Vaclav Kotesovec, Aug 30 2021: (Start)
Dirichlet g.f.: zeta(3*s-1) * Product_{p prime} (1 + p^(1 - s) + p^(1 - 2*s)).
Dirichlet g.f.: zeta(3*s-1) * zeta(s-1) * Product_{p prime} (1 - p^(2 - 3*s) + p^(1 - 2*s) - p^(2 - 2*s)).
Sum_{k=1..n} a(k) ~ c * zeta(5) * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.684286924186862318141968725791218083472312736723163777284618226290055... (End)

Corrected and extended by David W. Wilson

A053166 Smallest positive integer for which n divides a(n)^4.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38

Offset: 1

Henry Bottomley, Feb 29 2000

According to Broughan (2002, 2003, 2006), a(n) is the "upper 4th root of n". The "lower 4th root of n" is sequence A053164. - Petros Hadjicostas, Sep 15 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Cf. A000189, A000190, A008835, A013665, A015051.

f[p_, e_] := p^Ceiling[e/4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/4)); factorback(f); \\ Michel Marcus, Jun 09 2014

a(n) = n/A000190(n) = A019554(n)/(A008835(A019554(n)^2))^(1/4).
If n is 5th-power-free (i.e., not 32, 64, 128, 243, ...) then a(n) = A007947(n).
Multiplicative with a(p^e) = p^(ceiling(e/4)). - Christian G. Bower, May 16 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(7)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3528057925... . - Amiram Eldar, Oct 27 2022

A053149 Smallest cube divisible by n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 64, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 64, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648

Offset: 1

Henry Bottomley, Feb 28 2000

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

Cf. A000188, A000189, A007913, A008834, A019554, A048798, A050985.

Cf. A002117, A013667, A330596.

a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c]]]; Table[a[n], {n, 1, 44}] (* Jean-François Alcover, Sep 03 2012 *)
f[p_, e_] := p^(e + Mod[3 - e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
scdn[n_]:=Module[{c=Ceiling[Surd[n,3]]},While[!Divisible[c^3,n],c++];c^3]; Array[scdn,50] (* Harvey P. Dale, Jun 13 2020 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(f[i,2] + (3-f[i,2])%3));} \\ Amiram Eldar, Oct 27 2022

a(n) = (n/A000189(n))^3 = A008834(n)*A019554(A050985(n))^3 = n*A050985(n)^2/A000188(A050985(n))^3.
a(n) = n * A048798(n). - Franklin T. Adams-Watters, Apr 08 2009
From Amiram Eldar, Jul 29 2022: (Start)
Multiplicative with a(p^e) = p^(e + ((3-e) mod 3)).
Sum_{n>=1} 1/a(n) = Product_{p prime} ((p^3+2)/(p^3-1)) = 1.655234386560802506... . (End)
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(9)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013667*A330596/(4*A002117) = 0.1559906... . - Amiram Eldar, Oct 27 2022

A048798 Smallest k > 0 such that n*k is a perfect cube.

Original entry on oeis.org

1, 4, 9, 2, 25, 36, 49, 1, 3, 100, 121, 18, 169, 196, 225, 4, 289, 12, 361, 50, 441, 484, 529, 9, 5, 676, 1, 98, 841, 900, 961, 2, 1089, 1156, 1225, 6, 1369, 1444, 1521, 25, 1681, 1764, 1849, 242, 75, 2116, 2209, 36, 7, 20, 2601, 338, 2809, 4, 3025, 49, 3249

Offset: 1

Charles T. Le (charlestle(AT)yahoo.com)

Note that in general the smallest number k(>0) such that n*k is a perfect m-th power (rather obviously) = (the smallest m-th power divisible by n)/n and also (slightly less obviously) =n^(m-1)/(the number of solutions of x^m==0 mod n)^m. - Henry Bottomley, Mar 03 2000

			a(12) = a(2*2*3) = 2*3*3 = 18 since 12*18 = 6^3.
a(28) = a(2*2*7) = 2*7*7 = 98 since 28*98 = 14^3.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Peter Kagey)
Krassimir T. Atanassov, On the 22nd, the 23rd and 24th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5, No. 2 (1998), pp. 80-82.
Krassimir T. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Marcela Popescu and Mariana Nicolescu, About the Smarandache Complementary Cubic Function, Smarandache Notions Journal, Vol. 7, No. 1-2-3, 1996, pp. 54-62.
Florentin Smarandache, Only Problems, Not Solutions! (see Unsolved Problem: 28, p. 26).

Cf. A000189, A007913, A053149.
Cf. A254767 (analogous sequence with the restriction that k > n).
Cf. A002117, A013667.

a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c/n]]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Sep 03 2012 *)
f[p_, e_] := p^(Mod[-e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 10 2020 *)
With[{cbs=Range[3300]^3},Table[SelectFirst[cbs,Mod[#,n]==0&]/n,{n,60}]] (* Harvey P. Dale, May 10 2024 *)

a(n)=my(f=factor(n));prod(i=1,#f[,1],f[i,1]^(-f[i,2]%3)) \\ Charles R Greathouse IV, Feb 27 2013

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

from math import prod
from sympy import factorint
def A048798(n): return prod(p**(-e%3) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 05 2024

a(n) = A053149(n)/n = n^2/A000189(n)^3.
Multiplicative with a(p^e) = p^((-e) mod 3). - Mitch Harris, May 17 2005
Sum_{k=1..n} a(k) ~ c * n^3, where c = (zeta(9)/(3*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = 0.2079875504... . - Amiram Eldar, Oct 28 2022

More terms from Patrick De Geest, Feb 15 2000

A062378 n divided by largest cubefree factor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Henry Bottomley, Jun 18 2001

Numerator of n/rad(n)^2, where rad is the squarefree kernel of n (A007947), denominator: A055231. - Reinhard Zumkeller, Dec 10 2002

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

Cf. A000189, A000578, A007948, A008834, A019555, A048798, A050985, A053149, A053150, A056551, A056552. See A003557 for squares and A062379 for 4th powers.
Differs from A073753 for the first time at n=90, where a(90) = 1, while A073753(90) = 3.
Cf. A007947, A055231.

f[p_, e_] := p^Max[e-2, 0]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)

a(n)=my(f=factor(n));prod(i=1,#f~,f[i,1]^max(f[i,2]-2,0)) \\ Charles R Greathouse IV, Aug 08 2013

(define (A062378 n) (/ n (A007948 n)))
(definec (A007948 n) (if (= 1 n) n (* (expt (A020639 n) (min 2 (A067029 n))) (A007948 (A028234 n)))))
;; Antti Karttunen, Nov 28 2017

a(n) = n / A007948(n).
a(n) = A003557(A003557(n)). - Antti Karttunen, Nov 28 2017
Multiplicative with a(p^e) = p^max(e-2, 0). - Amiram Eldar, Sep 07 2020
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^s - 1/p^(2*s-1) + 1/p^(2*s)). - Amiram Eldar, Dec 07 2023

A056552 Powerfree kernel of cubefree part of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 1, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 3, 5, 26, 1, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 5, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 2, 55, 7, 57, 58, 59, 30, 61, 62, 21, 1, 65, 66, 67, 34, 69, 70, 71, 3, 73, 74, 15, 38, 77

Offset: 1

Henry Bottomley, Jun 25 2000

			a(32) = 2 because cubefree part of 32 is 4 and powerfree kernel of 4 is 2.

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

Cf. A000189, A000578, A007947, A008834, A013664, A019555, A048798, A050985, A053149, A053150, A056551.

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

a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/3), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

a(n) = A007947(A050985(n)) = A019555(A050985(n)) = n/(A053150(n)*A000189(n)) = A019555(n)/A053150(n) = A056551(n)^(1/3).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 3 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 3|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(6)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.3480772773... . - Amiram Eldar, Oct 28 2022
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-1)). - Amiram Eldar, Sep 16 2023

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

cp's OEIS Frontend

A003557 n divided by largest squarefree divisor of n; if n = Product p(k)^e(k) then a(n) = Product p(k)^(e(k)-1), with a(1) = 1.

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A000188 (1) Number of solutions to x^2 == 0 (mod n). (2) Also square root of largest square dividing n. (3) Also max_{ d divides n } gcd(d, n/d).

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A053150 Cube root of largest cube dividing n.

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A019555 Smallest number whose cube is divisible by n.

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A053166 Smallest positive integer for which n divides a(n)^4.

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