Partition function (number theory): Difference between revisions

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=== MacLaurinGenerating seriesfunction ===

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>

The corresponding generating function based on the [[MacLaurin series]] with the numbers q(n) as coefficients in front of xⁿ is as follows:

: <math display="block">(x;x\sum_{k=0}^2)_{\infty} q(k)x^k = \prod_{n = 1}^{\infty} (1 -+ x^{2nn) -= (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>\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>

: <math display="block">(x;x\sum_{k=0}^2)_{\infty} q(k)x^{-1}k = 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

The following first addends are obtained:

<math display="block">\sum_{k=0}^{\infty} q(k)x^k = \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>(x;x^2)_{\infty}^{-1} = 1+1x+1x^2+2x^3+2x^4+3x^5+4x^6+5x^7+6x^8+8x^9+10x^{10}...</math>

: <math display="block">\vartheta_{00}(x) = 1 + 2\sum_{n = 1}^{\infty} x^{n^2}</math>

and

In comparison, the generating function of the regular partition numbers p(n) has this identity with respect to the theta function:

: <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>\sum_{k=0}^{\infty} p(k)x^k = (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>

: <math display="block">\sum_{k=0}^{\infty} qp(k)x^k = (x;x^2)_{\infty}^{-1} = \vartheta_{00}(x)^{-1/6}\vartheta_{01}(x)^{-12/3}\biggl\{\frac{1}{16\,x}\bigl[\vartheta_{00}(x)^4 - \vartheta_{01}(x)^4\bigr]\biggr\}^{-1/24}.</math>

Important calculation formulas for the [[theta function]]:

: <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>

These are the definitions of the two mentioned theta functions:

: <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>

The products in the brackets are the so called [[Pochhammer symbol|Pochhammer products]] and they are defined as follows:

: <math>(a;b)_{\infty} = \prod_{k = 0}^{\infty} (1 - ab^{k})</math>

These are two examples:

: <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>

=== Identities about strict partition numbers ===