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Pentagonal number theorem

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From Wikipedia, the free encyclopedia
Theorem in number theory

Inmathematics,Euler'spentagonal number theorem relates the product and series representations of theEuler function. It states that

n=1(1xn)=k=(1)kxk(3k1)/2=1+k=1(1)k(xk(3k+1)/2+xk(3k1)/2).{\displaystyle \prod _{n=1}^{\infty }\left(1-x^{n}\right)=\sum _{k=-\infty }^{\infty }\left(-1\right)^{k}x^{k\left(3k-1\right)/2}=1+\sum _{k=1}^{\infty }(-1)^{k}\left(x^{k(3k+1)/2}+x^{k(3k-1)/2}\right).}

In other words,

(1x)(1x2)(1x3)=1xx2+x5+x7x12x15+x22+x26.{\displaystyle (1-x)(1-x^{2})(1-x^{3})\cdots =1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^{15}+x^{22}+x^{26}-\cdots .}

The exponents 1, 2, 5, 7, 12, ... on the right hand side are given by the formulagk =k(3k − 1)/2 fork = 1, −1, 2, −2, 3, ... and are called (generalized)pentagonal numbers (sequenceA001318 in theOEIS). (The constant term 1 corresponds tok=0{\displaystyle k=0}.)This holds as an identity of convergentpower series for|x|<1{\displaystyle |x|<1}, and also as an identity offormal power series.

A striking feature of this formula is the amount of cancellation in the expansion of the product.

Relation with partitions

The identity implies arecurrence for calculatingp(n){\displaystyle p(n)}, the number ofpartitions ofn:

p(n)=p(n1)+p(n2)p(n5)p(n7)+{\displaystyle p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots }

or more formally,

p(n)=k0(1)k1p(ngk){\displaystyle p(n)=\sum _{k\neq 0}(-1)^{k-1}p(n-g_{k})}

where the summation is over all nonzero integersk (positive and negative) andgk{\displaystyle g_{k}} is thekth generalized pentagonal number. Sincep(n)=0{\displaystyle p(n)=0} for alln<0{\displaystyle n<0}, the apparently infinite series on the right has only finitely many non-zero terms, enabling an efficient calculation ofp(n).

Franklin's bijective proof

The theorem can be interpretedcombinatorially in terms ofpartitions. In particular, the left hand side is agenerating function for the number of partitions ofn into an even number of distinct parts minus the number of partitions ofn into an odd number of distinct parts. Each partition ofn into an even number of distinct parts contributes +1 to the coefficient ofxn; each partition into an odd number of distinct parts contributes −1. (The article onunrestricted partition functions discusses this type of generating function.)

For example, the coefficient ofx5 is +1 because there are two ways to split 5 into an even number of distinct parts (4 + 1 and 3 + 2), but only one way to do so for an odd number of distinct parts (the one-part partition 5). However, the coefficient ofx12 is −1 because there are seven ways to partition 12 into an even number of distinct parts, but there are eight ways to partition 12 into an odd number of distinct parts, and 7 − 8 = −1.

This interpretation leads to a proof of the identity by canceling pairs of matched terms (involution method).[1] Consider theFerrers diagram of any partition ofn into distinct parts. For example, the diagram below showsn = 20 and the partition 20 = 7 + 6 + 4 + 3.

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***

Letm be the number of elements in the smallest row of the diagram (m = 3 in the above example). Lets be the number of elements in the rightmost 45 degree line of the diagram (s = 2 dots in red above, since 7 − 1 = 6, but 6 − 1 > 4). Ifm > s, take the rightmost 45-degree line and move it to form a new row, as in the matching diagram below.

******
*****
****
***
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If m ≤ s (as in our newly formed diagram wherem = 2,s = 5) we may reverse the process by moving the bottom row to form a new 45 degree line (adding 1 element to each of the firstm rows), taking us back to the first diagram.

A bit of thought shows that this process always changes the parity of the number of rows, and applying the process twice brings us back to the original diagram. This enables us to pair off Ferrers diagrams contributing 1 and −1 to thexn term of the series, resulting in a net coefficient of 0 forxn. This holds for every termexcept when the process cannot be performed on every Ferrers diagram withn dots. There are two such cases:

1)m =s and the rightmost diagonal and bottom row meet. For example,

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***

Attempting to perform the operation would lead us to:

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*

which fails to change the parity of the number of rows, and is not reversible in the sense that performing the operation again doesnot take us back to the original diagram. If there arem elements in the last row of the original diagram, then

n=m+(m+1)+(m+2)++(2m1)=m(3m1)2=k(3k1)2{\displaystyle n=m+(m+1)+(m+2)+\cdots +(2m-1)={\frac {m(3m-1)}{2}}={\frac {k(3k-1)}{2}}}

where the new indexk is taken to equalm. Note that the sign associated with this partition is (−1)s, which by construction equals (−1)m and (−1)k.

2)m =s + 1 and the rightmost diagonal and bottom row meet. For example,

******
*****
****

Our operation requires us to move the right diagonal to the bottom row, but that would lead to two rows of three elements, forbidden since we're counting partitions into distinct parts. This is the previous case but with one fewer row, so

n=m+(m+1)+(m+2)++(2m2)=(m1)(3m2)2=k(3k1)2,{\displaystyle n=m+(m+1)+(m+2)+\cdots +(2m-2)={\frac {(m-1)(3m-2)}{2}}={\frac {k(3k-1)}{2}},}

where we takek = 1−m (a negative integer). Here the associated sign is (−1)s withsm − 1 = −k, therefore the sign is again (−1)k.

In summary, it has been shown that partitions into an even number of distinct parts and an odd number of distinct parts exactly cancel each other, producing null terms 0xn, except ifn is a generalized pentagonal numbern=gk=k(3k1)/2{\displaystyle n=g_{k}=k(3k-1)/2}, in which case there is exactly one Ferrers diagram left over, producing a term (−1)kxn. But this is precisely what the right side of the identity says should happen, so we are finished.

Partition recurrence

We can rephrase the above proof, usinginteger partitions, which we denote as:n=λ1+λ2++λ{\displaystyle n=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{\ell }}, whereλ1λ2λ>0{\displaystyle \lambda _{1}\geq \lambda _{2}\geq \ldots \geq \lambda _{\ell }>0}.The number of partitions ofn is the partition functionp(n) having generating function:

n=0p(n)xn=k=1(1xk)1{\displaystyle \sum _{n=0}^{\infty }p(n)x^{n}=\prod _{k=1}^{\infty }(1-x^{k})^{-1}}

Note that is the reciprocal of the product on the left hand side of our identity:

(n=0p(n)xn)(n=1(1xn))=1{\displaystyle \left(\sum _{n=0}^{\infty }p(n)x^{n}\right)\cdot \left(\prod _{n=1}^{\infty }(1-x^{n})\right)=1}

Let us denote the expansion of our product byn=1(1xn)=n=0anxn,{\displaystyle \prod _{n=1}^{\infty }(1-x^{n})=\sum _{n=0}^{\infty }a_{n}x^{n},}so that

(n=0p(n)xn)(n=0anxn)=1.{\displaystyle \left(\sum _{n=0}^{\infty }p(n)x^{n}\right)\cdot \left(\sum _{n=0}^{\infty }a_{n}x^{n}\right)=1.}

Multiplying out the left hand side and equating coefficients on the two sides, we obtaina0p(0) = 1 andi=0np(ni)ai=0{\displaystyle \sum _{i=0}^{n}p(n-i)a_{i}=0} for alln1{\displaystyle n\geq 1}. This gives a recurrence relation definingp(n) in terms ofan, and vice versa a recurrence foran in terms ofp(n). Thus, our desired result:

ai:={1 if i=12(3k2±k) and k is even1 if i=12(3k2±k) and k is odd 0 otherwise {\displaystyle a_{i}:={\begin{cases}1&{\text{ if }}i={\frac {1}{2}}(3k^{2}\pm k){\text{ and }}k{\text{ is even}}\\-1&{\text{ if }}i={\frac {1}{2}}(3k^{2}\pm k){\text{ and }}k{\text{ is odd }}\\0&{\text{ otherwise }}\end{cases}}}

fori1{\displaystyle i\geq 1} is equivalent to the identityi(1)ip(ngi)=0,{\displaystyle \sum _{i}(-1)^{i}p(n-g_{i})=0,} wheregi:=12(3i2i){\displaystyle g_{i}:=\textstyle {\frac {1}{2}}(3i^{2}-i)} andi ranges over all integers such thatgin{\displaystyle g_{i}\leq n} (this range includes both positive and negative i, so as to use both kinds of generalized pentagonal numbers). This in turn means:

i evenp(ngi)=i oddp(ngi).{\displaystyle \sum _{i\mathrm {\ even} }p(n-g_{i})=\sum _{i\mathrm {\ odd} }p(n-g_{i}).}

In terms of sets of partitions, this is equivalent to saying that the following sets are of equal cardinality:

X:=i evenP(ngi){\displaystyle {\mathcal {X}}:=\bigcup _{i\mathrm {\ even} }{\mathcal {P}}(n-g_{i})}         and        Y:=i oddP(ngi),{\displaystyle {\mathcal {Y}}:=\bigcup _{i{\text{ odd}}}{\mathcal {P}}(n-g_{i}),}

whereP(n){\displaystyle {\mathcal {P}}(n)} denotes the set of all partitions ofn{\displaystyle n}.All that remains is to give a bijection from one set to the other, which is accomplished by the functionφ fromX toY which maps the partitionP(ngi)λ:ngi=λ1+λ2++λ{\displaystyle {\mathcal {P}}(n-g_{i})\ni \lambda :n-g_{i}=\lambda _{1}+\lambda _{2}+\dotsb +\lambda _{\ell }} to the partitionλ=φ(λ){\displaystyle \lambda '=\varphi (\lambda )} defined by:

φ(λ):={λ:ngi1=(+3i2)+(λ11)++(λ1) if +3i>λ1λ:ngi+1=(λ2+1)++(λ+1)+1++1λ13i if +3iλ1.{\displaystyle \varphi (\lambda ):={\begin{cases}\lambda ':n-g_{i-1}=(\ell +3i-2)+(\lambda _{1}-1)+\dotsb +(\lambda _{\ell }-1)&{\text{ if }}\ell +3i>\lambda _{1}\\\\\lambda ':n-g_{i+1}=(\lambda _{2}+1)+\dotsb +(\lambda _{\ell }+1)+\underbrace {1+\dotsb +1} _{\lambda _{1}-\ell -3i}&{\text{ if }}\ell +3i\leq \lambda _{1}.\end{cases}}}

This is an involution (a self-inverse mapping), and thus in particular a bijection, which proves our claim and the identity.

See also

The pentagonal number theorem occurs as a special case of theJacobi triple product.

Q-series generalize Euler's function, which is closely related to theDedekind eta function, and occurs in the study ofmodular forms. Themodulus of theEuler function (see there for picture) shows thefractalmodular group symmetry and occurs in the study of the interior of theMandelbrot set.

References

  1. ^Franklin, F. (1881). "Sur le developpement du produit (1 – x)(1 – x2)(1 − x3) ...".Comptes Rendus de l'Académie des Sciences de Paris, Série A.92:448–450.

External links

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