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Integers of the form n/tau(n)^2 as n runs through the integers, where tau(n) is the number of divisors of n.
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%I #14 Aug 07 2019 05:00:26

%S 1,1,2,25,3,2,5,9,25,81,6,4,16,3,5,28,36,12,128,81,81,24,7,40,8,21,28,

%T 50,12,11,16,12,96,49,13,35,44,2401,52,45,17,36,19,160,225,68,63,23,

%U 76,30,28,36,72,21,224,92,29,77,121,31,18,27,30,91,99,116,128,37,124

%N Integers of the form n/tau(n)^2 as n runs through the integers, where tau(n) is the number of divisors of n.

%H Amiram Eldar, <a href="/A226325/b226325.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A046754(n) / A000005(A046754(n))^2.

%e A046754(3) = 128, A000005(128) = 8, so a(3) = 128 / 8^2 = 2.

%p for n from 1 to 1000000 do

%p r := n/(numtheory[tau](n))^2 ;

%p if type(r,'integer') then

%p printf("%d,",r);

%p end if;

%p end do: # _R. J. Mathar_, Jun 07 2013

%t Select[Table[n/DivisorSigma[0,n]^2,{n,10^6}],IntegerQ] (* _Harvey P. Dale_, Jan 01 2015 *)

%Y Cf. A000005, A033950, A046754.

%Y Cf. A036762 (if d(n) divides n, then n/d(n) is appended to the sequence).

%K nonn

%O 1,3

%A _Alex Ratushnyak_, Jun 04 2013