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I had postulated following sequences by depth n(hereby d${n\in\Bbb N}$) which are gaps between adjacent terms where the first starts from given (d1):

$1 ,x , {1\over 2}x^2, {1\over6}x^3, {1\over 24}x^4 \dots {1\over n!}x^n\tag {d1}$

$x-1,\space {1\over 2}x^2 -x,\space {1\over 6}x^3-{1\over 2}x^2, \space{1\over 24}x^4-{1\over 6}x^3\cdots \frac{1}{n!}x^n-\frac{1}{(n-1)!}x^{n-1}\tag {d2}$

${1\over 2}x^2 -2x+1,\space {1\over 6}x^3-x^2+x,\space \cdots\tag{d3}\space, \frac{1}{n!}x^n-\frac{2}{(n-1)!}x^{n-1}+\frac{1}{(n-2)}x^{n-2}$

$$\cdots \tag{d4} $$ $$\cdots$$

Let $n$-th term of d$n$ sequence $f(d,n)$ then the following holds:

$$f(n,d) = f(n, d-1) - f(n-1, d-1)$$ by definition.

I would like to solve this recurrence realtion so that I could find most explicit formula of $f$ expressed in $n, d$ arguments.

How could I do that?

Beverlie
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  • What are the initial conditions for the proposed sequence? Have you tried induction? You'll readily see why I inquire about the initial conditions. – Cye Waldman Dec 19 '17 at 18:08

1 Answers1

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I assume that terms of $d$-th sequence are numbered from $d-1$. Then it is easy to conjecture that $$f(n,d)= \sum_{i=0}^{d-1}{d-1 \choose i}\frac{(-1)^ix^{n-i}}{(n-i)!}$$

and prove this conjecture by induction with respect to $d$.

Alex Ravsky
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