[seqfan] possible new G.F. for A000593 "sum of odd divisors of N"

Peter Lawrence peterl95124 at sbcglobal.net
Sun Jun 5 22:44:27 CEST 2011


here is what might be a new G.F. for A000593 "sum of odd divisors of N",
would folks like to review it for correctness, and also let me know if
it is worthy of adding to the OEIS.

note, as I don't yet read PARI, I cannot tell if this is simply the same
as Joerg Arndt's program of May 03, 2008. It too involves "log", but he
uses a convolution, and I can't tell if I am unwittingly doing so as  
well.



proposed G.F. for A000593

             a(n)/n = sum{k=1,...} LN( 1 + x^k )

I'm not sure what "type" of G.F. to call this, if it were a(n)/n! it  
would
be an exponential-G.F., but it is simply a(n)/n, and there doesn't  
seem to
be a category for this ?



k)             -----  the coefficients of LN( 1 + x^k )  -----
1)  +1  -1/2  +1/3  -1/4  +1/5  -1/6  +1/7  -1/8  +1/9  -1/10 +1/11  
-1/12 +1/13
2)      +1          -1/2        +1/3        -1/4        +1/5        -1/6
3)            +1                -1/2              +1/3              -1/4
4)                  +1                      -1/2                    +1/3
5)                        +1                            -1/2
6)                              +1                                  -1/2
etc...

k)   -----   multiply coefficient of x^k by k to get rid of  
factions   -----
1)  +1  -1    +1    -1    +1    -1    +1    -1    +1    -1    +1     
-1    +1
2)      +2          -2          +2          -2          +2          -2
3)            +3                -3                +3                -3
4)                  +4                      -4                      +4
5)                        +5                            -5
6)                              +1                                  -2
etc...

                    -----   finally sum the columns   -----
      1   1     4     1     6     4     8     1    13     6    12      
4    14

                 -----   which turns out to be A000593 -----



which is fairly easy to see since the columns agree with the already  
published
formula of Jovovic:

         a(n) = Sum_{d divides n} (-1)^(d+1)*n/d. - Vladeta Jovovic
          (vladeta(AT)eunet.rs), Sep 06 2002

once you realise that, sum{d|n}  (-1)^(d+1)*n/d  ==  sum{d|n} (-1)^(n/ 
d+1)*d,
because the set of divisors equals the set of quotients.



sincerely,
Peter A. Lawrence


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