[seqfan] Re: GF for A147843?

[William Keith]

On Mar 15, 2010, at 10:11 AM, Joerg Arndt wrote:

> * Joerg Arndt <arndt at jjj.de> [Mar 15. 2010 15:07]:
>> (PARI/GP):
>> deriv(-eta(x))
>> 1 + 2*x - 5*x^4 - 7*x^6 + 12*x^11 + ...
>> 
>> here:
>> eta(x) = prod(n=1, \infty, 1-x^n)
>> 
>> I get
>> d/dx eta(x) = eta(x) / x * sum(n=1,\infty, -x^(n-1)/(1-x^n))
>> this appears to check with the comment
>> 
>>  Convolved with the partition numbers A000041 = sigma(n) prefaced
>>  with an 0: (0, 1, 3, 4, 7, 6, 12, 8, 15, 13,...).

You can remove the "appears to have" in your comment on A147843.  (Is that what you were requesting comment on?  I wasn't entirely sure.)  Euler's Pentagonal Number Theorem tells us that eta(x) is \sum_{-\infty}^{\infty} (-1)^m q^{m(3m-1)/2} , and of course when taking the derivative each term (now one lower) will be multiplied by the power it represented.

Regarding your comment above, d/dx -eta(x) is eta(x) (sum n x^(n-1)/(1-x^n)), by the product rule with the lost factor tossed back in.  There's no extra x there, unless you meant to increase the exponent in the numerator of the sum from n-1 to n, and there should be a factor of n in the numerator.

(Without the x (which gives you a term x^{-1} to start with), and brute-expanding what you wrote down there, I get 
-1-x+x^2+x^3+3x^4+2x^6-x^7-2x^8-x^9-2x^10+..., which doesn't match anything currently.)

> So a GF for A000203 is:
> ? -x*deriv(eta(x))/eta(x)
> x + 3*x^2 + 4*x^3 + 7*x^4 + 6*x^5 + ...

(df/dx) / f = d/dx (ln f), which in the case of the eta function is d/dx sum ln (1-x^i).  So again we see that this is sum -n x^(n-1)/(1-x^n). Multiplied by -x, the coefficient on x^k will have a contribution of n for every n that divides k, so the coefficient will be the sum of the divisors of k.  Again, not sure if this is the comment you were requesting; the statement you wrote is true.

William Keith