[seqfan] A question of Donovan Johnson
shevelev at bgu.ac.il
Sun Oct 27 14:03:58 CET 2013
Donovan Johnson posed a question with respect to A227891
(Numbers for which number of odious proper divisors (A000069) equals number of evil proper divisors (A001969)):
"Are all terms odd? I searched up to (10^7)^2 and found no even terms."
We prove that the question is solved in affirmation.
Since the sequence contains only perfect squares, let
m^2 be even with prime power factorization
Denote by L the number of all divisors of odd part M of m^2 (since m^2 is even, then all they are proper divisors of m^2). We have L=(2*m_1+1)*...*(2m_r+1). Let difference
between numbers of odious and evil divisors of M be d.
Since L is odd, then |d|>=1.
1) Let d>=1. Let us calculate the number of proper divisors of m^2. Calculating it as sum for all divisors of M, new divisors of 2*M (which are not divisors of M), ..., new divisors of 2^(2*k)*M and minus 1, we have
L + L + L +...+ L - 1 =(2*k+1)*L-1
Since odiousness-evilness of a number does not change after multiplication by a power of 2, then we obtain that
the difference between odious and evil proper divisors of m^2 is (2^k+1)*d - 1>=2*k>=2.
2) The case of d<= -1 is considered analogously. Here such a difference <= -3-1=-4.
Thus any even m^2 is not in the sequence.
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