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Integers whose variance of their divisors is also an integer.
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%I #27 Apr 11 2021 22:17:37

%S 1,3,5,7,11,13,15,17,19,20,21,23,24,27,29,31,33,35,37,39,41,43,44,47,

%T 49,51,53,55,57,59,60,61,65,67,68,69,71,73,77,79,83,85,87,88,89,91,92,

%U 93,95,97,101,103,105,107,109,111,113,115,116,119,120,123

%N Integers whose variance of their divisors is also an integer.

%C The sequence contains all odd prime numbers p, since they have N=2 divisors whose mean xbar is an integer and the sum of their squared mean deviations SS is even; therefore, their variance Var(p)=SS/N is an integer.

%C k is a term if and only if A000005(k)^2 divides (A000005(k)*A001157(k)-A000203(k)^2). - _Chai Wah Wu_, Apr 11 2021

%H Oystein Ore, <a href="https://www.jstor.org/stable/2305616/">On the averages of the divisors of a number</a>, Amer. Math. Monthly, Vol. 55, No. 10 (1948), pp. 615-619.

%e a(10)=20 has six divisors d={1,2,4,5,10,20} whose variance is Var(d)=42, an integer.

%t Select[Range[500],IntegerQ[Mean[(Divisors[#] - Mean[Divisors[#]])^2]]&]

%o (PARI) isok(m) = my(d=divisors(m)); denominator(norml2(d)/#d - (vecsum(d)/#d)^2) == 1; \\ _Michel Marcus_, Apr 03 2021

%o (Python)

%o from sympy import divisor_sigma

%o A343011_list = [n for n in range(1,10**3) if (divisor_sigma(n,0)*divisor_sigma(n,2)-divisor_sigma(n,1)**2) % divisor_sigma(n,0)**2 == 0] # _Chai Wah Wu_, Apr 11 2021

%Y Cf. A000005, A000203, A001157, A003601, A065091.

%K nonn

%O 1,2

%A _Alexander M. Kerr_, Apr 02 2021