A019555 -id:A019555 - cp's OEIS frontend

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).

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)

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

A000189 Number of solutions to x^3 == 0 (mod n).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

Shadow transform of the cubes A000578. - Michel Marcus, Jun 06 2013

			a(4) = 2 because 0^3 == 0, 1^3 == 1, 2^3 == 0, and 3^3 == 3 (mod 4); also, a(9) = 3 because 0^3 = 0, 3^3 == 0, and 6^3 = 0 (mod 9), while x^3 =/= 0 (mod 9) for x = 1, 2, 4, 5, 7, 8. - _Petros Hadjicostas_, Sep 16 2019

T. D. Noe, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
Lorenz Halbeisen and Norbert Hungerbuehler, Number theoretic aspects of a combinatorial function, Notes on Number Theory and Discrete Mathematics 5(4) (1999), 138-150. (ps, pdf); see Definition 7 for the shadow transform.
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms).
OEIS Wiki, Shadow transform.
N. J. A. Sloane, Transforms.

Cf. A000578, A019555.

Array[ Function[ n, Count[ Array[ PowerMod[ #, 3, n ]&, n, 0 ], 0 ] ], 100 ]
f[p_, e_] := p^Floor[2*e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 19 2020 *)

a(n)=my(f=factor(n));prod(i=1,#f[,1],f[i,1]^(2*f[i,2]\3)) \\ Charles R Greathouse IV, Jun 06 2013

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

Multiplicative with a(p^e) = p^[2e/3]. - David W. Wilson, Aug 01 2001
a(n) = n/A019555(n). - Petros Hadjicostas, Sep 15 2019
Dirichlet g.f.: zeta(3*s-2) * Product_{p prime} (1 + 1/p^s + 1/p^(2*s-1)). - Amiram Eldar, Sep 09 2023
From Vaclav Kotesovec, Sep 09 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-1) * zeta(3*s-2) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s-1) - 1/p^(4*s-2) + 1/p^(4*s-1) + 1/p^(5*s-2)).
Let f(s) = Product_{primes p} (1 - 1/p^(2*s) - 1/p^(3*s-1) - 1/p^(4*s-2) + 1/p^(4*s-1) + 1/p^(5*s-2)).
Sum_{k=1..n} a(k) ~ (f(1)*n/6) * (log(n)^2/2 + (6*gamma - 1 + f'(1)/f(1))*log(n) + 1 - 6*gamma + 11*gamma^2 - 14*sg1 + (6*gamma - 1)*f'(1)/f(1) + f''(1)/(2*f(1))), where
f(1) = Product_{primes p} (1 - 3/p^2 + 2/p^3) = A065473 = 0.2867474284344787341078927127898384464343318440970569956414778593366522431...,
f'(1) = f(1) * Sum_{primes p} 9*log(p) / (p^2 + p - 2) = f(1) * 4.1970213428422788650375569145777616746065054412058004220013841318980729375...,
f''(1) = f'(1)^2/f(1) + f(1) * Sum_{primes p} (-29*p^2 - 17*p + 1) * log(p)^2 / (p^2 + p - 2)^2 = f'(1)^2/f(1) + f(1) * (-21.3646716550082193262514333696570765444176783899223644201265894338042468...),
gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). (End)

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

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

A015052 a(n) is the smallest positive integer m such that m^5 is divisible by n.

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, 2, 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, 77, 78

Offset: 1

R. Muller

A multiplicative companion function n/a(n) = 1,1,1,2,1,1,1,4,3,1,1,2,1,1,1,8,1,... can be defined using the 5th instead of the 4th power in A000190, which differs from A000190 and also from A003557. - R. J. Mathar, Jul 14 2012

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, 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.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
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), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015053 (outer 6th root).

Cf. A013667.

f[p_, e_] := p^Ceiling[e/5]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/5)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^(ceiling(e/5)). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(9)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8) = 0.3523622369... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

Name reworded by Jon E. Schoenfield, Oct 28 2022

A015053 Smallest positive integer for which n divides a(n)^6.

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, 2, 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, 2, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

R. Muller (Research37(AT)aol.com)

Differs from A007947 as follows: A007947(128)=2, a(128)=4; A007947(256)=2, a(256)=4; A007947(384)=6, a(384)=12; A007947(512)=2, a(512)=4; A007947(640)=10, a(640)=20, etc. - R. J. Mathar, Oct 28 2008

Harvey P. Dale, Table of n, a(n) for n = 1..1000
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.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
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), A053166 (outer 4th root), A015052 (5th outer root).

Cf. A013669.

spi[n_]:=Module[{k=1},While[PowerMod[k,6,n]!=0,k++];k]; Array[spi,80] (* Harvey P. Dale, Feb 29 2020 *)
f[p_, e_] := p^Ceiling[e/6]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/6)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^ceiling(e/6). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(11)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8 + 1/p^9 - 1/p^10) = 0.3522558764... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

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

A365488 The number of divisors of the smallest number whose cube is divisible by n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 3, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 3, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4

Offset: 1

Amiram Eldar, Sep 05 2023

First differs from A365171 at n = 32.

The number of divisors of the smallest cube divisible by n, A053149(n), is A365489(n).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)

Cf. A000005, A005117, A019555, A053149, A365171, A365489, A365498.

f[p_, e_] := Ceiling[e/3] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
With[{c=Range[200]^3},Table[DivisorSigma[0,Surd[SelectFirst[c,Mod[#,n]==0&],3]],{n,90}]] (* Harvey P. Dale, Sep 15 2024 *)

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

a(n) = A000005(A019555(n)).
Multiplicative with a(p^e) = ceiling(e/3) + 1.
a(n) <= A000005(n) with equality if and only if n is squarefree  (A005117).
Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)).
From Vaclav Kotesovec, Sep 06 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * zeta(3*s) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ zeta(3) * f(1) * n * (log(n) + 2*gamma - 1 + 3*zeta'(3)/zeta(3) + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.5358961538283379998085026313185459506482223745141452711510108346133288...,
f'(1) = f(1) * Sum_{p prime} (-4 + 3*p + 2*p^2) * log(p) / (1 - p - p^2 + p^4) = f(1) * 1.4525924794451595590371439593828547341482465114411929136723476679...
and gamma is the Euler-Mascheroni constant A001620. (End)

cp's OEIS Frontend

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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A053150 Cube root of largest cube dividing n.

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A000189 Number of solutions to x^3 == 0 (mod n).

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

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A062378 n divided by largest cubefree factor of n.

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