Semi Integer Multiply Perfect

[koh]

    Neil

    I submit generalized Multiply Perfect Number sequence
    I think these are finite.
    Could anyone compute more terms?

    %I A000001
    %S A000001 24,91963648,10200236032
    %N A000001 Semi Integer Multiply Perfect Number such that Sigma(m)=k*m , where k=n/2 for some integer n.
               For k=1/2 , no example.
               For k=3/2 , only one m=2.
               Sequence gives the case of k=5/2
               Factorization : 2^3*3
                               2^8*7*19*37*73
                               2^14*7*19*31*151
    %Y A000001 A000002,A000003
    %K A000001 none
    %O A000001 1,1
    %A A000001 Yasutoshi Kohmoto   zbi74583 at boat.zero.ad.jp

  
    %I A000002
    %S A000002 4680,26208,197064960,20427264,57575890944,21857648640,230361837156847526055247872
    %N A000002 Semi Integer Multiply Perfect Number such that Sigma(m)=k*m , where k=n/2 for some integer n.
               Sequence gives the case of k=7/2
               Factorization :2^3*3^2*5*13  
                              2^5*3^2*7*13  
                              2^8*3*5*19*37*73 
                              2^9*3^2*11*13*31 
                              2^13*3^2*11*13*43*127
                              2^14*3*5*19*31*151  
                              2^25*3^4*11^2*19^2*127*683*2731*8191
    %Y A000002 A000001,A000003
    %K A000002 none
    %O A000002 1,1
    %A A000002 Yasutoshi Kohmoto   zbi74583 at boat.zero.ad.jp

  
   
    %I A000003
    %S A000003 8841105,8583644160,206166804480
    %N A000003 Semi Integer Multiply Perfect Number such that Sigma(m)=k*m , where k=n/2 for some integer n.
               Sequence gives the case of k=9/2
               Factorization :2^7*3^2*5*7*13*17
　                            2^10*3^2*5*7*13*23*89
                              2^11*3^2*5*7*13^2*31*61 
    %Y A000003 A000001,A000002
    %K A000003 none
    %O A000003 1,1
    %A A000003 Yasutoshi Kohmoto   zbi74583 at boat.zero.ad.jp



    Yasutoshi