A019554 Smallest number whose square is divisible by 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, 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

Author

R. Muller

Keywords

• nonn
• easy
• mult
• nice

Comments

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

Links

• 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.

Crossrefs

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.

Programs

Haskell

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

Maple

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

Mathematica

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 *)

PARI

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

Formula

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)