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A triple of positive integers $(a,b,c)$ with $a<b$, $a+b=c$, $\gcd (a,b)=1$, and $\mathrm{rad}(abc)<c$, where $\mathrm{rad}$ is the radical of an integer. The example $(1,8,9)$ has $\mathrm{rad}(72)=6<9$.

A positive integer whose proper-divisor sum exceeds itself; equivalently, its sum-of-divisors function satisfies $\sigma (n)>2n$. For example, $s(12)=16>12$.

An addition chain for a positive integer $k$ is $1=a_0<a_1<\cdots <a_r=k$, where each $a_t$ is the sum of two earlier entries, which may coincide. It computes $x^k$ in $r$ multiplications by reusing previously computed powers. For example, $1,2,3,6,12,15$ computes $x^{15}$ in five multiplications.

A positive integer $A=pqr$ with nonzero integers $p,q,r$ satisfying $1/A=1/p+1/q+1/r$. Equivalently, $pq+pr+qr=1$. A useful parametrization is $A=n(n+d)(n+e)$ for positive integers $n,d,e$ with $de=n^2+1$, coming from the signed triple $(n,-n-d,-n-e)$.

Two distinct positive integers $a,b$ are amicable when the sum of the proper divisors of $a$ is $b$ and the sum of the proper divisors of $b$ is $a$. The pair $220,284$ is an example. Iterating the proper-divisor sum can also produce longer directed cycles.

A pair of distinct positive integers in which each is the proper-divisor sum of the other. The integers $220$ and $284$ form an amicable pair.

The arithmetic derivative on positive integers is defined by $p\prime =1$ for every prime $p$ and the product rule $(ab)\prime =a\prime b+ab\prime$. For $n={\prod }_i{p}_i^{e_i}$, these rules give $n\prime =n{\sum }_i{e}_i/p_i$. In particular, $(p^e)\prime =e{p}^{e-1}$.

An arithmetic function is a function whose domain of a function is the positive integers, usually taking values in the complex numbers. Thus $f(1),f(2),\dots$ are its values; $f(0)$ is outside this domain of a function. Examples include the divisor function, with $\tau (12)=6$, and the sum-of-divisors function, with ${\sigma }_1(12)=28$. Studying an arithmetic function packages a question about each positive integer into one object whose values can be combined using Dirichlet convolution. ^[ref]

Every red-blue finite coloring of the pairs of positive integers has either a blue complete arithmetic progression of any prescribed finite length $m$, or two disjoint $m$-term arithmetic progressions with all cross-pairs red. To see the geometry, assign $(a,d)$ a chosen red pair of indices $(r,s)$ in its progression, assuming the first alternative fails. Apply the Gallai theorem for an integer lattice to the fixed finite patternIn a monochromatic copy $(a_0,d_0)+qF$ with common label $(r,s)$, the selected red pairs are preciselyThe copy lies in the positive quadrant, so $d_0>q(m-1)$. Consequently the two displayed arithmetic progressions are disjoint. Translating the finite pattern before applying the lattice theorem permits its initially negative coordinates.

A balanced Dirichlet hyperbola method split takes real thresholds $A=B=\sqrt{N}$. An asymmetric split takes $1\le A\le N$ and $B=N/A$. Every lattice point with $ab\le N$ then has $a\le A$ or $b\le B$, and every point in their overlap rectangle is allowed because $AB=N$. For arithmetic functions $f,g$, define their partial sums by $F(t)={\sum }_{a\le t}f(a)$ and $G(t)={\sum }_{b\le t}g(b)$, always over positive integers. The same inclusion-exclusion principle, now with weight $f(a)g(b)$, yieldsNo multiplicativity assumption is needed: this is a finite rearrangement of Dirichlet convolution. The constant-one arithmetic function in both positions recovers the divisor summatory function. For $N=12$, $A=2$, $B=6$ givesIt uses two columns and six rows, with a twelve-point overlap.

For general positive real thresholds, $AB\ge N$ is sufficient for coverage, since $a>A$ and $b>B$ would imply $ab>AB\ge N$. If $AB>N$, however, the overlap can include points above the hyperbola. The correct subtraction is then ${\sum }_{a\le A,b\le B,ab\le N}f(a)g(b)$; the product $F(A)G(B)$ is justified only when the whole integer rectangle lies below it. For example, $N=12$, $A=B=4$ covers every allowed pair, but its rectangle contains the forbidden pair $(4,4)$, so subtracting all sixteen points would give $34$ instead of $35$.

Conversely, the balanced integer thresholds $r,r$ have $r^2\le N$ and still cover every allowed pair because $(r+1{)}^2>N$. More generally, positive integer thresholds $U,V$ give coverage exactly when $(U+1)(V+1)>N$, and their overlap is a full rectangle exactly when $UV\le N$. These discrete conditions explain why the square-root identity works even when $r^2<N$.

A positive integer whose first $\lceil k/2\rceil$ decimal numerical digits have the same sum as its last $\lceil k/2\rceil$ digits, where $k$ is the number of digits. For odd $k$, the shared middle digit cancels from the equality. Counts and value sums can therefore be assembled from two half-length digit dynamic programming tables indexed by base-p digit sum.

The Basel problem asks for the exact sum of the reciprocal squares of the positive integers. The answer isThe integral test proves that the series converges, but does not identify this exact value. The Euler evaluation of even zeta values does: the equally spaced poles of cotangent connect reciprocal integer powers to Bernoulli numbers and $\pi$. This separates three different questions: whether a sum exists, how well a partial sum approximates it, and what its exact value is.

For example, the ten-term partial sum is about $1.549768$, while the exact value is about $1.644934$. The zeta series truncation error bounds the missing amount between $1/11$ and $1/10$; it is much larger than the first omitted term $1/121$. More terms improve the approximation, but a numerical coincidence with ${\pi }^2/6$ would not prove the identity.

In On the sums of series of reciprocals by Leonhard Euler, §2 identifies the reciprocal-square sum with one-sixth of the squared circumference of a unit-diameter circle. The translation's publication note dates the presentation to 5 December 1735 and publication to 1740.^[ref] The cotangent coefficient proof of the even zeta values gives a short modern outline, while zeta series truncation error addresses the distinct approximation question.

The sequence $\lfloor n\alpha \rfloor$ for $n=1,2,\dots$, with a fixed positive irrational number $\alpha$. Complementary Beatty sequences for $\alpha ,\beta >1$ partition the positive integers when $1/\alpha +1/\beta =1$.

If $\alpha ,\beta >1$ are irrational numbers with $1/\alpha +1/\beta =1$, the Beatty sequences $\lfloor n\alpha \rfloor$ and $\lfloor n\beta \rfloor$ partition the positive integers.

The Bernoulli numbers are a sequence of rational numbers that turns integer power sums into exact polynomials. The same numbers describe corrections when a discrete summation is compared with an integral. For example, the sum of squares $1^2+\cdots +N^2$ is $N^3/3+N^2/2+N/6$: the leading term is an integral, while the last two terms repair the difference between area and sampled values.

We use $B_1=-\frac{1}{2}$ throughout. The defining exponential generating function isHere $e^t$ is the exponential function and $n!$ is a factorial; in particular, the coefficient of $t^2$ is $B_2/2!$, rather than $B_2$. At $t=0$ the quotient has limiting value one. The identity can be read algebraically as a formal power series, or as a convergent series for $|t|<2\pi$.^[ref]

To compute the numbers, multiply by $e^t-1$ and compare coefficients. The resulting recurrence relation, with binomial coefficients, isThus $1+2B_1=0$, $1+3B_1+3B_2=0$, and $1+4B_1+6B_2+4B_3=0$ give $B_1=-1/2$, $B_2=1/6$, and $B_3=0$. Continuing givesThe recurrence relation agrees with equation 24.5.3 after replacing its index $n$ by $m+1$.^[ref]

Why do so many values vanish? Put $F(t)=t/(e^t-1)$. Direct substitution gives $F(-t)=F(t)+t$, so $F(t)+t/2$ is an even function. Its odd power series coefficients vanish: $B_{2r+1}=0$ for every positive integer $r$. The exception $B_1$ matters. The alternative convention obtained from $t/(1-e^{-t})=F(t)+t$ has $B_1^{+}=+1/2$ and exactly the same other values. A formula copied between conventions must also retain its summation endpoints.

Historically, integer power sums were studied before the modern Bernoulli number notation. A historical study examines formulas published in 1631 and distinguishes those separate formulas from encoding all exponents using one sequence of constants. This is a useful motivation for the generating function, without assigning all discovery to a single moment. ^[ref]

Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

A positive integer whose sequence of numerical digits is neither weakly increasing nor weakly decreasing from left to right. Equivalently, one adjacent pair strictly rises and another strictly falls. Numbers in both monotone families have all digits equal, so their overlap consists of the repdigits.

For positive integers $k,s,\ell$, some $N$ ensures that every $k$-color finite coloring of $[N]$ contains a monochromatic setIt follows from the Van der Waerden theorem by mathematical induction on the number of colors. If $M$ is a bound for $k-1$ colors, take a long one-color arithmetic progression of length $(\ell -1)M+1$ and step $d$. Either one of $sd,2sd,\dots ,Msd$ has its color, giving the result immediately, or these multiples use only $k-1$ colors. Pull back their finite coloring to $[M]$ and use mathematical induction, then multiply the resulting configuration by $sd$. Enlarge the ambient finite integer interval to include all these multiples. The cases $k=1$ and $\ell =1$ are immediate. The parameter $s$ makes this useful for positive monochromatic solutions of a partition regular equation with arbitrary integer coefficients.

Every colouring of the positive integers, even with infinitely many colours, has an arithmetic progression of any prescribed finite length on which the colouring is constant or injective. Encode a progression by the equivalence relation of equality of its colours. This gives a finite colouring of its starting point and step. A Gallai theorem for an integer lattice copy of a suitable two-dimensional pattern makes either all indices inequivalent, or one equal-colour pair range independently over two whole progressions. In the second case, one of those progressions is monochromatic.

A triple of positive integers $(a,b,c)$ satisfying $\sqrt[3]{a+b\sqrt{c}}+\sqrt[3]{a-b\sqrt{c}}=1$, with real cube roots. Cubing the equality yields the equivalent integer identity $27b^{2}c=(a+1{)}^2(8a-1)$. In particular $a=3k-1$ for a positive integer $k$, and the condition becomes $b^{2}c=k^2(8k-3)$.