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Numbers whose divisors have an integer mean number of divisors.
5

%I #8 Jul 23 2024 20:46:56

%S 1,4,9,12,16,18,20,25,28,36,44,45,49,50,52,63,64,68,72,75,76,81,92,98,

%T 99,100,108,116,117,121,124,144,147,148,153,164,169,171,172,175,180,

%U 188,192,196,200,207,212,225,236,242,244,245,252,256,261,268,275,279

%N Numbers whose divisors have an integer mean number of divisors.

%C Numbers k such that A000005(k) | A007425(k).

%C Numbers k such that A374903(k) = 1.

%C If k is a term then all the numbers with the same prime signature as k are terms. The least terms of each prime signature are in A374905.

%C If {e_i} are the exponents in the prime factorization of k, then k is a term if and only if Product_{i} (e_i/2 + 1) is an integer.

%C 1 is the only squarefree (A005117) term.

%C All the squares are terms.

%H Amiram Eldar, <a href="/A374904/b374904.txt">Table of n, a(n) for n = 1..10000</a>

%e 4 is a term since it has 3 divisors, 1, 2 and 4, their numbers of divisors are 1, 2 and 3, and their mean is (1 + 2 + 3)/3 = 2 which is an integer.

%t f[p_, e_] := (e + 2)/2; q[1] = True; q[n_] := IntegerQ[Times @@ f @@@ FactorInteger[n]]; Select[Range[300], q]

%o (PARI) is(n) = denominator(vecprod(apply(x -> x/2 +1, factor(n)[, 2]))) == 1;

%Y Cf. A000005, A005117, A007425, A374902, A374903.

%K nonn,easy

%O 1,2

%A _Amiram Eldar_, Jul 23 2024