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I came up with a recursive formula for a problem I was working on. It is as follows. $$a_n = \Big(\frac{1-q^{f \cdot n}}{1-q^n}\Big)\displaystyle\Big(1+\sum_{i=0}^{n-1}\binom{n}{i}p^{n-i}q^ia_i\Big)$$

Here $a_0 = 0$ $p, q\in [0,1],$ $p + q = 1,$ and $f$ is a positive integer. Is there anyway to give a closed form solution to this equation?

Off the top of my head this seems related to Bell Numbers.

Other thought were to try setting $$\alpha(n) = \frac{1-q^{f \cdot n}}{1-q^n}$$ and then simplifying $$\frac{a_n}{\alpha(n)} - p\cdot\frac{a_{n-1}}{\alpha(n-1)}$$ using $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}.$$ I think I keep getting errors, but this should give something close to $$\frac{a_n}{\alpha(n)} - p\cdot\frac{a_{n-1}}{\alpha(n-1)} = (1-p) + n\cdot p^1q^{n-1}a_{n-1} + \displaystyle\sum_{i=0}^{n-2}\binom{n-i}{i}p^{n-i}q^ia_i.$$

I think $$\displaystyle\sum_{i=0}^{n-2}\binom{n-i}{i}p^{n-i}q^ia_i = p\cdot\displaystyle\sum_{i=0}^{n-2}\binom{n-i}{i}p^{(n-1) - i}q^ia_i = p\Big(\frac{a_{n-i}}{\alpha(n)} - 1\Big).$$

I should then be able to stick this back in and solve.

However, I seem to get a different answer, and regardless, this still is far from being a closed form solution.

Any suggestions? Thank you!

rabbitsLikeCarrots
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  • When you say "closed form", what exactly are you expecting? Simply plugging into the recurrence gives $$a_1 = \frac{1-q^{f}}{1-q} \ a_2 = \frac{(1-q^{2f})(1+2q-2q^{f+1})}{1-q^2}$$ – Peter Taylor Feb 19 '18 at 14:51
  • @PeterTaylor For example, Bell numbers have a form $$b_n = \displaystyle\sum_{k=0}^{n-1}\binom{n}{k}b_k$$ with $b_k = 1$. I consider this ``not closed" because it involves all of the previous terms. However, I consider the alternative form, $$b_n = \frac{1}{e}\displaystyle\sum_{k=0}^{\infty}\frac{k^n}{k!}$$ to be a closed form because it only involves the $b_n$ term. – rabbitsLikeCarrots Feb 19 '18 at 14:55

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