The base-ten Kaprekar routine takes a fixed-width digit string, sorts its digits into descending and ascending order, and subtracts the second integer from the first. Leading zeros are retained so the next step uses the same width.

For four digits, $3524\to 5432-2345=3087\to 8730-0378=8352\to 8532-2358=6174$. The next step returns $7641-1467=6174$, a Kaprekar constant. Equal-digit inputs instead reach zero. Since there are finitely many fixed-width strings, every orbit eventually reaches a fixed point or a cycle. This finite-state observation does not by itself prove which attractor occurs.

For $n\ge 1$, let $r_4(n)$ count the integer quadruples $(a,b,c,d)$ with $a^2+b^2+c^2+d^2=n$, counting order, signs, and zeros. Jacobi's four-square theorem givesThe right side is positive, so this also implies that every positive integer is a sum of four squares.

For $n=1$, the formula gives eight representations: choose the position of $\pm 1$. For $n=2$, it gives $8(1+2)=24$, matching the six choices of two nonzero positions and four sign choices. Specifying the counting convention matters: unordered nonnegative representations have a different count.

A cube-full number is a positive integer whose every occurring prime factor has prime exponent at least three. Equivalently, if a prime number $p$ divides $n$, then $p^3$ divides $n$. Examples are $8=2^3$, $16=2^4$ and $216=2^3{3}^3$. The number $72=2^3{3}^2$ fails because its factor three occurs only twice. The number $1$ is included, since it has no prime factors violating the condition.

Every cube-full number is a powerful number, but the converse fails: $36=2^2{3}^2$ is powerful without being cube-full. Being cube-full also differs from being a perfect cube; the exponents need only be at least three, rather than multiples of three. Thus $16$ is cube-full but not a perfect cube.

A pronic number is $n(n+1)$ for a nonnegative integer $n$, the product of two consecutive integers. The sequence begins $0,2,6,12,20,30$. Arranging points in a rectangle with side lengths $n$ and $n+1$ explains the product, while $n(n+1)=2\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)$ shows that each is twice a triangular number.

There is an exact recognition test: a nonnegative integer $m$ is a pronic number if and only if $4m+1$ is an perfect square. Indeed, $4n(n+1)+1=(2n+1{)}^2$; conversely an integer square equal to $4m+1$ has an odd positive square root $2n+1$, recovering $m=n(n+1)$. For $m=20$, the test gives $81=9^2$ and $n=4$.

The four exponentials conjecture asserts that if $x_1,x_2$ are complex numbers linearly independent over $\mathbb{Q}$, and $y_1,y_2$ are likewise linearly independent over $\mathbb{Q}$, then at least one of the four numbers $e^{x_i{y}_j}$ is a transcendental number.

Both independence hypotheses are essential to the statement. It is a problem in transcendental number theory, asking how a two-by-two array of products constrains exponentiation. The analogous statement with two independent $x_i$ and three independent $y_j$ is the six exponentials theorem; that stronger supply of inputs is known to force at least one transcendental output. The four-number assertion is a conjecture, not a consequence of checking individual exponentials.

A sum of two squares representation of a nonnegative integer $n$ is an equality $n=a^2+b^2$ with integers $a,b$. For example, $65=1^2+8^2=4^2+7^2$ has two representations even before changing signs or interchanging the two terms.

Counting conventions matter. If ordered pairs and signs are distinguished, the representation $(1,8)$ generates eight pairs, whereas $(0,5)$ generates only four and $(0,0)$ only one. Every perfect square is zero or one modulo four, so an integer congruent to three modulo four is impossible. That quick obstruction is insufficient by itself: $6$ also has no representation. The exact prime factorization criterion is the sum of two squares theorem.

A number-theoretic sieve that visits each composite number exactly once, through its smallest prime factor. Keep a growing list of prime numbers; for each $i$, generate $ip$ only for stored primes $p\le \mathrm{spf}(i)$. Each composite $m$ has a unique representation $m=\mathrm{spf}(m)(m/\mathrm{spf}(m))$ admitted by this rule, giving $O(N)$ arithmetic updates up to $N$.

A number-theoretic sieve that processes a large search range in smaller intervals. To find primes in $[L,H]$, first generate the primes at most $\sqrt{H}$, then mark their multiples in the current segment. Memory for the segment depends on its width rather than on $H$.

The Min_25 sieve computes summatory multiplicative functions by isolating weighted prime sums and then restoring composite contributions. A useful basic form applies when $f(p)={\sum }_{r=0}^d{a}_r{p}^r$ for primes $p$, the degree $d$ is small, and values on prime powers can be computed efficiently. The formulas below explain the two stages; optimized implementations differ in scheduling and cost.^[ref]

Use quotient arguments $Q_N=\{\lfloor N/i\rfloor :1\le i\le N\}$. There are $O(\sqrt{N})$ distinct values: indices at most $\sqrt{N}$ contribute that many, and larger indices yield values below $\sqrt{N}$. The floor function identity $\lfloor \lfloor N/i\rfloor /p\rfloor =\lfloor N/(ip)\rfloor$ keeps later quotients in this table.

For each exponent $r$, initialize $G_r(v)={\sum }_{m=2}^v{m}^r$. Process primes $p$ in increasing order, and eligible $v\ge p^2$ in decreasing order, applyingThis removes composites whose smallest prime factor is $p$. The subtraction inside parentheses excludes earlier primes from the possible cofactors. Decreasing $v$ ensures each smaller argument still has its pre-$p$ value. After sieving, $H(v)={\sum }_r{a}_r{G}_r(v)$ is the desired sum of $f$ over primes up to $v$.

To restore composites, let $p_j$ be the $j$th prime and let $T(x,j)$ sum $f(n)$ over $2\le n\le x$ with every prime factor greater than $p_j$. Set $T(x,j)=0$ for $x\le p_j$. Otherwise its prime contribution is $H(x)-H(p_j)$, and addUnique decomposition by smallest prime factor and its full exponent proves this recurrence: mixed numbers use the first term, pure powers use the second. With $p_0=1$, $H(1)=0$ and $f(1)=1$, the answer is $1+T(N,0)$. No complete-multiplicativity assumption is needed.

For the Euler totient function up to six, the prime contribution is $\varphi (2)+\varphi (3)+\varphi (5)=7$. The composites add $\varphi (4)+\varphi (6)=4$, and $\varphi (1)=1$ gives total twelve. This independently chosen small example checks that prime powers and mixed products are both counted once.

A Du Jiao sieve computes a summatory arithmetic function using a Dirichlet convolution identity and memoization of repeated quotient arguments. For the Mertens function $M(n)={\sum }_{k\le n}\mu (k)$, the identity ${\sum }_{d\mid k}\mu (d)=[k=1]$ gives $M(n)=1-{\sum }_{i=2}^{n}M(\lfloor n/i\rfloor )$. Grouping equal quotients and precomputing small arguments makes this recurrence relation useful for large inputs.

Choose a modulus $W$ that is a product of small primes. Every prime not dividing $W$ must be coprime to $W$, so a search can restrict candidates to the corresponding residue classes modulo $W$. For several linear polynomials, discard each residue class in which any required value is divisible by a selected prime. The surviving residue classes can be combined using the Chinese remainder theorem.

A number-theoretic sieve used to enumerate prime numbers by excluding composite numbers. The sieve of Eratosthenes marks multiples of each discovered prime, beginning at its square.

A smallest prime factor sieve stores the least prime divisor $\mathrm{spf}(n)$ for integers $n\ge 2$ up to a bound $N$. Initialize entries unset; scan primes in increasing order and fill their still-unset multiples. A composite $n$ first marked by $p$ has no smaller prime divisor, so the stored value is correct. For example, repeatedly dividing $84$ by its stored factors gives $84\to 42\to 21\to 7\to 1$, hence $84=2^2\cdot 3\cdot 7$. A table uses $O(N)$ entries and accelerates repeated factorizations; the values at zero and one need a separate sentinel convention.

A totient sieve computes the Euler totient function for many integers at once. Initialize $\varphi (n)=n$ and, for each prime $p$, replace $\varphi (n)$ by $\varphi (n)(1-1/p)$ for every multiple $n$ of $p$.

An odd-only sieve stores only odd candidates, handling the prime two separately. This reduces the candidate storage and avoids marking even composites repeatedly.

Hypoxanthine is the 6-oxopurine base of inosine.

Purine is the fused six- and five-membered nitrogen heterocyclic framework underlying adenine and hypoxanthine. Its numbered ring nitrogens distinguish protonation and hydrogen-bonding sites.

guanosine diphosphate, abbreviated GDP, is the diphosphate form of a guanine nucleotide. Exchange between GDP and GTP, rather than receptor-catalyzed phosphorylation of GDP, controls the usual G protein switch.

An RNA-type nucleotide containing ribose. Chemical modification can affect its contacts, pairing or recognition without changing its place in the transcript.

A nucleoside carrying three phosphate groups. The relevant ribonucleoside triphosphates supply substrates for RNA polymerase.