A053150 -id:A053150 - cp's OEIS frontend

A355263 a(n) = largest-nth-power(n, 3) * radical(n) = A053150(n) * A007947(n), where the largest-nth-power(n, e) is the largest positive integer b such that b^e divides n.

1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 20, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 18, 55, 28, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69

Offset: 1

Peter Luschny, Jul 12 2022

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

Cf. A000188, A013663, A053150, A064549, A355261.

with(NumberTheory): seq(LargestNthPower(n, 3)*Radical(n), n=1..69);

f[p_, e_] := p^(1 + Floor[e/3]); a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 13 2022 *)

from math import prod
from sympy import factorint
def A355263(n): return prod(p**(e//3+1) for p, e in factorint(n).items()) # Chai Wah Wu, Jul 13 2022

Multiplicative with a(p^e) = p^(1 + floor(e/3)). - Amiram Eldar, Jul 13 2022

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(5)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 2/p^5 + 1/p^6) = 0.3643121583... . - Amiram Eldar, Nov 13 2022

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

A019554 Smallest number whose square is divisible by n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 8, 33, 34, 35, 6, 37, 38, 39, 20, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 18, 55, 28, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69, 70, 71, 12, 73, 74, 15, 38, 77

Offset: 1

R. Muller

A note on square roots of numbers: 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.
Instead of the terms "inner square root" and "outer square root", we may use the terms "lower square root" and "upper square root", respectively. Upper k-th roots have been studied by Broughan (2002, 2003, 2006). - Petros Hadjicostas, Sep 15 2019
The number of times each number k appears in this sequence is A034444(k). The first time k appears is at position A102631(k). - N. J. A. Sloane, Jul 28 2021

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, Acta Arithmetica, 101(2) (2002), 105-114. See also here for another copy.
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.
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), 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. also A007913, A007947, A008833, A015049, A034444, A053143, A102631, A346602, A005117, A133466, A195085.

a019554 n = product $ zipWith (^)
            (a027748_row n) (map ((`div` 2) . (+ 1)) $ a124010_row n)
-- Reinhard Zumkeller, Apr 13 2013
(Python 3.8+)
from math import prod
from sympy import factorint
def A019554(n): return n//prod(p**(q//2) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021

with(numtheory):A019554 := proc(n) local i: RETURN(op(mul(i,i=map(x->x[1]^ceil(x[2]/2),ifactors(n)[2])))); end;

Flatten[Table[Select[Range[n],Divisible[#^2,n]&,1],{n,100}]] (* Harvey P. Dale, Oct 17 2011 *)
f[p_, e_] := p^Ceiling[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n)=n/core(n,1)[2] \\ Charles R Greathouse IV, Feb 24 2011

Replace any square factors in n by their square roots.
Multiplicative with a(p^e) = p^ceiling(e/2).
Dirichlet series:
   Sum_{n>=1} a(n)/n^s = zeta(2*s-1)*zeta(s-1)/zeta(2*s-2), (Re(s) > 2);
   Sum_{n>=1} (1/a(n))/n^s = zeta(2*s+1)*zeta(s+1)/zeta(2*s+2), (Re(s) > 0).
a(n) = n/A000188(n).
a(n) = denominator of n/n^(3/2). - Arkadiusz Wesolowski, Dec 04 2011
a(n) = Product_{k=1..A001221(n)} A027748(n,k)^ceiling(A124010(n,k)/2). - Reinhard Zumkeller, Apr 13 2013
Sum_{k=1..n} a(k) ~ 3*zeta(3)*n^2 / Pi^2. - Vaclav Kotesovec, Sep 18 2020
Sum_{k=1..n} 1/a(k) ~ 3*log(n)^2/(2*Pi^2) + (9*gamma/Pi^2 - 36*zeta'(2)/Pi^4)*log(n) + 6*gamma^2/Pi^2 - 108*gamma*zeta'(2)/Pi^4 + 432*zeta'(2)^2/Pi^6 - 36*zeta''(2)/Pi^4 - 15*sg1/Pi^2, where gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Jul 27 2021
a(n) = sqrt(n*A007913(n)). - Jianing Song, May 08 2022
a(n) = sqrt(A053143(n)). - Amiram Eldar, Sep 02 2023
From Mia Boudreau, Jul 17 2025: (Start)
a(n^2) = n.
a(A005117(n)) = A005117(n).
a(A133466(n)) = A133466(n)/2.
a(A195085(n)) = A195085(n)/3. (End)

A061704 Number of cubes 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, 2, 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, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 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, Jun 18 2001

			a(128) = 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
Index entries for sequences computed from exponents in factorization of n

Cf. A000005, A000578, A046951, A053150.

N:= 1000: # to get a(1)..a(N)
G:= add(x^(n^3)/(1-x^(n^3)),n=1..floor(N^(1/3))):
S:= series(G,x,N+1):
seq(coeff(S,x,j),j=1..N); # Robert Israel, Jul 28 2017
# alternative
A061704 := proc(n)
    local a,pe ;
    a := 1 ;
    for pe in ifactors(n)[2] do
        op(2,pe) ;
        a := a*(1+floor(%/3)) ;
    end do:
    a ;
end proc:
seq(A061704(n),n=1..80) ; # R. J. Mathar, May 10 2023

nn = 100; f[list_, i_]:= list[[i]]; Table[ DirichletConvolve[ f[ Boole[ Map[ IntegerQ[#] &, Map[#^(1/3) &, Range[nn]]]], n],f[Table[1, {nn}], n], n, m], {m, 1, nn}] (* Geoffrey Critzer, Feb 07 2015 *)
Table[DivisorSum[n, 1 &, IntegerQ[#^(1/3)] &], {n, 105}] (* Michael De Vlieger, Jul 28 2017 *)
f[p_, e_] := 1 + Floor[e/3]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)

a(n) = sumdiv(n, d, ispower(d, 3)); \\ Michel Marcus, Jan 31 2015

from math import prod
from sympy import factorint
def A061704(n): return prod(e//3+1 for e in factorint(n).values()) # Chai Wah Wu, Jun 05 2025

Multiplicative with a(p^e) = floor(e/3) + 1. - Mitch Harris, Apr 19 2005
G.f.: Sum_{n>=1} x^(n^3)/(1-x^(n^3)). - Joerg Arndt, Jan 30 2011
a(n) = A000005(A053150(n)).
Dirichlet g.f.: zeta(3*s)*zeta(s). - Geoffrey Critzer, Feb 07 2015
Sum_{k=1..n} a(k) ~ zeta(3)*n + zeta(1/3)*n^(1/3). - Vaclav Kotesovec, Dec 01 2020
a(n) = Sum_{k=1..n} (1 - ceiling(n/k^3) + floor(n/k^3)). - Wesley Ivan Hurt, Jan 28 2021

A053164 4th root of largest 4th power dividing n.

Original entry on oeis.org

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

Offset: 1

Henry Bottomley, Feb 29 2000

Multiplicative with a(p^e) = p^[e/4]. - Mitch Harris, Apr 19 2005

			a(32) = 2 since 2 = 16^(1/4) and 16 is the largest 4th power dividing 32.

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

Cf. A000188, A000190, A008835, A053150.

A053164 := proc(n) local a,f,e,p ; for f in ifactors(n)[2] do e:= op(2,f) ; p := op(1,f) ; a := a*p^floor(e/4) ; end do ; a ; end proc: # R. J. Mathar, Jan 11 2012

f[list_] := list[[1]]^Quotient[list[[2]], 4]; Table[Apply[Times, Map[f,FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)

a(n) = A000188(A000188(n)) = A008835(n)^(1/4).
Multiplicative with a(p^e) = p^[e/4].
Dirichlet g.f.: zeta(4s-1)*zeta(s)/zeta(4s). - R. J. Mathar, Apr 09 2011
Sum_{k=1..n} a(k) ~ 90*zeta(3)*n/Pi^4 + 3*zeta(1/2)*sqrt(n)/Pi^2. - Vaclav Kotesovec, Dec 01 2020
a(n) = Sum_{d^4|n} phi(d). - Ridouane Oudra, Dec 31 2020
G.f.: Sum_{k>=1} phi(k) * x^(k^4) / (1 - x^(k^4)). - Ilya Gutkovskiy, Aug 20 2021

More terms from Antti Karttunen, Sep 13 2017

A008834 Largest cube dividing 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, 8, 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, 64, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 27

Offset: 1

N. J. A. Sloane

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Vaclav Kotesovec, Graph - the asymptotic ratio.
Eric Weisstein's World of Mathematics, Cubic Part.

Cf. A000203, A050985, A053150.

with(numtheory): [ seq( expand(nthpow(i,3)),i=1..200) ];
# alternative:
A008834 := proc(n)
    local p;
    a := 1 ;
    for p in ifactors(n)[2] do
        e := floor(op(2,p)/3) ;
        a := a*op(1,p)^(3*e) ;
    end do:
    a ;
end proc:
seq(A008834(n),n=1..40) ; # R. J. Mathar, Dec 08 2015

a[n_] := Times @@ (#[[1]]^(#[[2]] - Mod[#[[2]], 3]) & ) /@ FactorInteger[n]; Table[a[n], {n, 1, 81}]
(* Jean-François Alcover, Jul 31 2011, after PARI prog. *)
  upto=1000;Flatten[With[{c=Range[Floor[Surd[upto,3]],1,-1]^3}, Table[ Select[ c,Divisible[n,#]&,1],{n,upto}]]](* Harvey P. Dale, Apr 07 2013 *)

a(n)=n=factor(n);prod(i=1,#n[,1],n[i,1]^(n[i,2]\3*3)) \\ Charles R Greathouse IV, Jul 28 2011

from math import prod
from sympy import factorint
def A008834(n): return prod(p**(e-e%3) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 08 2024

Multiplicative with a(p^e) = p^(3[e/3]). - Mitch Harris, Apr 19 2005
a(n) = A053150(n)^3. - R. J. Mathar, May 27 2011
Dirichlet g.f.: zeta(s)*zeta(3s-3)/zeta(3s). The Dirichlet convolution of this sequence with A050985 generates A000203. - R. J. Mathar, Apr 05 2011
Sum_{k=1..n} a(k) ~ 45 * zeta(4/3) * n^(4/3) / (2*Pi^4). - Vaclav Kotesovec, Jan 31 2019
a(n) = n/A050985(n). - Amiram Eldar, Aug 15 2023

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

A295659 Number of exponents larger than 2 in the prime factorization of n.

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, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Antti Karttunen, Nov 28 2017

nonn

			For n = 120 = 2^3 * 3^1 * 5^1 there is only one exponent larger than 2, thus a(120) = 1.
For n = 216 = 2^3 * 3^3 there are two exponents larger than 2, thus a(216) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A001221, A003557, A053150, A056169, A056170, A085541, A162642, A295657, A295662, A295883, A295884.

Cf. A004709 (positions of zeros), A046099 (of nonzeros), A212793.

Array[Count[FactorInteger[#][[All, -1]], ?(# > 2 &)] &, 105] (* _Michael De Vlieger, Nov 28 2017 *)

a(n) = { my(v = factor(n)[, 2], i=0); for(x=1, length(v), if(v[x]>2, i++)); i; } \\ Iain Fox, Nov 29 2017

Additive with a(p^e) = 1 if e>2, 0 otherwise.
a(n) = 0 iff A212793(n) = 1.
a(n) = A001221(A053150(n)).
a(n) = A056170(A003557(n)).
a(n) >= A295662(n) = A162642(n) - A056169(n).
a(n) = A295883(n) + A295884(n).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^3 = 0.174762... (A085541). - Amiram Eldar, Nov 01 2020

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

cp's OEIS Frontend

A355263 a(n) = largest-nth-power(n, 3) * radical(n) = A053150(n) * A007947(n), where the largest-nth-power(n, e) is the largest positive integer b such that b^e divides n.

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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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A019554 Smallest number whose square is divisible by n.

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A061704 Number of cubes dividing n.

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A053164 4th root of largest 4th power dividing n.

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A008834 Largest cube dividing n.

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