A048943 - OEIS

%I #37 Aug 05 2023 23:46:37

%S 1,6,8,10,14,15,16,21,22,24,26,27,30,33,34,35,38,39,40,42,46,51,54,55,

%T 56,57,58,60,62,65,66,69,70,72,74,77,78,81,82,84,85,86,87,88,90,91,93,

%U 94,95,96,102,104,105,106,108,110,111,114,115,118,119,120

%N Product of divisors of n is a square.

%C From _Gerard P. Michon_, Oct 10 2010: (Start)

%C If d is the number of divisors of N, a prime factor of N with multiplicity k in N has a multiplicity kd/2 in the product of all divisors of N (including N itself). Therefore the latter is a square if and only if kd/2 is always even (which is to say that kd is a multiple of 4 for any multiplicity k of a prime factor of N). This happens when one of the following three conditions hold:

%C - N is a fourth power (all the multiplicities are then multiples of 4 and d is odd).

%C - N has at least two prime factors with odd multiplicities.

%C - N has (at least) one prime factor with a multiplicity congruent to 3 modulo 4.

%C (End)

%C It follows from the comment above that if two terms are: a) even and odd, or b) terms of A006881, or c) terms of A000583, then their product is also a term. - _Ivan N. Ianakiev_, Jul 02 2023

%H T. D. Noe, <a href="/A048943/b048943.txt">Table of n, a(n) for n = 1..1000</a>

%H G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#productdiv">Divisor Product</a>, Numericana.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DivisorProduct.html">Divisor Product</a>

%e From _Gerard P. Michon_, Oct 10 2010: (Start)

%e a(1) = 1 because it's a fourth power. The product of all divisors of 1 is 1, which is a square.

%e a(2) = 6 because 2^1.3^1 is the product of two primes with odd multiplicities (1 in both cases). Indeed, the divisor product 1.2.3.6 = 36 is a square.

%e a(3) = 8 because 2 is a prime factor of 8 with multiplicity 3. Indeed, 1.2.4.8 = 64 is a square.

%e a(7) = 16 because it's a fourth power; 1.2.4.8.16 = 1024 is the square of 32. (End)

%t Select[Range[125], IntegerQ[Sqrt[Times @@ Divisors[#]]] &] (* _T. D. Noe_, Apr 30 2012 *)

%o (PARI) {for(k=1, 126, mpc=1;

%o M=divisors(k);

%o for(i=1, matsize(M)[2], mpc=mpc*M[i]);

%o if(issquare(mpc), print1(k, ", ")))} \\\ _Douglas Latimer_, Apr 30 2012

%o (PARI) is(n)=my(f=factor(n)[,2]); gcd(f)%4==0 || #select(k->k%2, f)>1 || #select(k->k%4==3, f) \\ _Charles R Greathouse IV_, Sep 18 2015

%o (Sage) [n for n in (1..125) if prod(divisors(n)).is_square()] # _Giuseppe Coppoletta_, Dec 16 2014

%o (Python)

%o from sympy import divisor_count

%o from gmpy2 import iroot

%o A048943_list = [i for i in range(1,10**3) if iroot(i,4)[1] or not divisor_count(i) % 4] # _Chai Wah Wu_, Mar 10 2016

%Y Supersequence of A229153.

%K nonn

%O 1,2

%A _Eric W. Weisstein_