[seqfan] Re: Are the partials sums of A025147 equal to A038348?
jvospost3 at gmail.com
Fri Apr 2 04:48:16 CEST 2010
Great. I suggest that this be added as a comment to each of the two
named sequences, the partial sum and the difference set respectively.
Thank you! Would you, William Keith, please submit that comment to
each, with your proof, citing us as co-authors?
-- Jonathan Vos Post
On Thu, Apr 1, 2010 at 7:04 PM, William Keith <wjk26 at drexel.edu> wrote:
> On Apr 1, 2010, at 9:18 PM, Jonathan Post wrote:
>> Are the partials sums of A025147 Number of partitions of n into
>> distinct parts >= 2.
>> equal to A038348 Expansion of (1/(1-x^2))*Product((1/(1-x^(2m+1)),
>> m=0..inf. (equivalently Number of partitions of n with at most one
>> even part)?
> Yes. The g.f. for partitions of n into distinct parts >= 2 is Product(1+x^m), m >= 2. The g.f. for its partial sums is then
> (1+x+x^2+x^3+...) Product(1+x^m), m >= 2
> = 1/(1-x) Product(1+x^m), m >= 2.
> Transform thus:
> 1/(1-x) Product(1+x^m), m >= 2
> = 1/(1-x) Product((1-x^(2m))/(1-x^m)) , m>= 2 (multiply by Product((1-x^m)/(1-x^m)), m>=2)
> = 1/(1-x^2) Product(1-x^(2m+1), m >= 0 (pull out 1/(1-x^2), push in 1/(1-x), cancel all even powers >= 4 in denominator)
> There is probably a straightforward bijective proof as well.
> William Keith
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