[seqfan] 1/(UnitarySigma(m))^(1/2)=1/(UnitarySigma(n))^(1/2)=k^(1/2)*(1/m^(1/2)-1/n^(1/2))

[koh]

    Hi,Seqfans
    I am not so sure if it is complete up to the third term.
    Could anyone search them using a computer?

    Two NG examples : 
    k=91/6
    x=3^4*5^2*17*41*2^2*13^2
    y=3^4*5^2*17*41*2^4*7^2

    k=47^2/2^5
    x=3^3*5^6*7^2*11*17*43*301*601*2^6*13^2
    y=3^3*5^6*7^2*11*17*43*301*601*2^2*47^2

    k didn't become integer.



    %I A000001
    %S A000001 4500,61200,4393701 
    %N A000001 Numbers 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 n number. 
    %e A000001 Factorization :
               5^3*2^2*3^2
               5^2*17*2^4*3^2
               13*17*3^2*47^2
    %Y A000001 A000002, A000003
    %K A000001 none
    %O A000001 0,1
    %A A000001 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp

    %I A000002
    %S A000002 6125,71825,4331600 
    %N A000002 Numbers 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 m number. 
    %e A000002 Factorization :
               5^3*7^2
               5^2*17*13^2
               13*17*2^4*5^2*7^2
    %Y A000002 A000001,A000003
    %K A000002 none
    %O A000002 0,1
    %A A000002 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp
    
    %I A000003
    %S A000003 35,130,15463 
    %N A000003 Numbers 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 k number. 
    %Y A000003 A000001,A000002
    %K A000003 none
    %O A000003 0,1
    %A A000003 Yasutsohi Kohmoto zbi74583.boat at orange.zero.jp



    Yasutoshi