QBinomial

The Gaussian (q-)binomial coefficient $\binom{n}{k}_q$, the generating polynomial in $q$ for partitions fitting in a $k \times (n-k)$ box. Provided by @enumeratio/analytic.

✓ checked against that system · ! checked, with disagreements (hover a chip)Wolfram LanguageQBinomial✓↗

QBinomial(n, k, q)the Gaussian binomial coefficient $\binom{n}{k}_q$.

SageMathq_binomial(n, k, q)

Typed signatures

Type	Package	Applies
(complex, complex, complex) -> number	analytic	always

Domain: Special functions

Details

• Built by the Pascal-like recurrence $\binom{n}{k}_q = \binom{n-1}{k-1}_q + q^k \binom{n-1}{k}_q$, with $\binom{n}{0}_q = \binom{n}{n}_q = 1$ — pure addition and multiplication, never division, so it stays a genuine polynomial for symbolic $q$ that Expand can open up. The equivalent quotient $[n]_q!/([k]_q![n-k]_q!)$ is exact numerically but Expand alone can't cancel it down to a polynomial when $q$ is symbolic, which is why the recurrence is used instead.
• At $q = 1$ it reduces to the ordinary $\binom{n}{k}$ by the same recurrence Pascal's triangle uses.
• A symbolic, non-integer or negative $n$ or $k$ is a held call (still canonical, so it sits in a sum or difference). At $q = 0$ with a non-integer rational $n$ or $k$, Series expands the product $(q^{k+1};q)_\infty (q^{n-k+1};q)_\infty / \big((q;q)_\infty (q^{n+1};q)_\infty\big)$ as a series in $q^{1/d}$, with Wolfram's truncation. A factor with a negative exponent contributes a leading negative power, so the series is a Laurent–Puiseux one; only a shape whose denominator reflects with no numerator doing so, or a pole, stays unevaluated. Where Wolfram itself falls back to Derivative coefficients (a family of non-integer shapes, and its order-0 1 + O(q)), ours is the exact expansion.

Plot

the complex plot cannot lower QBinomial

Examples

7 more kept as data, checked against the oracles

Basic #

$\frac{(2^4-1)(2^3-1)}{(2^2-1)(2-1)} = 35$

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In$\mathrm{QBinomial}(4, 2, 2)$Out$35$

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In$\operatorname{Series}(\mathrm{QBinomial}(\frac{1}{2}, 1, x), x, 0, 2)$Out$-\sqrt{x}-\sqrt{x}^{3}+1+x+x^2+O\left(\sqrt{x}^{5}\right)$

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In$\operatorname{Normal}(\operatorname{Series}(\mathrm{QBinomial}(\frac{1}{2}, 2, x), x, ∞, 2))$Out$\operatorname{Series}(\mathrm{QBinomial}(\frac{1}{2}, 2, x), x, ∞, 2)$

Scope #

A polynomial in q whose coefficients count partitions in a 2 × 2 box

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In$\mathrm{Expand}(\mathrm{QBinomial}(4, 2, q))$Out$q^4+q^3+2q^2+q+1$

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In$N(\mathrm{QBinomial}(\frac{15}{17}, 5, 1), 50)$Out$0.006\,383\,037\,165\,010\,279\,204\,173\,378\,023\,279\,809\,163\,880\,587\,974\,704\,5$

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In$N(\mathrm{QBinomial}(\frac{23}{47}, 5-i, 2))$Out$-0.000\,071\,801\,621\,838\,786\,11-0.000\,065\,724\,259\,755\,274\,13i$

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In$\mathrm{FullSimplify}(\operatorname{Normal}(\operatorname{Series}(\mathrm{QBinomial}(n, x, q), x, 0, 2)))$Out$\frac{1}{2}\mathrm{QBinomial}^{(0, 2, 0)}(n, 0, q)x^2+x\mathrm{QBinomial}^{(0, 1, 0)}(n, 0, q)+\mathrm{QBinomial}(n, 0, q)$

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In$\mathrm{FullSimplify}(\operatorname{Normal}(\operatorname{Series}(\mathrm{QBinomial}(n, x, q), x, \mathrm{x0}, 2)))$Out$\frac{1}{2}\frac{∂^{2}}{∂\mathrm{x0}^{2}} \mathrm{QBinomial}(n, \mathrm{x0}, q)(x-\mathrm{x0})^2+(x-\mathrm{x0})\frac{∂}{∂\mathrm{x0}} \mathrm{QBinomial}(n, \mathrm{x0}, q)+\mathrm{QBinomial}(n, \mathrm{x0}, q)$

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In$\mathrm{QBinomial}(7, 0, 5)$Out$1$

Past the top there is nothing to choose

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In$\mathrm{QBinomial}(3, 5, 2)$Out$0$

Applications #

The number of 2-dimensional subspaces of $\mathbb{F}_3^5$

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In$\mathrm{QBinomial}(5, 2, 3)$Out$1\,210$

Properties #

At q = 1 it is Binomial(6, 3)

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In$\mathrm{QBinomial}(6, 3, 1)$Out$20$

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In$\operatorname{Series}(\mathrm{QBinomial}(5, \frac{5}{2}, q), q, 0, 4)$Out$-4\sqrt{q}^{9}-2\sqrt{q}^{7}+1+q+2q^2+3q^3+5q^4+O\left(q^5\right)$

A negative exponent in the product starts the series below $q^0$

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In$\operatorname{Series}(\mathrm{QBinomial}(0, \frac{3}{2}, q), q, 0, 3)$Out$-9\sqrt{q}^{3}-3\sqrt{q}-\frac{1}{\sqrt{q}}+2+5q+15q^2+O\left(\sqrt{q}^{5}\right)$

$\binom{5}{3}_q=\binom{5}{2}_q$, the same 1210 as above

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In$\mathrm{QBinomial}(5, 3, 3)$Out$1\,210$

About

Examples	22
Checked against	Wolfram Language 21 agree1 unevaluatedSageMath 11 agree1 unevaluated10 error
Crosswalk	Wolfram LanguageSageMath

Implementation

primitive · kernela kernel — an algorithm over mutable state that would not be readable as a tree

nativeenginetypescriptpackages/symbols/analysis/analytic/src/q-series.ts

the Pascal-like recurrence, built at canonicalization time so Expand sees the tree.

mappedexternalwolframpackages/oracle/src/mappings.ts

QBinomial[n, k, q].

mappedsagepackages/oracle/src/mappings.ts

See also: QPochhammer, QFactorial, Binomial