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Consider pairs m,n such that 1/(UnitarySigma(m))^(1/2)=1/(UnitarySigma(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2)), n<m. Sequence gives values of n.
2

%I #2 Apr 19 2016 01:26:37

%S 4500,61200,4393701

%N Consider pairs m,n such that 1/(UnitarySigma(m))^(1/2)=1/(UnitarySigma(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2)), n<m. Sequence gives values of n.

%e Factorizations: 5^3*2^2*3^2 5^2*17*2^4*3^2 13*17*3^2*47^2

%Y A144491, A144492

%K nonn,bref

%O 1,1

%A _Yasutoshi Kohmoto_, Dec 11 2008