Editing Generating function (section)

===Generating functions prove congruences===
We say that two generating functions (power series) are congruent modulo {{mvar|m}}, written {{math|''A''(''z'') ≡ ''B''(''z'') (mod ''m'')}} if their coefficients are congruent modulo {{mvar|m}} for all {{math|''n'' ≥ 0}}, i.e., {{math|''a<sub>n</sub>'' ≡ ''b<sub>n</sub>'' (mod ''m'')}} for all relevant cases of the integers {{mvar|n}} (note that we need not assume that {{mvar|m}} is an integer here—it may very well be polynomial-valued in some indeterminate {{mvar|x}}, for example). If the "simpler" right-hand-side generating function, {{math|''B''(''z'')}}, is a rational function of {{mvar|z}}, then the form of this sequence suggests that the sequence is [[periodic function|eventually periodic]] modulo fixed particular cases of integer-valued {{math|''m'' ≥ 2}}. For example, we can prove that the [[Euler numbers]],
<math display="block">\langle E_n \rangle = \langle 1, 1, 5, 61, 1385, \ldots \rangle \longmapsto \langle 1,1,2,1,2,1,2,\ldots \rangle \pmod{3}\,,</math>
satisfy the following congruence modulo 3:<ref>{{harvnb|Lando|2003|loc=§5}}</ref>
<math display="block">\sum_{n = 0}^\infty E_n z^n = \frac{1-z^2}{1+z^2} \pmod{3}\,. </math>

One useful method of obtaining congruences for sequences enumerated by special generating functions modulo any integers (i.e., not only prime powers {{math|''p<sup>k</sup>''}}) is given in the section on continued fraction representations of (even non-convergent) ordinary generating functions by {{mvar|J}}-fractions above. We cite one particular result related to generating series expanded through a representation by continued fraction from Lando's ''Lectures on Generating Functions'' as follows:
{{math theorem
| name = Theorem: congruences for series generated by expansions of continued fractions
| math_statement = Suppose that the generating function {{math|''A''(''z'')}} is represented by an infinite [[continued fraction]] of the form
<math display="block">A(z) = \cfrac{1}{1-c_1z - \cfrac{p_1z^2}{1-c_2z - \cfrac{p_2 z^2}{1-c_3z - {\ddots}}}}</math>
and that {{math|''A<sub>p</sub>''(''z'')}} denotes the {{mvar|p}}th convergent to this continued fraction expansion defined such that {{math|''a<sub>n</sub>'' {{=}} [''z<sup>n</sup>''] ''A<sub>p</sub>''(''z'')}} for all {{math|0 ≤ ''n'' < 2''p''}}. Then:

# the function {{math|''A<sub>p</sub>''(''z'')}} is rational for all {{math|''p'' ≥ 2}} where we assume that one of divisibility criteria of {{math|''p'' {{!}} ''p''<sub>1</sub>, ''p''<sub>1</sub>''p''<sub>2</sub>, ''p''<sub>1</sub>''p''<sub>2</sub>''p''<sub>3</sub>}} is met, that is, {{math|''p'' {{!}} ''p''<sub>1</sub>''p''<sub>2</sub>⋯''p''<sub>''k''</sub>}} for some {{math|''k'' ≥ 1}}; and 
# if the integer {{mvar|p}} divides the product {{math|''p''<sub>1</sub>''p''<sub>2</sub>⋯''p''<sub>''k''</sub>}}, then we have {{math|''A''(''z'') ≡ ''A<sub>k</sub>''(''z'') (mod ''p'')}}.
}}

Generating functions also have other uses in proving congruences for their coefficients. We cite the next two specific examples deriving special case congruences for the [[Stirling numbers of the first kind]] and for the [[partition function (number theory)|partition function {{math|''p''(''n'')}}]] which show the versatility of generating functions in tackling problems involving [[integer sequences]].

====The Stirling numbers modulo small integers====

The [[Stirling numbers of the first kind#Congruences|main article]] on the Stirling numbers generated by the finite products
<math display="block">S_n(x) := \sum_{k=0}^n \begin{bmatrix} n \\ k \end{bmatrix} x^k = x(x+1)(x+2) \cdots (x+n-1)\,,\quad n \geq 1\,, </math>

provides an overview of the congruences for these numbers derived strictly from properties of their generating function as in Section 4.6 of Wilf's stock reference ''Generatingfunctionology''.
We repeat the basic argument and notice that when reduces modulo 2, these finite product generating functions each satisfy

<math display="block">S_n(x) = [x(x+1)] \cdot [x(x+1)] \cdots = x^{\left\lceil \frac{n}{2} \right\rceil} (x+1)^{\left\lfloor \frac{n}{2} \right\rfloor}\,, </math>

which implies that the parity of these [[Stirling numbers]] matches that of the binomial coefficient

<math display="block">\begin{bmatrix} n \\ k \end{bmatrix} \equiv \binom{\left\lfloor \frac{n}{2} \right\rfloor}{k - \left\lceil \frac{n}{2} \right\rceil} \pmod{2}\,, </math>

and consequently shows that {{math|{{resize|150%|[}}{{su|p=''n''|b=''k''|a=c}}{{resize|150%|]}}}} is even whenever {{math|''k'' < ⌊ {{sfrac|''n''|2}} ⌋}}.

Similarly, we can reduce the right-hand-side products defining the Stirling number generating functions modulo 3 to obtain slightly more complicated expressions providing that
<math display="block">\begin{align}
\begin{bmatrix} n \\ m \end{bmatrix} & \equiv
 [x^m] \left(
 x^{\left\lceil \frac{n}{3} \right\rceil} (x+1)^{\left\lceil \frac{n-1}{3} \right\rceil}
 (x+2)^{\left\lfloor \frac{n}{3} \right\rfloor}
 \right) && \pmod{3} \\
 & \equiv
 \sum_{k=0}^{m} \begin{pmatrix} \left\lceil \frac{n-1}{3} \right\rceil \\ k \end{pmatrix}
 \begin{pmatrix} \left\lfloor \frac{n}{3} \right\rfloor \\ m-k - \left\lceil \frac{n}{3} \right\rceil \end{pmatrix} \times
 2^{\left\lceil \frac{n}{3} \right\rceil + \left\lfloor \frac{n}{3} \right\rfloor -(m-k)} && \pmod{3}\,.
\end{align}</math>

====Congruences for the partition function====

In this example, we pull in some of the machinery of infinite products whose power series expansions generate the expansions of many special functions and enumerate partition functions. In particular, we recall that ''the'' [[partition function (number theory)|partition function]] {{math|''p''(''n'')}} is generated by the reciprocal infinite [[q-Pochhammer symbol|{{mvar|q}}-Pochhammer symbol]] product (or {{mvar|z}}-Pochhammer product as the case may be) given by
<math display="block">\begin{align}
\sum_{n = 0}^\infty p(n) z^n & = \frac{1}{\left(1-z\right)\left(1-z^2\right)\left(1-z^3\right) \cdots} \\[4pt]
& = 1 + z + 2z^2 + 3 z^3 + 5z^4 + 7z^5 + 11z^6 + \cdots.
\end{align}</math>

This partition function satisfies many known [[Ramanujan's congruences|congruence properties]], which notably include the following results though there are still many open questions about the forms of related integer congruences for the function:<ref>{{harvnb|Hardy|Wright|Heath-Brown|Silverman|2008|loc=§19.12}}</ref>
<math display="block">\begin{align}
p(5m+4) & \equiv 0 \pmod{5} \\
p(7m+5) & \equiv 0 \pmod{7} \\
p(11m+6) & \equiv 0 \pmod{11} \\
p(25m+24) & \equiv 0 \pmod{5^2}\,.
\end{align}</math>

We show how to use generating functions and manipulations of congruences for formal power series to give a highly elementary proof of the first of these congruences listed above.

First, we observe that in the binomial coefficient generating function
<math display="block">\frac{1}{(1-z)^5} = \sum_{i=0}^\infty \binom{4+i}{4}z^i\,,</math>
all of the coefficients are divisible by 5 except for those which correspond to the powers {{math|1, ''z''<sup>5</sup>, ''z''<sup>10</sup>, ...}} and moreover in those cases the remainder of the coefficient is 1 modulo 5.  Thus, 
<math display="block">\frac{1}{(1-z)^5} \equiv \frac{1}{1-z^5} \pmod{5}\,,</math> 
or equivalently
<math display="block"> \frac{1-z^5}{(1-z)^5} \equiv 1 \pmod{5}\,.</math>
It follows that
<math display="block">\frac{\left(1-z^5\right)\left(1-z^{10}\right)\left(1-z^{15}\right) \cdots }{\left((1-z)\left(1-z^2\right)\left(1-z^3\right) \cdots \right)^5} \equiv 1 \pmod{5}\,. </math>

Using the infinite product expansions of 
<math display="block">z \cdot \frac{\left(1-z^5\right)\left(1-z^{10}\right) \cdots }{\left(1-z\right)\left(1-z^2\right) \cdots } =
z \cdot \left((1-z)\left(1-z^2\right) \cdots \right)^4 \times \frac{\left(1-z^5\right)\left(1-z^{10}\right) \cdots }{\left(\left(1-z\right)\left(1-z^2\right) \cdots \right)^5}\,,</math>
it can be shown that the coefficient of {{math|''z''<sup>5''m'' + 5</sup>}} in {{math|''z'' · ((1 − ''z'')(1 − ''z''<sup>2</sup>)⋯)<sup>4</sup>}} is divisible by 5 for all {{mvar|m}}.<ref>{{cite book |last1=Hardy |first1=G.H. |last2=Wright |first2=E.M.|title=An Introduction to the Theory of Numbers}} p.288, Th.361</ref> Finally, since
<math display="block">\begin{align}
\sum_{n = 1}^\infty p(n-1) z^n & = \frac{z}{(1-z)\left(1-z^2\right) \cdots} \\[6px]
& = z \cdot \frac{\left(1-z^5\right)\left(1-z^{10}\right) \cdots }{(1-z)\left(1-z^2\right) \cdots } \times \left(1+z^5+z^{10}+\cdots\right)\left(1+z^{10}+z^{20}+\cdots\right) \cdots
\end{align}</math>
we may equate the coefficients of {{math|''z''<sup>5''m'' + 5</sup>}} in the previous equations to prove our desired congruence result, namely that {{math|''p''(5''m'' + 4) ≡ 0 (mod 5)}} for all {{math|''m'' ≥ 0}}.