Binomial Coefficient Pattern

Let $a_n = \dbinom{p_n}{q}$ For the proof, let $b,c$ denote arbitrary positive integers, such that $b < c$ .

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Arithmetic Progression

If $a_n$ is the arithmetic progression, then $a_c = a_b + (c - b) \times d$

which shows that the following must hold $a_b + a_{c - b} = a_c$ Evaluating, $\begin{array}{rlccl} \dbinom{p_b}{q} + \dbinom{p_{c - b}}{q} &= \dbinom{p_c}{q}\\ \dfrac{\left(p_b\right)!}{q!\left(p_b - q\right)!} + \dfrac{\left(p_{c - b}\right)!}{q!\left(p_{c - b} - q\right)!} &= \dfrac{\left(p_c\right)!}{q!\left(p_c - q\right)!} & & & {\color{#3D99F6}\text{By definition}}\\ \dfrac{\left(p_b\right)!}{q!\left(p_b - q\right)!} + \dfrac{\left(p_{c - b}\right)!}{q!\left(p_{c - b} - q\right)!} &= \dfrac{\left(p_c\right)!}{q!\left(p_c - q\right)!} & & & {\color{#3D99F6}\text{Multiply both sides by }q!} \end{array}$ Either $q = 1$ or $q > 1$ .

Case 1

If $q = 1$ , then $\begin{array}{rl} p_b + p_{c - b} &= p_c\\ p_b + \left(c - b\right)d_p &= p_c \end{array}$ which satisfies the given that $p_n$ is an arithmetic progression. Therefore, $a_n$ is an arithmetic progression.

Case 2

Otherwise, $q > 1$ , which however violates that $p_n$ is an arithmetic progression. Consider $\dfrac{\left(p_b\right)!}{\left(p_b - q\right)!} + \dfrac{\left(p_{c - b}\right)!}{\left(p_{c - b} - q\right)!} = \dfrac{\left(p_c\right)!}{\left(p_c - q\right)!}$ Evaluating, $p_b \times \left(p_b - 1\right) \times \dots \times \left(p_b - q + 1\right) + p_{c - b} \times \left(p_{c - b} - 1\right) \times \dots \times \left(p_{c - b} - q + 1\right) = p_c \times \left(p_c - 1\right) \times \dots \times \left(p_c - q + 1\right)$ which is impossible to express in the closed arithmetic form for $q > 1$ since there occurs multiple $p$ 's after expanding the above equation. We can also note that by expanding and comparing both sides, $p_b \times \left(p_b - 1\right) \times \dots \times \left(p_b - q + 1\right) + p_{c - b} \times \left(p_{c - b} - 1\right) \times \dots \times \left(p_{c - b} - q + 1\right) < p_c \times \left(p_c - 1\right) \times \dots \times \left(p_c - q + 1\right)$

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Geometric Progression

If this were to be true, then $a_c = a_b \times r^{c - b}$ where $r > 1$ . If $q = 1$ , then $p_c = p_b \times r^{c - b}$ where $r = \left(\dfrac{p_c}{p_b}\right)^{\frac{1}{c - b}}$ . If suppose we let $e$ be the positive integer, such that $e \neq b,c$ , $p_e = p_b \times r^{e - b}$ However, $\begin{array}{rl} p_b \times \left(\dfrac{p_c}{p_b}\right)^{\frac{e - b}{c - b}} &= p_e\\ &= p_b + p_{e - b} \end{array}$ which shows that this cannot happen since $p_n$ is arithmetic. If $q > 1$ , $a_n$ can't be geometric by the similar reasoning.

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Harmonic Progression

Because $p_n$ is an arithmetic sequence of strictly increasing integers, $a_n$ is increasing as well. Thus, it impossible for $a_n$ to be harmonic for arbitrary $p$ 's and $q$ 's.

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Conclusion

$\dbinom{2}{1}, \dbinom{4}{1}, \dbinom{6}{1}, \dbinom{8}{1}, \dots, \dbinom{2 + 2d}{1}, \dots$ If $p_n$ is arithmetic progression, so is $a_n$ . However, this only works for $p = 1$ . There is no other way for $a_n$ to be arithmetic for any other $p$ .

Likewise, if $p_n$ is geometric progression, so is $a_n$ for $p = 1$ . Proving this is routine, which is much like how arithmetic progression is determined for this problem.