

A013929


Numbers that are not squarefree. Numbers that are divisible by a square greater than 1. The complement of A005117.


463



4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 28, 32, 36, 40, 44, 45, 48, 49, 50, 52, 54, 56, 60, 63, 64, 68, 72, 75, 76, 80, 81, 84, 88, 90, 92, 96, 98, 99, 100, 104, 108, 112, 116, 117, 120, 121, 124, 125, 126, 128, 132, 135, 136, 140, 144, 147, 148, 150, 152, 153, 156, 160
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OFFSET

1,1


COMMENTS

Sometimes misnamed squareful numbers, but officially those are given by A001694.
This is different from the sequence of numbers k such that A007913(k) < phi(k). The two sequences differ at the values: 420, 660, 780, 840, 1320, 1560, 4620, 5460, 7140, ..., which is essentially A070237.  Ant King, Dec 16 2005
Numbers k such that Sum_{dk} (d/phi(d))*mu(k/d) = 0.  Benoit Cloitre, Apr 28 2002
Numbers k for which there exists a partition into two parts p and q such that p + q = k and p*q is a multiple of k.  Amarnath Murthy, May 30 2003
Numbers k such that there is a solution 0 < x < k to x^2 == 0 (mod k).  Franz Vrabec, Aug 13 2005
Numbers k such that moebius(k) = 0.
a(n) = k such that phi(k)/k = phi(m)/m for some m < k.  Artur Jasinski, Nov 05 2008
Appears to be numbers such that when a column with index equal to a(n) in A051731 is deleted, there is no impact on the result in the first column of A054525.  Mats Granvik, Feb 06 2009
Numbers k such that the number of prime divisors of (k+1) is less than the number of nonprime divisors of (k+1).  JuriStepan Gerasimov, Nov 10 2009
Orders for which at least one noncyclic finite abelian group exists: A000688(a(n)) > 1. This follows from the fact that not all exponents in the prime factorization of a(n) are 1 (moebius(a(n)) = 0). The number of such groups of order a(n) is A192005(n) = A000688(a(n))  1.  Wolfdieter Lang, Jul 29 2011
It appears that terms are the numbers m such that Product_{k=1..m} (prime(k) mod m) <> 0. See Maple code.  Gary Detlefs, Dec 07 2011
Numbers k such that A001222(k) > A001221(k), since in this case at least one prime factor of k occurs more than once, which implies that k is divisible by at least one perfect square > 1.  Carlos Eduardo Olivieri, Aug 02 2015
Lexicographically least sequence such that each term has a positive even number of proper divisors not occurring in the sequence, cf. the sieve characterization of A005117.  Glen Whitney, Aug 30 2015
There are arbitrarily long runs of consecutive terms. Record runs start at 4, 8, 48, 242, ... (A045882).  Ivan Neretin, Nov 07 2015
Every squareful number > 1 is nonsquarefree, but the converse is false and the nonsquarefree numbers that are not squareful (see first comment) are in A332785.  Bernard Schott, Apr 11 2021
Integers m where at least one k < m exists such that m divides k^m.  Richard R. Forberg, Jul 31 2021
Consider the Diophantine equation S(x,y) = (x+y) + (xy) + (x*y) + (x/y) = z, when x and y are both positive integers with y  x. Then, there is a solution (x,y) iff z is a term of this sequence; in this case, if x = K*y, then z = S(K*y,y) = K*(y+1)^2 (see A351381, link and references Perelman); example: S(12,4) = 75 = a(28). The number of solutions for S(x,y) = a(n) is A353282(n).  Bernard Schott, Mar 29 2022
For each positive integer m, the number of unitary divisors of m = the number of squarefree divisors of m (see A034444); but only for the terms of this sequence does the set of unitary divisors differ from the set of squarefree divisors. Example: the set of unitary divisors of 20 is {1, 4, 5, 20}, while the set of squarefree divisors of 20 is {1, 2, 5, 10}.  Bernard Schott, Oct 15 2022


REFERENCES

I. Perelman, L'Algèbre récréative, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101102.
Ya. I. Perelman, Algebra can be fun, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131132.


LINKS

Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131132.


FORMULA



EXAMPLE

For the terms up to 20, we compute the squares of primes up to floor(sqrt(20)) = 4. Those squares are 4 and 9. For every such square s, put the terms s*k^2 for k = 1 to floor(20 / s). This gives after sorting and removing duplicates the list 4, 8, 9, 12, 16, 18, 20.  David A. Corneth, Oct 25 2017


MAPLE

a := n > `if`(numtheory[mobius](n)=0, n, NULL); seq(a(i), i=1..160); # Peter Luschny, May 04 2009
t:= n> product(ithprime(k), k=1..n): for n from 1 to 160 do (if t(n) mod n <>0) then print(n) fi od; # Gary Detlefs, Dec 07 2011
with(NumberTheory): isQuadrateful := n > irem(Radical(n), n) <> 0:
select(isQuadrateful, [`$`(1..160)]); # Peter Luschny, Jul 12 2022


MATHEMATICA

Union[ Flatten[ Table[ n i^2, {i, 2, 20}, {n, 1, 400/i^2} ] ] ]
Select[ Range[2, 160], (Union[Last /@ FactorInteger[ # ]][[ 1]] > 1) == True &] (* Robert G. Wilson v, Oct 11 2005 *)


PROG

(PARI) {a(n)= local(m, c); if(n<=1, 4*(n==1), c=1; m=4; while( c<n, m++; if(!issquarefree(m), c++)); m)} /* Michael Somos, Apr 29 2005 */
(PARI) for(n=1, 1e3, if(omega(n)!=bigomega(n), print1(n, ", "))) \\ Felix Fröhlich, Aug 13 2014
(PARI) upto(n)=my(res = List()); forprime(p = 2, sqrtint(n), for(k = 1, n \ p^2, listput(res, k * p^2))); listsort(res, 1); res \\ David A. Corneth, Oct 25 2017
(Magma) [ n : n in [1..1000]  not IsSquarefree(n) ];
(Haskell)
a013929 n = a013929_list !! (n1)
a013929_list = filter ((== 0) . a008966) [1..]
(Python)
from sympy.ntheory.factor_ import core
def ok(n): return core(n, 2) != n
(Python)
from math import isqrt
from sympy import mobius
def f(x): return n+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
m, k = n, f(n)
while m != k:
m, k = k, f(k)


CROSSREFS



KEYWORD

nonn,easy,changed


AUTHOR



EXTENSIONS



STATUS

approved



