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{{Short description|The number of partitions of an integer}}
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[[File:Ferrer partitioning diagrams.svg|thumb|The values <math>p(1), \dots, p(8)</math> of the partition function (1, 2, 3, 5, 7, 11, 15, and 22) can be determined by counting the [[Young diagram]]s for the partitions of the numbers from 1 to 8.]]
In [[number theory]], the '''partition function''' {{math|''p''(''n'')}} represents the [[number]] of possible [[PartitionInteger (number theory)partition|partitions]] of a non-negative integer {{mvar|n}}. For instance, {{math|1=''p''(4) = 5}} because the integer 4 has the five partitions {{math|1 + 1 + 1 + 1}}, {{math|1 + 1 + 2}}, {{math|1 + 3}}, {{math|2 + 2}}, and {{math|4}}.
 
No [[closed-form expression]] for the partition function is known, but it has both [[asymptotic analysis|asymptotic expansions]] that accurately approximate it and [[recurrence relation]]s by which it can be calculated exactly. It grows as an [[exponential function]] of the [[square root]] of its argument. The [[multiplicative inverse]] of its [[generating function]] is the [[Euler function]]; by Euler's [[pentagonal number theorem]] this function is an alternating sum of [[pentagonal number]] powers of its argument.
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==Definition and examples==
For a positive integer {{math|''n''}}, {{math|''p''(''n'')}} is the number of distinct ways of representing {{mvar|n}} as a [[summation|sum]] of positive integers. For the purposes of this definition, the order of the terms in the sum is irrelevant: two sums with the same terms in a different order (e.g., {{math|1 + 1 + 2}} and {{math|1 + 2 + 1}}) are not considered to be distinct.
 
By convention {{math|1=''p''(0) = 1}}, as there is one way (the [[empty sum]]) of representing zero as a sum of positive integers. Furthermore {{math|1=''p''(''n'') = 0}} when {{mvar|n}} is negative.
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\end{align}</math>
 
==Generating function==
{{As of|2022|June}}, the largest known [[prime number]] among the values of {{math|''p''(''n'')}} is {{math|''p''(1289844341)}}, with 40,000 decimal digits.{{r|caldwell}}<ref>{{citation |title=PrimePage Primes: p(1289844341) |url=https://primes.utm.edu/primes/page.php?id=130336 |website=primes.utm.edu |access-date=30 June 2022}}</ref> Until March 2022, this was also the largest prime that has been proved using [[elliptic curve primality proving]].<ref>{{citation |title=The Top Twenty: Elliptic Curve Primality Proof |url=https://primes.utm.edu/top20/page.php?id=27 |website=primes.utm.edu |access-date=30 June 2022}}</ref>
 
== Generating function==
[[File:Euler_partition_function.svg|thumb|upright|link={{filepath:Euler_partition_function.svg}}|Using Euler's method to find {{math|''p''(40)}}: A ruler with plus and minus signs (grey box) is slid downwards, the relevant terms added or subtracted. The positions of the signs are given by differences of alternating natural (blue) and odd (orange) numbers. In [{{filepath:Euler_partition_function.svg}} the SVG file,] hover over the image to move the ruler.]]
{{main|Pentagonal number theorem}}
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<math display="block">\begin{align}
\sum_{n=0}^\infty p(n)x^n &= \prod_{k=1}^\infty \left(\frac {1}{1-x^k} \right)\\
&=\left(1+x+x^2+x^3+\cdots\right) \left(1+x^2+x^4+x^6+\cdots\right)
\left(1+x^3+x^6+x^9+\cdots\right) \cdots \\
&=\frac{1}{1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} - \cdots}\\
&=1 \Big/ \sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k-1)/2}.
\end{align}</math>The equality between the products on the first and second lines of this formula is obtained by expanding each factor <math>1/(1-x^k)</math> into the [[geometric series]] <math>(1+x^k+x^{2k}+x^{3k}+\cdots).</math> To see that the expanded product equals the sum on the first line, apply the [[distributive law]] to the product. This expands the product into a sum of [[monomial]]s of the form <math>x^{a_1} x^{2a_2} x^{3a_3} \cdots</math> for some sequence of coefficients <math>a_i</math>, only finitely many of which can be non-zero. The exponent of the term is <math display="inline">n = \sum i a_i</math>, and this sum can be interpreted as a representation of <math>n</math> as a partition into <math>a_i</math> copies of each number <math>i</math>. Therefore, the number of terms of the product that have exponent <math>n</math> is exactly <math>p(n)</math>, the same as the coefficient of <math>x^n</math> in the sum on the left. Therefore, the sum equals the product.
\end{align}</math>
The equality between the products on the first and second lines of this formula
is obtained by expanding each factor <math>1/(1-x^k)</math> into the [[geometric series]] <math>(1+x^k+x^{2k}+x^{3k}+\cdots).</math>
To see that the expanded product equals the sum on the first line,
apply the [[distributive law]] to the product. This expands the product into a sum of [[monomial]]s of the form <math>x^{a_1} x^{2a_2} x^{3a_3} \cdots</math> for some sequence of coefficients
<math>a_i</math>, only finitely many of which can be non-zero.
The exponent of the term is <math display="inline">n = \sum i a_i</math>, and this sum can be interpreted as a representation of <math>n</math> as a partition into <math>a_i</math> copies of each number <math>i</math>. Therefore, the number of terms of the product that have exponent <math>n</math> is exactly <math>p(n)</math>, the same as the coefficient of <math>x^n</math> in the sum on the left.
Therefore, the sum equals the product.
 
The function that appears in the denominator in the third and fourth lines of the formula is the [[Euler function]]. The equality between the product on the first line and the formulas in the third and fourth lines is Euler's [[pentagonal number theorem]].
The exponents of <math>x</math> in these lines are the [[pentagonal number]]s <math>P_k = k(3k-1)/2</math> for <math>k \in \{ 0, 1,-1,2,-2,\dots\}</math> (generalized somewhat from the usual pentagonal numbers, which come from the same formula for the positive values {{nowrap|of <math>k</math>).}} The pattern of positive and negative signs in the third line comes from the term <math>(-1)^k</math> in the fourth line: even choices of <math>k</math> produce positive terms, and odd choices produce negative terms.
 
More generally, the generating function for the partitions of <math>n</math> into numbers selected from a set <math>A</math> of positive integers can be found by taking only those terms in the first product for which <math>k\in A</math>. This result is due to [[Leonhard Euler]].{{r|e}} The formulation of Euler's generating function is a special case of a [[q-Pochhammer symbol|<math>q</math>-Pochhammer symbol]] and is similar to the product formulation of many [[modular form]]s, and specifically the [[Dedekind eta function]].
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coming from the nonzero values of <math>k</math> in the range
<math display="block"> - \frac{\sqrt{24n+1}-1}{6} \leq k \leq \frac{\sqrt{24n+1}+1}{6}.</math>
The recurrence relation can also be written in the equivalent form
<math display="block">
p(n) = \sum_{k = 1}^\infty (-1)^{k+1} \big(p(n-k(3k-1)/2) + p(n-k(3k+1)/2)\big) .
</math>
 
Another recurrence relation for <math>p(n)</math> can be given in terms of the [[divisor function|sum of divisors function]] {{math|''σ''}}:{{r|wilf}}
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For instance the number of partitions is divisible by five whenever the decimal representation of <math>n</math> ends in the digit 4 or 9, as expressed by the congruence{{r|hw}}
<math display="block">p(5k+4) \equiv 0 \pmod 5</math>
For instance, the number of partitions for the integer 4 is 5.
For the integer 9, the number of partitions is 30; for 14 there are 135 partitions. This congruence is implied by the more general identity
<math display="block"> \sum_{k=0}^\infty p(5k+4)x^k = 5~ \frac{ (x^5)^5_\infty } {(x)^6_\infty},</math>
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=== Definition and properties ===
 
A partition in which no part occurs more than once is called ''strict'', or is said to be a partition ''into distinct parts''. The function ''q''(''n'') gives the number of these strict partitions of the given sum ''n''. For example, ''q''(3) = 2 because the partitions 3 and 1 + 2 are strict, while the third partition 1 + 1 + 1 of 3 has repeated parts. The number ''q''(''n'') is also equal to the number of partitions of ''n'' in which only odd summands are permitted.<ref>{{cite book|first=Richard P.|last=Stanley|author-link=Richard P. Stanley|title=Enumerative Combinatorics 1 |series=Cambridge Studies in Advanced Mathematics|volume=49|publisher=Cambridge University Press|isbn=0-521-66351-2 |year=1997|at=Proposition 1.8.5}}</ref>
If no summand occurs repeatedly<ref>{{cite web|title=code golf - Strict partitions of a positive integer|periodical=|publisher=|url=https://codegolf.stackexchange.com/questions/71941/strict-partitions-of-a-positive-integer|url-status=|format=|access-date=2022-03-09|archive-url=|archive-date=|last=|date=|year=|language=|pages=|quote=}}</ref> in the affected partition sums, then the so called strict partitions are present. The function Q(n) gives the number of these strict partitions in relation to the given sum n. Therefore the strict partition sequence Q(n) satisfies the criterion Q(n) ≤ P(n) for all <math>n \isin \mathbb{N}_0</math>. The same result<ref>{{cite web|title=A000009 - OEIS|periodical=|publisher=|url=https://oeis.org/A000009|url-status=|format=|access-date=2022-03-09|archive-url=|archive-date=|last=|date=|year=|language=|pages=|quote=}}</ref> results if only odd summands<ref>{{cite web|title=Partition Function Q|periodical=|publisher=|url=https://mathworld.wolfram.com/|url-status=|format=|access-date=2022-03-09|archive-url=|archive-date=|last=Eric W. Weisstein|date=|year=|language=en|pages=|quote=}}</ref> may appear in the partition sum, but these may also occur more than once.
 
=== Exemplary values of strict partition numbers ===
 
Representations of the partitions:
{| class="wikitable"
|+ ExemplaryExample values of Q''q''(''n'') and associated number partitions
|-
! ''n'' || Q''q''(''n'') || NumberStrict partitions without repeated addends
!Number partitionsPartitions with only odd addendsparts
|-
| 0 || 1 || () empty partition
|() empty partition
|-
| 1 || 1 || (1)
|(1)
|-
| 2 || 1 || (2)
|(1+1)
|-
| 3 || 2 || (1+2), (3)
|(1+1+1), (3)
|-
| 4 || 2 || (1+3), (4)
|(1+1+1+1), (1+3)
|-
| 5 || 3 || (2+3), (1+4), (5)
|(1+1+1+1+1), (1+1+3), (5)
|-
| 6 || 4|| (1+2+3), (2+4), (1+5), (6)
|(1+1+1+1+1+1), (1+1+1+3), (3+3), (1+5)
|-
|7
|5
|(1+2+4), (3+4), (2+5), (1+6), (7)
|(1+1+1+1+1+1+1), (1+1+1+1+3), (1+3+3), (1+1+5), (7)
|-
|8
|6
|(1+3+4), (1+2+5), (3+5), (2+6), (1+7), (8)
|(1+1+1+1+1+1+1+1), (1+1+1+1+1+3), (1+1+3+3), (1+1+1+5), (3+5), (1+7)
|-
|9
|8
|(2+3+4), (1+3+5), (4+5), (1+2+6), (3+6), (2+7), (1+8), (9)
|1+1+1+1+1+1+1+1+1, 1+1+1+1+1+1+3, 1+1+1+3+3, 3+3+3, 1+1+1+1+5, 1+3+5, 1+1+7, 9
|......
|-
|10
|10
|(1+2+3+4), (2+3+5), (1+4+5), (1+3+6), (4+6), (1+2+7), (3+7), (2+8), (1+9), (10)
|......
|}
=== MacLaurin series ===
The corresponding generating function based on the [[MacLaurin series]] with the numbers Q(n) as coefficients in front of x<sup>n</sup> is as follows:
 
=== Generating function ===
: <math>\sum_{k=0}^{\infty} Q(k)x^k = (x;x^2)_{\infty}^{-1} = \vartheta_{00}(x)^{1/6}\vartheta_{01}(x)^{-1/3}\biggl\{\frac{1}{16\,x}\bigl[\vartheta_{00}(x)^4 - \vartheta_{01}(x)^4\bigr]\biggr\}^{1/24}</math>
The generating function for the numbers ''q''(''n'') is given by a simple infinite product:<ref>{{cite book|first=Richard P.|last=Stanley|author-link=Richard P. Stanley|title=Enumerative Combinatorics 1 |series=Cambridge Studies in Advanced Mathematics|volume=49|publisher=Cambridge University Press|isbn=0-521-66351-2 |year=1997|at=Proof of Proposition 1.8.5}}</ref>
<math display="block">\sum_{n=0}^{\infty} q(n)x^n = \prod_{k = 1}^\infty (1 + x^k) = (x;x^2)_{\infty}^{-1},</math>
where the notation <math>(a;b)_{\infty}</math> represents the [[Pochhammer symbol]] <math>(a;b)_{\infty} = \prod_{k = 0}^{\infty} (1 - ab^{k}).</math> From this formula, one may easily obtain the first few terms {{OEIS|A000009}}:
<math display="block">\sum_{n=0}^{\infty} q(n)x^n = 1+1x+1x^2+2x^3+2x^4+3x^5+4x^6+5x^7+6x^8+8x^9+10x^{10}+\ldots.</math>
This series may also be written in terms of [[theta function]]s as
<math display="block">\sum_{n=0}^{\infty} q(n)x^n = \vartheta_{00}(x)^{1/6}\vartheta_{01}(x)^{-1/3}\biggl\{\frac{1}{16\,x}\bigl[\vartheta_{00}(x)^4 - \vartheta_{01}(x)^4\bigr]\biggr\}^{1/24},</math>
where
<math display="block">\vartheta_{00}(x) = 1 + 2\sum_{n = 1}^{\infty} x^{n^2}</math>
and
<math display="block">\vartheta_{01}(x) = 1 + 2\sum_{n = 1}^{\infty} (-1)^{n} x^{n^2}.</math>
In comparison, the generating function of the regular partition numbers ''p''(''n'') has this identity with respect to the theta function:
<math display="block">\sum_{n=0}^{\infty} p(n)x^n = (x;x)_{\infty}^{-1} = \vartheta_{00}(x)^{-1/6}\vartheta_{01}(x)^{-2/3}\biggl\{\frac{1}{16\,x}\bigl[\vartheta_{00}(x)^4 - \vartheta_{01}(x)^4\bigr]\biggr\}^{-1/24}.</math>
 
=== Identities about strict partition numbers ===
The following first addends are obtained:
Following identity is valid for the Pochhammer products:
 
: <math>(x;x^2)_{\infty}^{-1} = 1+1x+1x(x^2+2x;x^3+2x2)_{\infty}^4+3x{-1}(x;x^5+4x^6+5x^7+6x^8+8x^9+10x2)_{\infty}^{10-1}...</math>
 
From this identity follows that formula:
In comparison, the generating function of the regular partition numbers P(n) has this identity with respect to the theta function:
 
: <math>\biggl[\sum_{kn=0}^{\infty} Pp(kn)x^kn\biggr] = (x;x)_{\infty}^biggl[\sum_{-1} n= \vartheta_{000}(x)^{-1/6}\vartheta_{01}infty p(n)x)^{-2/32n}\biggr]\biggl\{\frac{1}{16\,x}\bigl[\vartheta_sum_{00n=0}(x)^4\infty - \vartheta_{01}q(n)x)^4n\bigrbiggr]\biggr\}^{-1/24}</math>
 
ImportantTherefore calculationthose two formulas are valid for the [[thetasynthesis of the number sequence function]]p(n):
: <math>(x;x)_{\infty}(x;x^2)_{\infty} = \vartheta_{01}(x)</math>
: <math>16\,x\,(x;x)_{\infty}^8 = (x;x^2)_{\infty}^{16}\,\vartheta_{00}(x)^4 \bigl[\vartheta_{00}(x)^4 - \vartheta_{01}(x)^4\bigr]</math>
 
: <math>p(2n) = \sum_{k=0}^{n} p(n - k)q(2k)</math>
These are the definitions of the two mentioned theta functions:
: <math>p(2n+1) = \sum_{k=0}^{n} p(n - k)q(2k + 1)</math>
 
In the following, two examples are accurately executed:
: <math>\vartheta_{00}(x) = 1 + 2\sum_{n = 1}^{\infty} x^{n^2}</math>
: <math>\vartheta_{01}(x) = 1 + 2\sum_{n = 1}^{\infty} (-1)^{n} x^{n^2}</math>
 
: <math>p(8) = \sum_{k=0}^{4} p(4 - k)q(2k) =</math>
The products in the brackets are the so called [[Pochhammer symbol|Pochhammer products]] and they are defined as follows:
: <math>= p(4)q(0) + p(3)q(2) + p(2)q(4) + p(1)q(6) + p(0)q(8) = </math>
: <math>= 5\times 1 + 3\times 1 + 2\times 2 + 1\times 4 + 1\times 6 = 22 </math>
: <math>p(9) = \sum_{k=0}^{4} p(4 - k)q(2k + 1) =</math>
: <math>= p(4)q(1) + p(3)q(3) + p(2)q(5) + p(1)q(7) + p(0)q(9) = </math>
: <math>= 5\times 1 + 3\times 2 + 2\times 3 + 1\times 5 + 1\times 8 = 30 </math>
 
== Restricted partition function ==
: <math>(a;b)_{\infty} = \prod_{k = 0}^{\infty} (1 - ab^{k})</math>
 
More generally, it is possible to consider partitions restricted to only elements of a subset ''A'' of the natural numbers (for example a restriction on the maximum value of the parts), or with a restriction on the number of parts or the maximum difference between parts. Each particular restriction gives rise to an associated partition function with specific properties. Some common examples are given below.
These are two examples:
 
=== Euler and Glaisher's theorem ===
: <math>(x;x)_{\infty} = \prod_{n = 1}^{\infty} (1 - x^{n})</math>
: <math>(x;x^2)_{\infty} = \prod_{n = 1}^{\infty} (1 - x^{2n - 1})</math>
 
Two important examples are the partitions restricted to only odd integer parts or only even integer parts, with the corresponding partition functions often denoted <math>p_o(n)</math> and <math>p_e(n)</math>.
=== Identities about strict partition numbers ===
Following identity is valid for the Pochhammer products:
 
A theorem from Euler shows that the number of strict partitions is equal to the number of partitions with only odd parts: for all ''n'', <math>q(n) = p_o(n)</math>. This is generalized as [[Glaisher's theorem]], which states that the number of partitions with no more than ''d-1'' repetitions of any part is equal to the number of partitions with no part divisible by ''d''.
: <math>(x;x)_{\infty}^{-1} = (x^2;x^2)_{\infty}^{-1}(x;x^2)_{\infty}^{-1}</math>
 
=== Gaussian binomial coefficient ===
From this identity follows that formula:
 
If we denote <math>p(N, M, n)</math> the number of partitions of ''n'' in at most ''M'' parts, with each part smaller or equal to ''N'', then the generating function of <math>p(N, M, n)</math> is the following [[Gaussian binomial coefficient]]:
: <math>\biggl[\sum_{n=0}^\infty P(n)x^n\biggr] = \biggl[\sum_{n=0}^\infty P(n)x^{2n}\biggr]\biggl[\sum_{n=0}^\infty Q(n)x^n\biggr]</math>
 
:<math>\sum_{n=0}^\infty p(N, M, n)q^n = {N+M \choose M}_q
Therefore those two formulas are valid for the synthesis of the number sequence P(n):
=
\frac{(1-q^{N+M})(1-q^{N+M-1})\cdots(1-q^{N+1})} {(1-q)(1-q^2)\cdots(1-q^M)}</math>
 
=== Asymptotics ===
: <math>P(2n) = \sum_{k=0}^{n} P(n - k)Q(2k)</math>
: <math>P(2n+1) = \sum_{k=0}^{n} P(n - k)Q(2k + 1)</math>
 
Some general results on the asymptotic properties of restricted partition functions are known. If ''p''<sub>''A''</sub>(''n'') is the partition function of partitions restricted to only elements of a subset ''A'' of the natural numbers, then:
In the following, two examples are accurately executed:
 
If ''A'' possesses positive [[natural density]] α then <math> \log p_A(n) \sim C \sqrt{\alpha n}</math>, with <math>C = \pi\sqrt\frac23</math>
: <math>P(8) = \sum_{k=0}^{4} P(4 - k)Q(2k) =</math>
 
: <math>= P(4)Q(0) + P(3)Q(2) + P(2)Q(4) + P(1)Q(6) + P(0)Q(8) = </math>
and conversely if this asymptotic property holds for ''p''<sub>''A''</sub>(''n'') then ''A'' has natural density α.{{sfn|Nathanson|2000|pp=475–85}} This result was stated, with a sketch of proof, by Erdős in 1942.<ref name=erdos42></ref>{{sfn|Nathanson|2000|p=495}}
: <math>= 5\times 1 + 3\times 1 + 2\times 2 + 1\times 4 + 1\times 6 = 22 </math>
 
: <math>P(9) = \sum_{k=0}^{4} P(4 - k)Q(2k + 1) =</math>
If ''A'' is a finite set, this analysis does not apply (the density of a finite set is zero). If ''A'' has ''k'' elements whose greatest common divisor is 1, then{{sfn|Nathanson|2000|pp=458–64}}
: <math>= P(4)Q(1) + P(3)Q(3) + P(2)Q(5) + P(1)Q(7) + P(0)Q(9) = </math>
 
: <math>= 5\times 1 + 3\times 2 + 2\times 3 + 1\times 5 + 1\times 8 = 30 </math>
:<math> p_A(n) = \left(\prod_{a \in A} a^{-1}\right) \cdot \frac{n^{k-1}}{(k-1)!} + O(n^{k-2}) . </math>
 
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