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The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

An integer equal to the third power of another integer, such as $8=2^3$ or $-27=(-3{)}^3$.

The ratio of the sum of the positive divisors of a positive integer to that integer. For coprime $a,b$, the ratio is multiplicative because the divisor function $\sigma$ is multiplicative. The contribution of a prime power $p^e$ is $(1+p+\cdots +p^e)/p^e$, so a prescribed quotient can be searched through its remaining numerator and denominator.

A perfect number is a positive integer equal to the sum of all its positive proper divisors. The divisor itself is excluded: including it gives the equivalent condition $\sigma (n)=2n$ for the sum-of-divisors function. The smallest examples are $6=1+2+3$ and $28=1+2+4+7+14$.

For comparison, $8$ has proper divisors $1,2,4$, whose sum is $7$: it is a deficient number. The proper divisors of $12$ are $1,2,3,4,6$, whose sum is $16$: it is an abundant number. The proper-divisor sum classifies these examples with the same rule, and gives $s(1)=0$, so one is not a perfect number.

The interesting question is how such balances can be constructed without testing every smaller integer as a possible divisor. The Euclid-Euler theorem answers it for even integers: a Mersenne prime $2^p-1$ gives $2^{p-1}(2^p-1)$, and every even perfect number arises this way. For $p=2,3,5,7$, this produces $6,28,496,8128$. The connection links divisor sums to primality testing for a particular family, while keeping the necessary hypotheses visible.^[ref]

Euclid's Elements already gives the construction in Book IX, Proposition 36: add a succession of powers of two, and if their sum is prime, multiply by the final power of two. The modern finite geometric series $1+2+\cdots +2^{p-1}=2^p-1$ makes the same pattern transparent. This original statement supplies the forward construction, rather than the converse classification of all even perfect numbers.^[ref]

For a nonnegative integer $n$, compute its integer square root $r=\lfloor \sqrt{n}\rfloor$ and accept exactly when $r^2=n$. An approximate square root must be corrected with exact integer comparisons before this equality is tested. This procedure distinguishes a squareness test from merely approximating a square root.

For positive integers $a,b$, partition the nonnegative integers by equality of $\lfloor n/a\rfloor$ and separately by equality of $\lfloor n/b\rfloor$. The intersection of equivalence relations gives nonempty intersections of these consecutive blocks. Its boundaries occur at multiples of either $a$ or $b$.

Let $\ell =\mathrm{lcm}(a,b)$. Translating by $\ell$ increases the two floor quotients by the fixed integers $\ell /a$ and $\ell /b$, so it preserves all within-block equalities and all block boundaries. Thus the refinement repeats in intervals of length $\ell$. If $0=c_0<c_1<\cdots <c_r=\ell$ lists the multiples of $a$ or $b$ within one period, its blocks are exactly $\{j\ell +c_i,\dots ,j\ell +c_{i+1}-1\}$ for $j\ge 0$ and $0\le i<r$. The pattern repeats, but blocks in different periods remain distinct equivalence classes.

For $T>0$, values of a function $f$ on a half-open interval $[a,a+T)$ determine its periodic extension to the real line: choose the unique integer $k$ with $x-kT\in [a,a+T)$ and set $F(x)=f(x-kT)$. The resulting periodic function has period $T$. If the endpoint limits of $f$ differ, the extension has a jump discontinuity at every period boundary. Under the usual piecewise-smooth convergence hypotheses, its Fourier series takes the average of the two one-sided limits there; an assigned value at an isolated endpoint does not change the Fourier coefficients.

A block $x_0,\dots ,x_{N-1}$ determines a repeating sequence by setting $x_n=x_{n\mathrm{mod}N}$ for every integer $n$. The inverse discrete Fourier transform gives this extension because each discrete Fourier mode satisfies $e^{2\pi ik(n+N)/N}=e^{2\pi ikn/N}$. The extension explains why circular convolution wraps indices around the block. It specifies values at discrete indices; a continuous curve between samples requires an additional choice of interpolation.

If a periodic unimodular function $a$ and its inverse have absolutely summable Fourier coefficients, replacing ${\hat{\phi }}_0$ by $\hat{\phi }=a{\hat{\phi }}_0$ preserves the orthonormal integer-translate span and integrability of an integrable scaling function. In physical space this is an absolutely summable combination of integer function translations. The new MRA low-pass filter is $m(t)=a(2t)m_0(t)/a(t)$. A rough phase can therefore preserve the multiresolution analysis while changing MRA low-pass filter regularity.

A sequence indexed from zero with an integer period $p\ge 1$ satisfying the displayed identity for all $n\ge 0$. Repeating a finite cycle is an example. The Cesàro mean article explains why its cumulative averages approach the one-cycle arithmetic mean.

The permanent of a matrix adds the products obtained by choosing one entry in each row and one in each column. For an $n\times n$ matrix $A=(a_{ij})$ with entries in the real numbers or complex numbers,Here each permutation $\sigma$ chooses column $\sigma (i)$ in row $i$, and the symmetric group $S_n$ lists every such bijection. The formula resembles the Leibniz formula for determinants, but omits the sign of a permutation. For a two-by-two matrix,The absence of alternating signs makes the permanent of a matrix particularly useful for counting allowed assignments.

For a zero-one matrix, let rows represent people, columns represent jobs, and $a_{ij}=1$ mean that person $i$ can do job $j$. This is the biadjacency matrix of a bipartite graph. A product in the permanent is one exactly when all the selected assignments are allowed; otherwise it is zero. Each surviving permutation assigns every person a different job, which is a perfect matching. Thus the permanent counts perfect matchings. This interpretation and the definition are given in Yufei Zhao, Probabilistic Methods in Combinatorics, §10.2, pp. 178–179.

For a fully enumerated example, takeAmong the six permutations of three columns, exactly three give nonzero products: $(1,2,3)$, $(1,3,2)$ and $(2,1,3)$. These tuples list the chosen column in each successive row. Each product is one, so $\mathrm{perm}A=3$. Their permutation signs are $+1,-1,-1$, giving $\det A=-1$. The diagram displays all three allowed assignments.

A zero determinant need not mean that no assignment exists. The two-by-two all-ones matrix has permanent $2$ and determinant $0$: its two assignments cancel in the determinant. The nonzero determinant criterion for a perfect matching is sufficient for existence, but the converse fails for a fixed zero-one matrix.

For nonnegative entries, the permanent instead sums assignment weights, with each assignment weighted by the product of its selected entries. It is then a total weight rather than necessarily an integer count. With arbitrary entries in the real numbers or complex numbers, even the permanent can have cancellation; the counting interpretation specifically uses zero-one entries. The permanent as probability of distinct independent choices gives another useful nonnegative interpretation.

There are $n!$ terms in the defining sum, linking direct evaluation to the factorial. For the all-ones $n\times n$ matrix, every term is one and the permanent is exactly $n!$; the identity matrix has permanent one. Finding one allowed assignment and counting all allowed assignments are different questions. Their relationship to computational complexity theory was studied in L. G. Valiant, The complexity of computing the permanent, Theoretical Computer Science 8(2), 189–201, 1979. The elementary sum is useful for small examples, while the existing dense permanent approximation describes a more specialized way of estimating large counts.

For a dividend-free Black-Scholes model with $r>0$, $\sigma >0$, $K>0$ and positive integer $n$, the optimal stopping value of the power put option isIn the continuation region the stationary Black-Scholes equation has powers $S$ and $S^{-\gamma }$. Boundedness at infinity removes $S$. Value matching gives the coefficient $(K-b{)}^n{b}^{\gamma }$, maximized at the displayed $b$; equivalently this is the smooth fit boundary.

For a verification beyond constant-trigger policies, let $g(S)=(K-S{)}_{+}^n$. The maximum property of $b^{\gamma }(K-b{)}^n$ gives $V\ge g$. For the infinitesimal generator $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$, $(\mathcal{L}-r)V=0$ above $b$. Below $b$, set $y=S/(K-S)\le \gamma /n$; thenThe candidate is continuously differentiable across $b$, so the generalized Itô formula has no derivative-jump local time of a semimartingale term. The discounted candidate is a nonnegative supermartingale, bounding every stopping reward by optional stopping. First hitting of $b$ attains the bound; the remainder at infinite time vanishes since $V\le K^n$ and $r>0$. For $n=1$ this recovers the perpetual put option. At $r=0$ the value is $K^n$ for $S>0$, approached but not attained at a finite exercise time. At $r<0$ the perpetual value is infinite, because exercise at deterministic times tending to infinity has unbounded discounted expected value.

Place $N$ objects into $m\ge 1$ boxes, assigning each object to exactly one box. Here $N$ is a nonnegative integer and $m$ is a positive integer. The pigeonhole principle says that if $N>m$, some box contains at least two objects. Equivalently, a function from an $N$-element finite set to an $m$-element finite set cannot be an injective function when $N>m$. The boxes may stand for categories rather than physical containers. ^[ref]

For thirteen objects and twelve boxes, a box containing at least two is unavoidable. Thinking of thirteen people and their birth months gives the same conclusion: at least two share a birth month. Twelve people could occupy the twelve months one apiece, so thirteen is the smallest number that forces a shared month. The generalized pigeonhole principle quantifies how many objects must share a box when there are still more objects.

The useful surprise is that the pigeonhole principle proves existence without needing to locate the objects involved. Choosing a good assignment rule is often the creative step: equal labels can mean equal remainders, nearby points, or repeated states of an algorithm. Once the rule is chosen, a short count can force a relationship among a much larger collection of objects. The pigeonhole principle itself does not name the crowded box. When the objects and their assignments are explicitly available, a simple algorithm finds one: scan the objects, keep the first object seen in each box, and stop when a later object enters an occupied box. At most $m+1$ scanned objects are needed when $N>m$.

A geometric example. Any five points in a square of side one contain two whose Euclidean distance is at most $1/\sqrt{2}$. Give each point Cartesian coordinates $(x,y)\in [0,1{]}^2$. Divide both ranges into the intervals $[0,1/2)$ and $[1/2,1]$, making four cells. Thus $x<1/2$ belongs to the left half and $x\ge 1/2$ to the right half; $y<1/2$ belongs to the lower half and $y\ge 1/2$ to the upper half. This convention assigns every boundary point to exactly one cell.

The pigeonhole principle puts two of the five points in the same cell. Their differences in Cartesian coordinates have magnitude at most $1/2$, so the Pythagorean theorem gives their Euclidean distance $d$ the boundFor the plotted points $(0.1,0.1)$, $(0.4,0.4)$, $(0.8,0.2)$, $(0.2,0.8)$ and $(0.8,0.8)$, the first two occupy the lower-left cell and have Euclidean distance $\sqrt{0.3^2+0.3^2}=\sqrt{0.18}\approx 0.424264$. These Cartesian coordinates exhibit the forced pair; the argument applies to every placement of five points.

An arithmetic example. Assign eight integers to their seven residue classes modulo the integer modulus $7$. The pigeonhole principle gives two with the same residue class, so $7$ is an integer divisor of their difference. In $\{2,5,9,12,16,20,24,27\}$, for example, $2$ and $9$ satisfy the integer congruence $9\equiv 2(\mathrm{mod}7)$, and $9-2=7$. Likewise, assigning fractional parts to short intervals gives the Dirichlet approximation theorem; its existing article explains the approximation bound.

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.

For an integer $n$ to be factored, John Pollard's method chooses an integer $a$ coprime to $n$ and an exponent $M$ divisible by many small prime powers. Compute $d=\gcd (a^M-1,n)$ by modular exponentiation and the Euclidean algorithm. If $1<d<n$, it is a proper factor. If a prime divisor $p$ of $n$ has $p-1\mid M$, Fermat's little theorem ensures $p\mid d$; success requires some other prime divisor not to divide $a^M-1$. Common choices are $M=B!$ and $M=\mathrm{lcm}(1,\dots ,B)$. The method exploits a smooth $p-1$, rather than a small factor $p$ itself.

For integers $s\ge 3$ and $n\ge 1$, the $n$th $s$-gonal number is $P_{s,n}=((s-2)n^2-(s-4)n)/2$. Its successive differences form an arithmetic progression. The cases $s=3,4,5,6$ give triangular numbers, square numbers, pentagonal numbers, and hexagonal numbers.

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.