hcju/proofwikips huggingface dataset:
Falling Factorial as Quotient of Factorials
:$x^{\underline n} = \dfrac {x!} {\paren {x - n}!} = \dfrac {\map \Gamma {x + 1} } {\map \Gamma {x - n + 1} }$
One to Integer Rising is Integer Factorial
:$1^{\overline n} = n!$
Number to Power of One Rising is Itself
:$x^{\overline 1} = x$
Integer to Power of Itself Falling is Factorial
:$n^{\underline n} = n!$
Number to Power of One Falling is Itself
:$x^{\underline 1} = x$
Number of Permutations of One Less
:${}^{n - 1} P_n = {}^n P_n$ where ${}^k P_n$ denotes the number of ordered selections of $k$ objects from $n$.
Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation
Let $R = \left({S, \preceq}\right)$ be a bounded below join semilattice. Let $\mathit{Ids}\left({R}\right)$ be the set of all ideals in $R$. Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$. Let $M = \left({F, \preccurlyeq}\right)$ be the ordered set of increasing mappings $g:S \to \mathit{Ids}\left({R}\right)$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$. Let $f \in F$. Then :there exists an auxiliary relation $\mathcal R$ on $S$ such that ::$\forall x \in S:f\left({x}\right) = x^{\mathcal R}$ where $x^{\mathcal R}$ denotes the $\mathcal R$-segment of $x$.
Binomial Coefficient with Two/Corollary
:$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$ where $T_n$ is the $n$th triangular number.
Increasing Mappings Satisfying Inclusion in Lower Closure is Isomorphic to Auxiliary Relations
Let $R = \left({S, \preceq}\right)$ be a bounded below join semilattice. Let $\mathit{Ids}\left({R}\right)$ be the set of all ideals in $R$. Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$. Let $M = \left({F, \preccurlyeq}\right)$ be the ordered set of increasing mappings $g:S \to \mathit{Ids}\left({R}\right)$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$. Let $\mathit{Aux}\left({R}\right)$ be the set of all auxiliary relations on $S$. Let $P = \left({ \mathit{Aux}\left({R}\right), \precsim'}\right)$ be an ordered set where $\precsim' \mathop = \subseteq\restriction_{\mathit{Aux}\left({R}\right) \times \mathit{Aux}\left({R}\right)}$. Then :there exists an order isomorphism between $P$ and $M$
Binomial Theorem/Abel's Generalisation
:$\displaystyle \left({x + y}\right)^n = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}$
Sum over k of r Choose m+k by s Choose n+k
Let $s \in \R, r \in \Z_{\ge 0}, m, n \in \Z$. Then: :$\displaystyle \sum_k \binom r {m + k} \binom s {n + k} = \binom {r + s} {r - m + n}$
Sum over k of r Choose k by s+k Choose n by -1^r-k
Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$. Then: :$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$
Sum over k of r-k Choose m by s+k Choose n
Let $m, n, r, s \in \Z_{\ge 0}$ such that $n \ge s$. Then: :$\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$
Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
Let $r, s, t \in \R, n \in \Z$. Then: :$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
Sum over k of r Choose k by s Choose k by k
:$\displaystyle \sum_k \binom r k \binom s k k = \binom {r + s - 1} {r - 1} s$
Relation Segment is Increasing
Let $S$ be a set. Let $\RR, \QQ$ be relations on $S$ such that :$\RR \subseteq \QQ$ Let $x \in S$. Then :$x^\RR \subseteq x^\QQ$ where $x^\RR$ denotes the $\RR$-segment of $x$.
Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1
:$\displaystyle \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$
Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice. Let $x \in S$. Then :$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$ where $\mathit{Ids}$ denotes the set of all ideals in $L$.
Way Below Closure is Ideal in Bounded Below Join Semilattice
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice. Let $x \in S$. Then :$x^\ll$ is ideal in $L$.
Sum over k of r Choose k by -1^r-k by Polynomial
Let $r \in \Z_{\ge 0}$. Then: :$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$ where: :$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
Sum over k of r Choose k by s-kt Choose r by -1^k
Let $r \in \Z_{\ge 0}$. Then: :$\displaystyle \sum_k \binom r k \binom {s - k t} r \left({-1}\right)^k = t^r$ where $\dbinom r k$ etc. are binomial coefficients.
Multinomial Coefficient expressed as Product of Binomial Coefficients
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$ where: :$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a multinomial coefficient :$\dbinom {k_1 + k_2} {k_1}$ etc. denotes binomial coefficients.
Universal Closures are Semantically Equivalent
Let $\mathbf A$ be a WFF of predicate logic. Let $\mathbf B, \mathbf B'$ be universal closures of $\mathbf A$. Then $\mathbf B$ and $\mathbf B'$ are semantically equivalent.
Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset
Let $L = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice. Then :$L$ is meet-continuous  :$\forall x \in S$, directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$
Continuous Lattice is Meet-Continuous
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice. Then $L$ is meet-continuous.
Number to Power of Zero Rising is One
:$x^{\overline 0} = 1$
Number to Power of Zero Falling is One
:$x^{\underline 0} = 1$
Product of Number by its Falling Factorial
Let $x^{\underline n}$ denote the $n$th falling factorial power of $x$. Then: :$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$
Product of Number by its Rising Factorial
Let $x^{\overline n}$ denote the $n$th rising factorial power of $x$. Then: :$x x^{\overline n} = x^{\overline {n + 1} } - n x^{\overline n}$
Value of Formula under Assignment Determined by Free Variables
Let $\mathbf A$ be a WFF of predicate logic. Let $\mathcal A$ be a structure for predicate logic. Let $\sigma, \sigma'$ be assignments for $\mathbf A$ in $\mathcal A$ such that: :For each free variable $x$ of $\mathbf A$, $\sigma \left({x}\right) = \sigma' \left({x}\right)$ Then: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]$ where $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right]$ is the value of $\mathbf A$ under $\sigma$.
First Inversion Formula for Stirling Numbers
For all $m, n \in \Z_{\ge 0}$: :$\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$ where: :$\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind :$\displaystyle \left\{ {k \atop m}\right\}$ denotes a Stirling number of the second kind :$\delta_{m n}$ denotes the Kronecker delta.
Unsigned Stirling Number of the First Kind of 1
:$\displaystyle {1 \brack n} = \delta_{1 n}$
Signed Stirling Number of the First Kind of 1
:$s \left({1, n}\right) = \delta_{1 n}$
Stirling Number of the Second Kind of 1
:$\displaystyle {1 \brace n} = \delta_{1 n}$
Second Inversion Formula for Stirling Numbers
For all $m, n \in \Z_{\ge 0}$: :$\displaystyle \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$ where: :$\displaystyle \left\{ {n \atop k}\right\}$ denotes a Stirling number of the second kind :$\displaystyle \left[{k \atop m}\right]$ denotes an unsigned Stirling number of the first kind :$\delta_{m n}$ denotes the Kronecker delta.
Zero Choose n
:$\dbinom 0 n = \delta_{0 n}$
Unsigned Stirling Number of the First Kind of 0
:$\displaystyle \left[{0 \atop n}\right] = \delta_{0 n}$
Particular Values of Signed Stirling Numbers of the First Kind
This page gathers together some particular values of signed Stirling numbers of the first kind.
Particular Values of Unsigned Stirling Numbers of the First Kind
This page gathers together some particular values of unsigned Stirling numbers of the first kind.
Particular Values of Stirling Numbers of the Second Kind
This page gathers together some particular values of Stirling numbers of the second kind.
Stirling Number of the Second Kind of 0
:$\displaystyle {0 \brace n} = \delta_{0 n}$
Unsigned Stirling Number of the First Kind of Number with Self
:$\displaystyle \left[{n \atop n}\right] = 1$
Unsigned Stirling Number of the First Kind of Number with Greater
Let $\displaystyle {n \brack k}$ denote an unsigned Stirling number of the first kind. Then: :$\displaystyle {n \brack k} = 0$
Signed Stirling Number of the First Kind of Number with Greater
Let $\map s {n, k}$ denote a signed Stirling number of the first kind. Then: :$\map s {n, k} = 0$
Stirling Number of Number with Greater
Let $n, k \in \Z_{\ge 0}$. Let $k > n$.
Stirling Number of the Second Kind of Number with Self
:$\displaystyle {n \brace n} = 1$
Value of Term under Assignment Determined by Variables
Let $\tau$ be a term of predicate logic. Let $\mathcal A$ be a structure for predicate logic. Let $\sigma, \sigma'$ be assignments for $\tau$ in $\mathcal A$ such that: :For each variable $x$ occurring in $\tau$, $\sigma \left({x}\right) = \sigma' \left({x}\right)$ Then: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]$ where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the value of $\tau$ under $\sigma$.
Relation between Signed and Unsigned Stirling Numbers of the First Kind
Let $m, n \in \Z_{\ge 0}$ be positive integers. Then: :$\displaystyle {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$ where: :$\displaystyle {n \brack m}$ denotes an unsigned Stirling number of the first kind :$\map s {n, m}$ denotes a signed Stirling number of the first kind.
Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1
Let $n, k \in \Z_{\ge 0}$ such that $k > n$. {{:Unsigned Stirling Number of the First Kind of Number with Greater}}
Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2
Let $n, k \in \Z_{\ge 0}$ such that $k > n$. {{:Unsigned Stirling Number of the First Kind of Number with Greater}}
Stirling Number of the Second Kind of Number with Greater/Proof 1
Let $n, k \in \Z_{\ge 0}$ such that $k > n$. {{:Stirling Number of the Second Kind of Number with Greater}}
Stirling Number of the Second Kind of Number with Greater/Proof 2
Let $n, k \in \Z_{\ge 0}$ such that $k > n$. {{:Stirling Number of the Second Kind of Number with Greater}}
Signed Stirling Number of the First Kind of Number with Self
:$s \left({n, n}\right) = 1$
Unsigned Stirling Number of the First Kind of n with n-1
:$\displaystyle \left[{n \atop n - 1}\right] = \binom n 2$
Stirling Number of the Second Kind of n with n-1
:$\displaystyle {n \brace n - 1} = \binom n 2$
Signed Stirling Number of the First Kind of n with n-1
:$\map s {n, n - 1} = -\dbinom n 2$
Unsigned Stirling Number of the First Kind of n+1 with 0
:$\displaystyle {n + 1 \brack 0} = 0$
Stirling Number of the Second Kind of n+1 with 0
:$\displaystyle {n + 1 \brace 0} = 0$
Unsigned Stirling Number of the First Kind of n+1 with 1
:$\displaystyle \left[{n + 1 \atop 1}\right] = n!$
Stirling Number of the Second Kind of n+1 with 1
:$\displaystyle {n + 1 \brace 1} = 1$
Signed Stirling Number of the First Kind of n+1 with 1
:$s \left({n + 1, 1}\right) = \left({-1}\right)^n n!$
Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$ where: :$\displaystyle \left[{n \atop k}\right]$ denotes an unsigned Stirling number of the first kind :$\dbinom k m$ denotes a binomial coefficient.
Semantically Equivalent Terms are Equal
Let $\tau_1, \tau_2$ be terms. Suppose that they are semantically equivalent with respect to the empty set. Then $\tau_1 = \tau_2$.
Down Mapping is Generated by Approximating Relation
Let $L = \left({S, \wedge, \preceq}\right)$ be a bounded below meet-continuous meet semilattice. Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$. Let $I$ be an ideal in $L$. Let $f: S \to \mathit{Ids}\left({L}\right)$ be a mapping such that :$\forall x \in S: x \preceq \sup I \implies f\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$ and :$\forall x \in S: x \npreceq \sup I \implies f\left({x}\right) = x^\preceq$ where $x^\preceq$ denotes the lower closure of $x$. Then :there exists an auxiliary approximating relation $\mathcal R$ on $S$ such that ::$\forall s \in S: f\left({s}\right) = s^{\mathcal R}$ where $s^{\mathcal R}$ denotes the $\mathcal R$-segment of $s$.
Substitution Instance of Term is Term
Let $\beta, \tau$ be terms of predicate logic. Let $x \in \operatorname {VAR}$ be a variable. Let $\map \beta {x \gets \tau}$ be the substitution instance of $\beta$ substituting $\tau$ for $x$. Then $\map \beta {x \gets \tau}$ is a term.
Substitution Theorem for Terms
Let $\beta, \tau$ be terms. Let $x \in \mathrm{VAR}$ be a variable. Let $\beta \left({x \gets \tau}\right)$ be the substitution instance of $\beta$ substituting $\tau$ for $x$. Let $\mathcal A$ be a structure for predicate logic. Let $\sigma$ be an assignment for $\beta$ and $\tau$. Suppose that: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = a$ where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the value of $\tau$ under $\sigma$. Then: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta}\right) } \left[{\sigma + \left({x / a}\right)}\right]$ where $\sigma + \left({x / a}\right)$ is the extension of $\sigma$ by mapping $x$ to $a$.
Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \left[{n + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{n \atop m}\right]$ where: :$\displaystyle \left[{n + 1 \atop k + 1}\right]$ etc. denotes an unsigned Stirling number of the first kind :$\dbinom k m$ denotes a binomial coefficient.
Substitution Theorem for Well-Formed Formulas
Let $\mathbf A$ be a WFF of predicate logic. Let $x \in \mathrm{VAR}$ be a variable. Let $\tau$ be a term of predicate logic which is freely substitutable for $x$ in $\mathbf A$. Let $\mathbf A \left({x \gets \tau}\right)$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$. Let $\mathcal A$ be a structure for predicate logic. Let $\sigma$ be an assignment for $\mathbf A$ and $\tau$. Suppose that: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = a$ where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the value of $\tau$ under $\sigma$. Then: :$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]$ where $\sigma + \left({x / a}\right)$ is the extension of $\sigma$ by mapping $x$ to $a$.
Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$ where: :$\displaystyle \left\{ {k \atop m}\right\}$ denotes a Stirling number of the second kind :$\dbinom n k$ denotes a binomial coefficient.
Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom n k \left({-1}\right)^{n - k} = \left\{ {n \atop m}\right\}$ where: :$\displaystyle \left\{ {k + 1 \atop m + 1}\right\}$ etc. denotes a Stirling number of the second kind :$\dbinom n k$ denotes a binomial coefficient.
Universal Instantiation/Informal Statement
Suppose we have a universal statement: :$\forall x: \map P x$ where $\forall$ is the universal quantifier and $\map P x$ is a propositional function. Then we can deduce: :$\map P {\mathbf a}$ where $\mathbf a$ is any arbitrary object we care to choose in the universe of discourse. In natural language: :''Suppose $P$ is true of everything in the universe of discourse.'' :''Let $\mathbf a$ be an element of the universe of discourse." :''Then $P$ is true of $\mathbf a$.''
Universal Instantiation/Model
Let $\map {\mathbf A} x$ be a WFF of predicate logic. Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$. Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology.
Existential Generalisation/Informal Statement
:$P \left({\mathbf a}\right) \vdash \exists x: P \left({x}\right)$ Suppose we have the following: : We can find an arbitrary object $\mathbf a$ in our universe of discourse which has the property $P$. Then we may infer that: : there exists in that universe ''at least one'' object $x$ which has that property $P$. This is called the '''Rule of Existential Generalisation''' and often appears in a proof with its abbreviation '''EG'''.
Way Below is Approximating Relation
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice. Then $\ll$ is an approximating relation on $S$.
Intersection of Relation Segments of Approximating Relations equals Way Below Closure
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below meet-continuous lattice. Let $\mathit{App}\left({L}\right)$ be the set of all auxiliary approximating relations on $S$. Let $x \in S$. Then : $\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = x^\ll$
Limit of Monotone Real Function/Increasing
Let $f$ be a real function which is increasing and bounded above on the open interval $\openint a b$. Let the supremum of $f$ on $\openint a b$ be $L$. Then: :$\displaystyle \lim_{x \mathop \to b^-} \map f x = L$ where $\displaystyle \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$.
Limit of Monotone Real Function/Decreasing
Let $f$ be a real function which is decreasing and bounded below on the open interval $\openint a b$. Let the infimum of $f$ on $\openint a b$ be $l$. Then: :$\displaystyle \lim_{x \mathop \to b^-} \map f x = l$ where $\displaystyle \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$.
Sum over k of m choose k by -1^m-k by k to the n
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$ where: :$\dbinom m k$ denotes a binomial coefficient :$\displaystyle {n \brace m}$ etc. denotes a Stirling number of the second kind :$m!$ denotes a factorial.
Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$ where: :$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients :$\displaystyle {m + k \brace k}$ denotes a Stirling number of the second kind :$\displaystyle {n \brack n - m}$ denotes an unsigned Stirling number of the first kind.
Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$ where: :$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients :$\displaystyle {m + k \brack k}$ denotes an unsigned Stirling number of the first kind :$\displaystyle {n \brace n - m}$ denotes a Stirling number of the second kind.
Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m
Let $m, n \in \Z_{\ge 0}$. :$\displaystyle \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$ where: :$\dbinom n m$ denotes a binomial coefficient :$\displaystyle \left[{k \atop m}\right]$ denotes an unsigned Stirling number of the first kind :$\displaystyle \left\{ {n + 1 \atop k + 1}\right\}$ denotes a Stirling number of the second kind.
Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below meet-continuous lattice. Let $\mathit{Ids}$ be the set of all ideals in $L$. Let for all $I \in \mathit{Ids}$: $m_I: S \to \mathit{Ids}$ be a mapping: :$\forall x \in S: x \preceq \sup I \implies m_I\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$ and :$\forall x \in S: x \npreceq \sup I \implies m_I\left({x}\right) = x^\preceq$ where $x^\preceq$ denotes the lower closure of $x$. Let $x \in S$. Then :$\displaystyle \bigcap \left\{ {m_I \left({x}\right): I \in \mathit{Ids} }\right\} = x^\ll$ where $x^\ll$ denotes the way below closure of $x$.
Zero Choose Zero
:$\dbinom 0 0 = 1$
Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal
Let $\struct {S, \preceq}$ be a meet semilattice. Let $I$ be an ideal in $\struct {S, \preceq}$. Let $x \in S$. Then: :$\paren {x^\preceq} \cap I = \set {x \wedge i: i \in I}$ where $x^\preceq$ denotes the lower closure of $x$.
Condition on Lower Coefficient for Binomial Coefficient to be Maximum
Let $n, k \in \Z_{\ge 0}$ be positive integers. Let $\dbinom n k$ denote the binomial coefficient of $n$ choose $k$. Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a maximum is: :$k_\max = \floor {\dfrac n 2}$ and also: :$k_\max = \ceiling {\dfrac n 2}$
N Choose Negative Number is Zero
Let $n \in \Z$ be an integer. Let $k \in \Z_{<0}$ be a (strictly) negative integer. Then: :$\dbinom n k = 0$
Kummer's Theorem
Let $p$ be a prime number. Let $a, b \in \Z_{\ge 0}$. Let: :$p^n \divides \dbinom {a + b} b$ but :$p^{n + 1} \nmid \dbinom {a + b} b$ where: :$\divides$ denotes divisibility :$\nmid$ denotes non-divisibility :$\dbinom {a + b} b$ denotes a binomial coefficient. Then $n$ equals the number of carries that occur when $a$ is added to $b$ using the classical addition algorithm in base $p$.
Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1
Let $n \in \Z_{\ge 0}$ be a positive integer. Then the non-zero elements of the $n$th row of Pascal's triangle are all odd : :$n = 2^m - 1$ for some $m \in \Z_{\ge 0}$. {{:Definition:Pascal's Triangle}} As can be seen, the entries in rows $0, 1, 3, 7$ are all odd.
Sum of Sequence of Fourth Powers
:$\displaystyle \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$
Complete Lattice is Bounded
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice. Then :$L$ is bounded
Constant Function is Primitive Recursive/General Case
The constant function of $k$ variables: $f_c^k: \N^k \to \N$, defined as: :$\map {f_c^k} {n_1, n_2, \ldots, n_k} = c$ where $c \in \N$ is primitive recursive.
Value of Multiplicative Function at One
Let $f: \N \to \C$ be a multiplicative function. If $f$ is not identically zero, then $\map f 1 = 1$.