hcju/proofwikiqa huggingface dataset:
First we show that this series converges when $n \in \Z_{\ge 0}$. Consider the coefficients: :$1, \paren {1 - \dfrac 1 {1!} }, \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} }, \ldots$ By Power Series Expansion for Exponential Function, they converge to $\dfrac 1 e$. Starting from the $\paren {n + 1}$th term, there is a factor of $\paren {n - n}$. In this case, all subsequent terms of the expansion equal $0$. Therefore the series converges. {{qed|lemma}} Then we show that this series diverges for $n \notin \Z_{\ge 0}$. We do this via Divergence Test. That is, we show that $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!} \ne 0$. {{AimForCont}} $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ exists. By Power Series Expansion for Exponential Function, we have that $\displaystyle \lim_{k \mathop \to \infty} \frac {k!} {!k} = e$. Hence $\displaystyle \lim_{k \mathop \to \infty} n^{\underline k}$ would exist as well. For $k > \floor {n + 2}$: {{begin-eqn}} {{eqn | l = n - k       | o = <       | r = -2 + n + 2 - \floor {n + 2} }} {{eqn | o = \le       | r = -2 }} {{eqn | ll = \leadsto       | l = \size {n^{\underline k} }       | r = \size {n^{\underline {k - 1} } } \size {n - k} }} {{eqn | o = >       | r = 2 \size {n^{\underline {k - 1} } } }} {{end-eqn}} Since $n - k \ne 0$, $\size {n^{\underline k} }$ increases without bound. Hence its limit does not exist, a contradiction. Therefore $\displaystyle \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ does not exist for $n \notin \Z_{\ge 0}$. By Divergence Test, our series diverges. {{qed}}
As $M$ is bounded, it has a finite diameter $R$. Consider arbitary $x, y \in \R^n$, expressed in their usual coordinates. {{begin-eqn}} {{eqn | l = \norm {y - x}       | r = \sqrt {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2} }} {{eqn | o = \le       | r = \sqrt {n \max_i \set {\paren {y_i - x_i}^2} } }} {{eqn | r = \sqrt n \max_i \size {y_i - x_i} }} {{end-eqn}} Thus, up to a positive constant multiple, the inequality: :$\displaystyle \sup_{x, y \mathop \in M} \max_i \size {y_i - x_i} \le R$ implies that $M$ is bounded with radius $R$. But this inequality is a characterization of an $n$-cube. Because $R \le 2 R \sqrt n$ for all $R > 0$, $M$ can be placed inside an $n$-cube of the form $\closedint {c - R} {c + R}^n$, by diameter of $n$-cube. For any $\epsilon > 0$, pick an integer $k > \sqrt n R \epsilon^{-1}$. Consider a normal subdivision of $\closedint {c - R} {c + R}$ into $k$ pieces. The length of each subinterval is then: :$\dfrac {\paren {c + R - \paren {c - R} } } k = 2 R k^{-1}$ For any subinterval chosen this way, the product of $n$ copies of such an interval will itself be a cube, because the subintervals were chosen as to all be the same length. By diameter of $n$-cube, each sub-cube has diameter $2 R k^{-1} \sqrt n$ By the choice of $k$, we have $2 R k^{-1} \sqrt n < 2 \epsilon$ That means that each sub-cube can fit inside an open ball $n$-ball of radius $\epsilon$. This in turn implies that $M$ can be covered by a finite number of balls with radius $\epsilon$. Thus $M$ is totally bounded by definition, as $\epsilon$ was arbitrary. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexive === {{begin-eqn}} {{eqn | o =        | r = \forall x \in \R: x - x = 0 \in \Q }} {{eqn | o = \leadsto        | r = x \sim x }} {{end-eqn}} So $\sim$ is reflexive. {{qed|lemma}} === Symmetric === {{begin-eqn}} {{eqn | l = x       | o = \sim       | r = y }} {{eqn | o = \leadsto       | r = x - y \in \Q }} {{eqn | o = \leadsto       | r = x - y = \frac p q       | c = for some integers $p, q$ }} {{eqn | o = \leadsto       | r = -\paren {x - y} = \frac {-p} q       | c = {{Defof|Diagonal Relation}} }} {{eqn | o = \leadsto       | r = y - x \in \Q       | c = as $-p$ is an integer }} {{eqn | o = \leadsto       | r = y \sim x }} {{end-eqn}} So $\sim$ is symmetric. {{qed|lemma}} === Transitive === {{begin-eqn}} {{eqn | o =        | r = \forall x, y, z \in \R: x \sim y \land y \sim z       | c =  }} {{eqn | o = \implies       | r = x - y = \frac p q \land y - z = \frac r s       | c = for some integers $p, q, r, s$ }} {{eqn | o = \leadsto       | r = x - z = \frac p q - \frac r s       | c = adding the second equality to the first }} {{eqn | r = \frac {s p - q r} {q s} \in \Q       | c = as $s p - q r$ and $q s$ are integers }} {{eqn | o = \leadsto       | r = x \sim z }} {{end-eqn}} So $\sim$ is transitive. {{qed|lemma}} Thus $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}} Category:Equivalence Relations ekwt4zcsd3nb81pjo6cjb5ubfhljval
First we note that if any of the $\beta$s is a negative integer, the {{LHS}} would have $0$ as its denominator, and so would be undefined. We have from the Euler form of the Gamma function that: :$\Gamma \left({1 + \beta_i}\right) = \displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\left({1 + \beta_i}\right) \left({2 + \beta_i}\right) \cdots \left({m + 1 + \beta_i}\right)}$ and so the {{RHS}} can be written as: {{begin-eqn}} {{eqn | o =        | r = \dfrac {\left({\displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_1} m^{1 + \beta_2} \cdots m^{1 + \beta_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)} }\right)} {\left({\displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \alpha_1} m^{1 + \alpha_2} \cdots m^{1 + \alpha_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} }\right)}       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \dfrac {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right) m^k m^{\beta_1 + \beta_2 + \cdots + \beta_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right) m^k m^{\alpha_1 + \alpha_2 + \cdots + \alpha_k} \left({m!}\right)^k}       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \dfrac {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right) m^{\beta_1 + \beta_2 + \cdots + \beta_k} } {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right) m^{\alpha_1 + \alpha_2 + \cdots + \alpha_k} }       | c = simplifying slightly }} {{eqn | r = \lim_{m \mathop \to \infty} \displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \dfrac {\left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)}       | c = as $\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$ }} {{eqn | r = \displaystyle \prod_{n \mathop \ge 1} \dfrac {\left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x^{\underline {m + n} }       | r = \prod_{j \mathop = 0}^{m + n - 1} \left({x - j}\right)       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \left({x - j}\right) \prod_{j \mathop = m}^{m + n - 1} \left({x - j}\right)       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \left({x - j}\right) \prod_{j \mathop = 0}^{n - 1} \left(x - {\left({m + j}\right)}\right)       | c = Translation of Index Variable of Product }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \left({x - j}\right) \prod_{j \mathop = 0}^{n - 1} \left({\left({x - m}\right) - j}\right)       | c =  }} {{eqn | r = x^{\underline m} \left({x - m}\right)^{\underline n}       | c = {{Defof|Falling Factorial}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x^{\overline {m + n} }       | r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x + j}       | c = {{Defof|Rising Factorial}} }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = m}^{m + n - 1} \paren {x + j}       | c =  }} {{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = 0}^{n - 1} \paren {x + \paren {m + j} }       | c = Translation of Index Variable of Product }} {{eqn | r = x^{\overline m} \paren {x + m}^{\overline n}       | c = {{Defof|Rising Factorial}} }} {{end-eqn}} {{qed}}
Denote: :$\displaystyle E = \bigcup_{k \mathop \in \N} E_k$ :$\displaystyle F = \bigcup_{k \mathop \in \N} F_k$ where: :$\displaystyle F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$ We first show that $E = F$. That $x \in E \implies x \in F$ follows from the construction of $F$ from subsets of $E$. Thus $E \subseteq F$. For the other implication: {{begin-eqn}} {{eqn | l = x \in \bigcup_{k \mathop \in \N} F_k       | o = \leadstoandfrom       | r = \exists k \in \N : x \in F_k }} {{eqn | o = \leadstoandfrom       | r = \exists k \in \N: x \in E_k \land x \notin \paren {E_0 \cup E_1 \cup E_2 \cup \cdots \cup E_{k - 1} } }} {{eqn | o = \leadsto       | r = \exists k \in \N: x \in E_k       | c = Rule of Simplification }} {{end-eqn}} so $F \subseteq E$. Thus $E = F$ from the definition of set equality. To show that the sets in $F$ are disjoint, consider an arbitrary $x \in F$.  Then $x \in F_k$ for some $F_k$. By the well-ordering principle, there is a smallest such $k$. Then for any $j < k$, $x \notin F_j$ Choose any distinct $\ell, m \in \N$.  If $m > \ell$, then: :$x \in F_\ell \implies x \in E_\ell$ :$x \in F_m \implies x \notin E_m$ If $m < \ell$, then: :$x \in F_m \implies x \in E_m$ :$x \in F_\ell \implies x \notin E_\ell$ So the sets $F_\ell, F_m$ are disjoint.  Thus $F$ is the disjoint union of sets equal to $E$: :$\displaystyle \bigcup_{k \mathop \in \N} E_k = \bigsqcup_{k \mathop \in \N} F_k$ {{qed}} === Proof of Corollary === This is a direct application of De Morgan: :$\displaystyle E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j} = \bigcap_{j \mathop = 0}^{k \mathop - 1} \paren {E_k \setminus E_j}$ {{qed}} Category:Indexed Families 4ptfmnnyb9fg90orzlhsukv1np4ucxl
By definition, a closed set is the relative complement of an open set. The result follows from Sigma-Algebra Generated by Complements of Generators. {{qed}}
The proof proceeds by induction over $m$. For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n \atop n - m}\right]$ is a polynomial in $n$ of degree $2 m$. === Basis for the Induction === $P \left({0}\right)$ is the case: :$\displaystyle \left[{n \atop n}\right] = 1$ which is a polynomial in $n$ of degree $0$ Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{n \atop n - k}\right]$ is a polynomial in $n$ of degree $2 k$. from which it is to be shown that: :$\displaystyle \left[{n \atop n - \left({k + 1}\right)}\right]$ is a polynomial in $n$ of degree $2 \left({k + 1}\right)$. === Induction Step === This is the induction step: Let $f \left({n, d}\right)$ denote an arbitrary polynomial in $n$ of degree $d$. Then: {{begin-eqn}} {{eqn | l = \left[{n \atop n - \left({k + 1}\right)}\right]       | r = \left[{n \atop n - k - 1}\right]       | c =  }} {{eqn | r = \left({n - 1}\right) \left[{n - 1 \atop n - k - 1}\right] + \left[{n - 1 \atop n - k - 2}\right]       | c = {{Defof|Stirling Numbers of the First Kind/Unsigned|Unsigned Stirling Number of the First Kind|index = 1}} }} {{eqn | r = \left({n - 1}\right) \left[{n - 1 \atop \left({n - 1}\right) - k}\right] + \left[{n - 1 \atop \left({n - 1}\right) - \left({k + 1}\right)}\right]       | c =  }} {{end-eqn}} {{finish|This doesn't look promising}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \left[{n \atop n - m}\right]$ is a polynomial in $n$ of degree $2 m$.
The proof proceeds by induction. === Basis for the Induction === For all $n \in \Z_{\ge 2}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n \atop n - 2}\right] = \binom n 4 + 2 \binom {n + 1} 4$ $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left[{2 \atop 0}\right]       | r = \delta_{2 0}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{k \atop k - 2}\right] = \binom k 4 + 2 \binom {k + 1} 4$ from which it is to be shown that: :$\displaystyle \left[{ {k + 1} \atop k - 1}\right] = \binom {k + 1} 4 + 2 \binom {k + 2} 4$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{k + 1 \atop k - 1}\right]       | r = k \left[{k \atop k - 1}\right] + \left[{k \atop k - 2}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = k \binom k 2 + \left[{k \atop k - 2}\right]       | c = Unsigned Stirling Number of the First Kind of n with n-1 }} {{eqn | r = k \binom k 2 + \binom k 4 + 2 \binom {k + 1} 4       | c = Induction Hypothesis }} {{eqn | r = k \frac {k \left({k - 1}\right)} 2 + \dfrac {k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right)} {4!} + \dfrac {\left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right)} {4!} + \binom {k + 1} 4       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \binom {k + 1} 4 + k \left({k - 1}\right) \frac {12 k + \left({k - 2}\right) \left({k - 3}\right) + \left({k + 1}\right) \left({k - 2}\right)} {4!}       | c =  }} {{eqn | r = \binom {k + 1} 4 + k \left({k - 1}\right) \frac {12 k + k^2 - 5 k + 6 + k^2 - k - 2} {4!}       | c =  }} {{eqn | r = \binom {k + 1} 4 + k \left({k - 1}\right) \frac {2 \left({k^2 + 3 k + 2}\right)} {4!}       | c =  }} {{eqn | r = \binom {k + 1} 4 + \frac {2 \left({k + 2}\right) \left({k + 1}\right) k \left({k - 1}\right)} {4!}       | c =  }} {{eqn | r = \binom {k + 1} 4 + 2 \binom {k + 2} 4       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 2}: \left[{n \atop n - 2}\right] = \binom n 4 + 2 \binom {n + 1} 4$ {{qed}}
We first show that $X$ is infinite. By the definition of a $\sigma$-algebra, $\MM$ is a subset of $\powerset X$. Were $X$ finite, by Cardinality of Power Set of Finite Set, the cardinality of $\MM$ would be at most $2^{\map \Card X}$ As $2^{\map \Card X}$ is finite if $X$ is finite, $X$ must be infinite by the assumption that $\MM$ is infinite. By the definition of $\sigma$-algebra, $X \in \MM$.  Also, by Sigma-Algebra Contains Empty Set, $\O \in \MM$. Construct a countable collection of sets $\family {F_1, F_2, F_3, \ldots}_\N$ as follows: :$F_1 = \O$ :$F_2 = X$ We can continue this construction using the axiom of choice: :$F_3 = \text{ any set in } \MM \setminus \set {\O, X}$ :$\ \ \vdots$ :$F_n = \text{ any set in } \MM \setminus \set {F_1, F_2, \ldots, F_{n - 1} }$ :$\ \ \vdots$ Consider an arbitrary $S \in \bigcup_{k \mathop \in \N} F_k$ Then $S \in F_k$ for some $F_k$. By the well-ordering principle, there is a smallest such $k$. Then for any $j < k$, $S \notin F_j$ Thus the sets in $\family {F_i}$ are disjoint. Recall $\MM$ is infinite. Then by Relative Difference between Infinite Set and Finite Set is Infinite, $\MM \setminus \set {F_1, F_2, \cdots , F_{k - 1} }$ is infinite. Thus this process can continue indefinitely, choosing an arbitary set in $\MM$ that hasn't already been chosen for an earlier $F_k$. By the definition of a $\sigma$-algebra, $\displaystyle \bigsqcup_{i \mathop \in \N} F_i$ is measurable. By the definition of an indexed family, $\family {F_i}_{i \mathop \in \N}$ corresponds to a mapping $\iota: \N \hookrightarrow \bigsqcup_{i \mathop \in \N} F_i$ Such a mapping $\iota$ is injective because: :each $F_i$ is disjoint from every other :each $F_i$ contains a distinct $x \in X$ :$X$ has infinitely many elements. Define: :$\iota^*: \powerset \N \to \MM$: :$\map {\iota^*} N = \bigsqcup_{i \mathop \in N} F_i$ That is, for every $N \subseteq \N$, $\map {\iota^*} N$ corresponds to a way to select a countable union of the sets in $\family {F_i}$.  Because any distinct $F_i, F_j$ are disjoint, any two distinct ways to create a union $S \mapsto \bigsqcup_{i \mathop \in S} F_i$ will result in a different union $\map {\iota^*} S$. Thus $\iota^*$ is an injection into $\MM$. Then the cardinality of $\MM$ is at least $\powerset \N$. From Continuum equals Cardinality of Power Set of Naturals, $\R \sim \powerset \N$. Thus $\MM$ is uncountable and: :$\map \Card \MM \ge \mathfrak c$ {{qed}} {{AoC}}
The proof proceeds by induction. === Basis for the Induction === For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\displaystyle {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$ $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = {2 \brace 0}       | r = \delta_{2 0}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k \brace k - 2} = \binom {k + 1} 4 + 2 \binom k 4$ from which it is to be shown that: :$\displaystyle {k + 1 \brace k - 1} = \binom {k + 2} 4 + 2 \binom {k + 1} 4$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 1 \brace k - 1}       | r = \paren {k - 1} {k \brace k - 1} + {k \brace k - 2}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = \paren {k - 1} \binom k 2 + {k \brace k - 2}       | c = Stirling Number of the Second Kind of n with n-1 }} {{eqn | r = \paren {k - 1} \binom k 2 + \binom {k + 1} 4 + 2 \binom k 4       | c = Induction Hypothesis }} {{eqn | r = \paren {k - 1} \frac {k \paren {k - 1} } 2 + \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!} + 2 \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4!}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = k \paren {k - 1} \paren {\frac {k - 1} 2 + \dfrac {\paren {k + 1} \paren {k - 2} } {4!} + 2 \dfrac {\paren {k - 2} \paren {k - 3} } {4!} }       | c =  }} {{eqn | r = k \paren {k - 1} \paren {\frac {12 k - 12 + k^2 - k - 2 + 2 k^2 - 10 k + 12} {4!} }       | c =  }} {{eqn | r = k \paren {k - 1} \paren {\frac {k^2 + 3 k + 2 + 2 k^2 - 2 k + 4} {4!} }       | c =  }} {{eqn | r = k \paren {k - 1} \paren {\frac {\paren {k + 2} \paren {k + 1} + 2 \paren {k + 1} \paren {k - 2} } {4!} }       | c =  }} {{eqn | r = \frac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} } {4!} + 2 \frac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!}       | c =  }} {{eqn | r = \binom {k + 2} 4 + 2 \binom {k + 1} 4       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 2}: {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$ {{qed}}
{{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the negative integers. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}}
{{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the complex plane. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}}
The statement: :$\dbinom n k$ is not divisible by $p$ is equivalent to: :$\dbinom n k \not \equiv 0 \pmod p$ The corollary to Lucas' Theorem gives: :$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$ where: : $n, k \in \Z_{\ge 0}$ and $p$ is prime : the representations of $n$ and $k$ to the base $p$ are given by: ::$n = a_r p^r + \cdots + a_1 p + a_0$ ::$k = b_r p^r + \cdots + b_1 p + b_0$ Consider the form of $n = a p^m - 1$ when represented to the base $p$: {{begin-eqn}} {{eqn | l = n       | r = a p^m - 1       | c =  }} {{eqn | r = a \paren {p^m - 1} + \paren {a - 1}       | c =  }} {{eqn | r = a \paren {p - 1} \sum_{j \mathop = 0}^{m - 1} p^i + \paren {a - 1}       | c = Sum of Geometric Sequence: Corollary 1 }} {{eqn | r = \paren {a - 1} p^m + \sum_{j \mathop = 0}^{m - 1} \paren {p - 1} p^i       | c = after algebra }} {{end-eqn}} That is, all of the digits of $n$ are $p - 1$ except perhaps the first one. For $k$ such that $0 \le k \le n$, the digits of $k$ will all range over $0$ to $p - 1$. === Necessary Condition === Let $n = a p^m - 1$. Then all the binomial coefficients $\dbinom {a_j} {b_j}$ (except perhaps the first) are of the form $\dbinom {p - 1} k$ for $0 \le k < p$. By Binomial Coefficient of Prime Minus One Modulo Prime: :$\dbinom {p - 1} k \equiv \paren {-1}^k \pmod p$ and so: :$\dbinom {a_j} {b_j} \equiv \paren {-1}^k \pmod p$ for $0 \le j < r$ For the first digit, we have $0 \le b_r \le a_r < p$. We have $\dbinom {a_r} {b_r} > 0$. Since $a_r!$ is not divisible by $p$, neither does $\dbinom {a_r} {b_r}$. Thus: :$\dbinom {a_r} {b_r} \not \equiv 0 \pmod p$ Hence: :$\displaystyle \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \equiv \pm \dbinom {a_r} {b_r} \not \equiv 0 \pmod p$ which means $\dbinom n k$ is not divisible by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$. {{qed|lemma}} === Sufficient Condition === We show the contrapositive. Suppose $n \ne a p^m - 1$. Then one of the digits $a_j$ (that is not the first one) will be less than $p - 1$. But the digits of all possible $k$ will range over $0$ to $p - 1$. Therefore there must be at least one $\dbinom {a_j} {b_j}$ such that $b_j > a_j$. Such a condition leads to $\dbinom {a_j} {b_j} = 0$. That is: :$\displaystyle \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \equiv 0 \pmod p$ which means $\dbinom n k$ is divisible by $p$ for some $k \le n$. {{qed}}