[seqfan] A set of sequences of perfect squares

Vladimir Shevelev shevelev at bgu.ac.il
Sat Oct 26 21:11:40 CEST 2013

Dear SeqFans,
Consider a partition of N by two sets N_1, N_2. Let {a(n)} lists numbers which have the same number of proper divisors from N_1 and N_2. Then {a(n)} contains only perfect squares. 
Indeed, let m belongs to this sequence and has prime power factorizationp_1^k_1*...*p_r^k_r. Then, as well known, the number of all its divisors equals (k_1+1)*...*(k_r+1) and, therefore, the number of its proper divisors is (k_1+1)*...*(k_r+1) - 1. By the condition, this number should be even. Thus all k_i should be even which concludes the proof. 
In particular, if N_1, N_2 are sets of odious and evil numbers,
we obtain submitted by me and Peter sequence A227891, all terms of which are squares.
Best regards,

 Shevelev Vladimir‎

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