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I have a particular polynomial

$$ 1-10x+35x^2-50x^3 $$

Which can be written nicely as

$$1-(1+2+3+4)x+(1\cdot2+1\cdot3+1\cdot4+2\cdot3+2\cdot4+3\cdot4)x^2$$ $$+(1\cdot2\cdot3+1\cdot2\cdot4+1\cdot3\cdot4+2\cdot3\cdot4)x^3$$

I can write the coefficients as a pattern using summation notation as

$$\sum_{1\le j_1<...<j_4\le4}\frac{4!}{j_1\cdot...\cdot j_4}-\sum_{1\le j_1<j_2<j_3\le4}\frac{4!}{j_1\cdot j_2\cdot j_3}x+\sum_{1\le j_1<j_2\le4}\frac{4!}{j_1\cdot j_2}x^2-\sum_{1\le j_1\le4}\frac{4!}{j_1}x^3$$

I would like to condense this down in to a single double summation. I think I can handle it by doing

$$4!\sum_{k=1}^{4}\sum_{1\le j_1\lt ...j_{5-k}\le 4}\frac{(-x)^{k-1}}{j_1\cdot...\cdot j_k}$$

a) Is this notation okay? What i mean is, it the notation correct mathematically?

b) Is there an easier way to write this?

EDIT: I made a slight error in the denominator. I think this fixes is.

$$4!\sum_{k=1}^{4}\sum_{1\le j_1\lt ...j_{5-k}\le 4}\frac{(-x)^{k-1}}{j_1\cdot...\cdot j_{5-k}}$$

EDIT 2: I came across the generalized harmonic numbers defined as $$H_{n,j}=\sum_{1\le k_1<...,<k_j\le n}\frac{1}{k_1\cdot...\cdot k_j}$$

Therefore I can write the above double sum using this definition as

$$4!\sum_{j=0}^3H_{4,4-j}x^j$$

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    $\begingroup$ Not an answer but in case it helps the polynomial is $(4x-1)(3x-1)(2x-1)(x-1)-24x^4$. Also the coefficients are elementary symmetric functions, and there are conventional ways of expressing those using efficient notation - would that be what you are looking for? $\endgroup$ – Mark Bennet Dec 5 '14 at 10:53
  • $\begingroup$ Yes, that would actually be quite useful. A quick question, is the summation correct? $\endgroup$ – Eleven-Eleven Dec 5 '14 at 11:06
  • $\begingroup$ So in the general case the polynomial would be $\prod_{k=1}^m(1-kx)-m!x^m$? $\endgroup$ – Eleven-Eleven Dec 5 '14 at 11:54
  • $\begingroup$ @Eleven-Eleven Define "okay"! $\endgroup$ – Aditya Hase Dec 5 '14 at 13:22
  • $\begingroup$ Ha ha. Yeah I guess that's not a great math word... I'm wondering if the notation is correct. I also am looking for other ways to write it because i feel there are. @MarkBennet showed another way which seems better, but I also was curious of other ways as well. $\endgroup$ – Eleven-Eleven Dec 5 '14 at 13:27

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