高等组合学笔记(十): 分拆数恒等式, 分拆函数同余性质

分拆数恒等式

定理6 : Gauss分拆恒等式
1(x:q)∞=∑n≥0xn(q:q)n.\frac1{(x:q)_\infty}=\sum_{n\geq0}\frac{x^n}{(q:q)_n.} (x:q)∞1=n≥0∑(q:q)n.xn

证明:

由
∑n,l≥0Pl(n)xlqn=∏k≥111−xqk=1(qx:q)∞;\sum_{n,l\geq0}P_l(n)x^lq^n=\prod_{k\geq1}\frac1{1-xq^k}=\frac1{(qx:q)_{\infty}}; n,l≥0∑Pl(n)xlqn=k≥1∏1−xqk1=(qx:q)∞1;
得到:
1(qx:q)∞=∑l≥0xl∑n≥0Pl(n)qn=∑l≥0xlql(q:q)l\begin{aligned} \frac1{(qx:q)_{\infty}}&=\sum_{l\geq0}x^l\sum_{n\geq0}P_l(n)q^n\\ &=\sum_{l\geq0}x^l\frac{q^l}{(q:q)_l} \end{aligned} (qx:q)∞1=l≥0∑xln≥0∑Pl(n)qn=l≥0∑xl(q:q)lql
令x→xqx\to \frac xqx→qx, 得到
1(x:q)∞=∑l≥0xl(q:q)l.(1(q:q)∞=∑k≥0qk(q:q)k.)\frac1{(x:q)_\infty}=\sum_{l\geq0}\frac{x^l}{(q:q)_l}.\quad\left(\frac1{(q:q)_\infty}=\sum_{k\geq0}\frac{q^k}{(q:q)_k}.\right) (x:q)∞1=l≥0∑(q:q)lxl.((q:q)∞1=k≥0∑(q:q)kqk.)

定理7 (Euler分拆恒等式)
(x:q)∞=∑n≥0(−1)nq(n2)(q:q)nxn.(x:q)_\infty=\sum_{n\geq0}\frac{(-1)^nq^{\binom n2}}{(q:q)_n}x^n. (x:q)∞=n≥0∑(q:q)n(−1)nq(2n)xn.

证明:

根据
∑n,l≥0Ql(n)xlqn=∏k≥1(1+xqk)=(−qx:q)∞,\sum_{n,l\geq0}Q_l(n)x^lq^n=\prod_{k\geq1}({1+xq^k})={(-qx:q)_{\infty}}, n,l≥0∑Ql(n)xlqn=k≥1∏(1+xqk)=(−qx:q)∞,
得到:
(−qx:q)∞=∑l≥0xl∑n≥0Ql(n)qn=∑l≥0xlq(l+12)(q:q)l\begin{aligned} {(-qx:q)_{\infty}}&=\sum_{l\geq0}x^l\sum_{n\geq0}Q_l(n)q^n\\ &=\sum_{l\geq0}x^l\frac{q^{\binom {l+1}2}}{(q:q)_l} \end{aligned} (−qx:q)∞=l≥0∑xln≥0∑Ql(n)qn=l≥0∑xl(q:q)lq(2l+1)
令x→−xqx\to-\frac xqx→−qx, 得到
(x:q)∞=∑l≥0(−1)lq(l2)(q:q)nxl.(x:q)_\infty=\sum_{l\geq0}\frac{(-1)^lq^{\binom l2}}{(q:q)_n}x^l. (x:q)∞=l≥0∑(q:q)n(−1)lq(2l)xl.

定理8: nnn的自共轭分拆的个数等于nnn的各部分都是奇数且两两不同分分拆的个数. (利用生成函数证明, Euler分拆恒等式)

证明:

由(Euler恒等式)
(−qx:q2)∞=∑n≥0qn2xn(q2:q2)n,(-qx:q^2)_\infty=\sum_{n\geq0}\frac{q^{n^2}x^n}{(q^2:q^2)_n}, (−qx:q2)∞=n≥0∑(q2:q2)nqn2xn,
令x=1x=1x=1,
(−q:q2)∞=∑n≥0qn2(q2:q2)n,(-q:q^2)_\infty=\sum_{n\geq0}\frac{q^{n^2}}{(q^2:q^2)_n}, (−q:q2)∞=n≥0∑(q2:q2)nqn2,
意义:

左端: 各部分都是奇数的分拆.
(−q:q2)∞=(1+q)(1+q3)⋯(-q:q^2)_\infty=(1+q)(1+q^3)\cdots (−q:q2)∞=(1+q)(1+q3)⋯

右端: 自共轭分拆, 利用上节定理5, 中间为qn2q^{n^2}qn2, 旁边为⩽n\leqslant n⩽n, 即1(q2:q2)∞\frac1{(q^2:q^2)_\infty}(q2:q2)∞1.

定义4:

Pl(n∣⩽m)P_l(n|\leqslant m)Pl(n∣⩽m): nnn的lll部分拆, 且每部分⩽m\leqslant m⩽m的分拆数;
Pl(n∣⩽m)P^l(n|\leqslant m)Pl(n∣⩽m): nnn的部分数⩽l\leqslant l⩽l, 且每部分⩽m\leqslant m⩽m的分拆数;
Ql(n∣⩽m)Q_l(n|\leqslant m)Ql(n∣⩽m): nnn的lll部互异分拆, 且每部分⩽m\leqslant m⩽m的分拆数.

定理9:
∑n,l≥0Pl(n∣⩽m)xlqn=1(qx:q)m∑n,l≥0Ql(n∣⩽m)xlqn=(−qx:q)m\sum_{n,l\geq0}P_l(n|\leqslant m)x^lq^n=\frac1{(qx:q)_m}\\ \ \sum_{n,l\geq0}Q_l(n|\leqslant m)x^lq^n={(-qx:q)_m} n,l≥0∑Pl(n∣⩽m)xlqn=(qx:q)m1 n,l≥0∑Ql(n∣⩽m)xlqn=(−qx:q)m
类比:

∑n,l≥0Pl(n)xlqn=∏k≥111−xqk=1(qx:q)∞;\sum_{n,l\geq0}P_l(n)x^lq^n=\prod_{k\geq1}\frac1{1-xq^k}=\frac1{(qx:q)_{\infty}}; n,l≥0∑Pl(n)xlqn=k≥1∏1−xqk1=(qx:q)∞1;
∑n,l≥0Ql(n)xlqn=∏k≥1(1+xqk)=(−qx:q)∞.\sum_{n,l\geq0}Q_l(n)x^lq^n=\prod_{k\geq1}({1+xq^k})={(-qx:q)_{\infty}}. n,l≥0∑Ql(n)xlqn=k≥1∏(1+xqk)=(−qx:q)∞.

以及:

∑l,n≥0Pl(n∣S)xlqn=∏k∈S11−qkx.\sum_{l,n\geq0}P_l(n|S)x^lq^n=\prod_{k\in S}\frac1{1-q^kx}. l,n≥0∑Pl(n∣S)xlqn=k∈S∏1−qkx1.
∑l,n≥0Ql(n∣S)xlqn=∏k∈S(1+xqk).\sum_{l,n\geq0}Q_l(n|S)x^lq^n=\prod_{k\in S}{(1+xq^k)}. l,n≥0∑Ql(n∣S)xlqn=k∈S∏(1+xqk).

定理10 :Jacobi 三重积恒等式
(q2:q2)∞(−qx:q2)∞(−q/x:q2)∞=∑nqn2xn,(q:q)∞(x:q)∞(q/x:q)∞=∑n(−1)nq(n2)xn(−qx→x,q2→q).(q^2:q^2)_\infty(-qx:q^2)_\infty(-q/x:q^2)_\infty=\sum_{n}q^{n^2}x^n,\\ (q:q)_\infty(x:q)_\infty(q/x:q)_\infty=\sum_{n}(-1)^nq^{\binom n2}x^n\quad(-qx\to x,q^2\to q). (q2:q2)∞(−qx:q2)∞(−q/x:q2)∞=n∑qn2xn,(q:q)∞(x:q)∞(q/x:q)∞=n∑(−1)nq(2n)xn(−qx→x,q2→q).

证明:
根据Euler分拆恒等式, 以及(x:q)∞=(x:q)n(qnx:q)∞⟺(q2:q2)∞=(q2:q2)n(q2n+2:q2)∞(x:q)_\infty=(x:q)_n(q^nx:q)_\infty\iff (q^2:q^2)_\infty=(q^2:q^2)_n(q^{2n+2}:q^2)_\infty(x:q)∞=(x:q)n(qnx:q)∞⟺(q2:q2)∞=(q2:q2)n(q2n+2:q2)∞,
(−qx:q2)∞=∑n≥0qn2xn(q2:q2)n=1(q2:q2)∞∑n≥0qn2xn(q2n+2:q2)∞=1(q2:q2)∞∑nqn2xn(q2n+2:q2)∞(n<0,(q2n+2:q2)=0)=1(q2:q2)∞∑nqn2xn∑m≥0(−1)mqm2+m+2mn(q2:q2)m(Euler)=1(q2:q2)∞∑m≥0(−1)mqmx−m(q2:q2)m∑nq(m+n)2xm+n=1(q2:q2)∞∑m≥0(−q/x)m(q2:q2)m∑nqn2xn(Gauss)=1(q2:q2)∞1(−q/x:q2)∞∑nqn2xn\begin{aligned} (-qx:q^2)_\infty&=\sum_{n\geq0}\frac{q^{n^2}x^n}{(q^2:q^2)_n}\\ &=\frac1{(q^2:q^2)_\infty}\sum_{n\geq0}q^{n^2}x^n(q^{2n+2}:q^2)_\infty\\ &=\frac1{(q^2:q^2)_\infty}\sum_{n}q^{n^2}x^n(q^{2n+2}:q^2)_\infty \qquad\big(n<0,(q^{2n+2}:q^2)=0\big)\\ &=\frac1{(q^2:q^2)_\infty}\sum_{n}q^{n^2}x^n\sum_{m\geq0}\frac{(-1)^mq^{m^2+m+2mn}}{(q^2:q^2)_m}\qquad(Euler)\\ &=\frac1{(q^2:q^2)_\infty}\sum_{m\geq0}\frac{(-1)^mq^{m}x^{-m}}{(q^2:q^2)_m}\sum_{n}q^{(m+n)^2}x^{m+n}\\ &=\frac1{(q^2:q^2)_\infty}\sum_{m\geq0}\frac{(-q/x)^m}{(q^2:q^2)_m}\sum_{n}q^{n^2}x^{n}\qquad(Gauss)\\ &=\frac1{(q^2:q^2)_\infty}\frac1{(-q/x:q^2)_\infty}\sum_{n}q^{n^2}x^{n}\\ \end{aligned} (−qx:q2)∞=n≥0∑(q2:q2)nqn2xn=(q2:q2)∞1n≥0∑qn2xn(q2n+2:q2)∞=(q2:q2)∞1n∑qn2xn(q2n+2:q2)∞(n<0,(q2n+2:q2)=0)=(q2:q2)∞1n∑qn2xnm≥0∑(q2:q2)m(−1)mqm2+m+2mn(Euler)=(q2:q2)∞1m≥0∑(q2:q2)m(−1)mqmx−mn∑q(m+n)2xm+n=(q2:q2)∞1m≥0∑(q2:q2)m(−q/x)mn∑qn2xn(Gauss)=(q2:q2)∞1(−q/x:q2)∞1n∑qn2xn

定理11: Jacobi恒等式
(q:q)∞3=∑n≥0(−1)n(2n+1)q(n+12).(q:q)_\infty^3=\sum_{n\geq0}(-1)^n(2n+1)q^{\binom {n+1}2}. (q:q)∞3=n≥0∑(−1)n(2n+1)q(2n+1).

证明:

由
(q:q)∞(x:q)∞(q/x:q)∞=∑n(−1)nq(n2)xn=∑n≥1(−1)nq(n2)xn+∑n≥0(−1)nq(n+12)x−n=−∑n≥0(−1)nq(n+12)xn+1+∑n≥0(−1)nq(n+12)x−n=∑n≥0(−1)n(x−n−xn+1)q(n+12)\begin{aligned} (q:q)_\infty(x:q)_\infty(q/x:q)_\infty&=\sum_{n}(-1)^nq^{\binom n2}x^n\\ &=\sum_{n\geq1}(-1)^nq^{\binom n2}x^n+\sum_{n\geq0}(-1)^nq^{\binom {n+1}2}x^{-n}\\ &=-\sum_{n\geq0}(-1)^nq^{\binom {n+1}2}x^{n+1}+\sum_{n\geq0}(-1)^nq^{\binom {n+1}2}x^{-n}\\ &= \sum_{n\geq0}(-1)^n\big(x^{-n}-x^{n+1}\big)q^{\binom {n+1}2}\\ \end{aligned} (q:q)∞(x:q)∞(q/x:q)∞=n∑(−1)nq(2n)xn=n≥1∑(−1)nq(2n)xn+n≥0∑(−1)nq(2n+1)x−n=−n≥0∑(−1)nq(2n+1)xn+1+n≥0∑(−1)nq(2n+1)x−n=n≥0∑(−1)n(x−n−xn+1)q(2n+1)
两边同时除以1−x1-x1−x, 并令x→1x\to1x→1, 得到:
(q:q)∞3=∑n≥0(−1)n(2n+1)q(n+12).(q:q)_\infty^3=\sum_{n\geq0}(-1)^n(2n+1)q^{\binom {n+1}2}. (q:q)∞3=n≥0∑(−1)n(2n+1)q(2n+1).

定理12 Gauss三角数定理(Jacobi恒等式的特例)
(q2:q2)∞(q:q2)∞=∑n≥0q(n+12).\frac{(q^2:q^2)_\infty}{(q:q^2)_\infty}=\sum_{n\geq0}q^{\binom {n+1}2}. (q:q2)∞(q2:q2)∞=n≥0∑q(2n+1).

三角形数:
(n+12)=1,3,6,10,15,⋯\binom{n+1}2=1,3,6,10,15,\cdots (2n+1)=1,3,6,10,15,⋯

证明:

由平方差公式以及(q:q)∞=(q:q2)∞(q2:q2)∞(q:q)_\infty=(q:q^2)_\infty(q^2:q^2)_\infty(q:q)∞=(q:q2)∞(q2:q2)∞, 得到
(q2:q2)∞(q:q2)∞=(q:q)∞(−q:q)∞(q:q2)∞=(q:q2)∞(q2:q2)∞(−q:q)∞(q:q2)∞=(q2:q2)∞(−q:q)∞=(q:q)∞(−q:q)∞(−q:q)∞=11+1(q:q)∞(−1:q)∞(−q:q)∞=12∑n(−1)nq(n2)(−1)n(Jacobi)=12(∑n≥1q(n2)+∑n≥0q(n+12))=∑n≥0q(n+12).\begin{aligned} \frac{(q^2:q^2)_\infty}{(q:q^2)_\infty}&=\frac{(q:q)_\infty(-q:q)_\infty}{(q:q^2)_\infty}\\ &=\frac{(q:q^2)_\infty(q^2:q^2)_\infty(-q:q)_\infty}{(q:q^2)_\infty}\\ &=(q^2:q^2)_\infty(-q:q)_\infty\\ &=(q:q)_\infty(-q:q)_\infty(-q:q)_\infty\\ &=\frac1{1+1}(q:q)_\infty(-1:q)_\infty(-q:q)_\infty\\ &=\frac12\sum_n(-1)^nq^{\binom n2}(-1)^n\quad(Jacobi)\\ &=\frac12\left(\sum_{n\geq1}q^{\binom n2}+\sum_{n\geq0}q^{\binom{n+1}2}\right)=\sum_{n\geq0}q^{\binom {n+1}2}. \end{aligned} (q:q2)∞(q2:q2)∞=(q:q2)∞(q:q)∞(−q:q)∞=(q:q2)∞(q:q2)∞(q2:q2)∞(−q:q)∞=(q2:q2)∞(−q:q)∞=(q:q)∞(−q:q)∞(−q:q)∞=1+11(q:q)∞(−1:q)∞(−q:q)∞=21n∑(−1)nq(2n)(−1)n(Jacobi)=21(n≥1∑q(2n)+n≥0∑q(2n+1))=n≥0∑q(2n+1).

定理13 Euler五角数定理
(q:q)∞=∑k(−1)kqk(3k−1)2=∑k(−1)kqk(3k+1)2=1+∑k≥1(−1)k(qk(3k−1)2+qk(3k+1)2)\begin{aligned} (q:q)_\infty&=\sum_k(-1)^kq^{\frac{k(3k-1)}2}=\sum_k(-1)^kq^{\frac{k(3k+1)}2}\\ &=1+\sum_{k\geq1}(-1)^k\left(q^{\frac{k(3k-1)}2}+q^{\frac{k(3k+1)}2}\right) \end{aligned} (q:q)∞=k∑(−1)kq2k(3k−1)=k∑(−1)kq2k(3k+1)=1+k≥1∑(−1)k(q2k(3k−1)+q2k(3k+1))

五角形数n(3n−1)2:1,5,12,22,⋯\dfrac{n(3n-1)}2:1,5,12,22,\cdots2n(3n−1):1,5,12,22,⋯

证明:利用Jacobi三重积恒等式
(q:q)∞=(1−q)(1−q2)(1−q3)(1−q4)(1−q5)(1−q6)⋯(间隔2项取)=(q:q3)∞(q2:q3)∞(q3:q3)∞=(q3:q3)∞(q2:q3)∞(q:q3)∞(令x=q2,或者x=q均可)=∑n(−1)nq3(n2)q2n=∑n(−1)nq3(n2)+2n.\begin{aligned} (q:q)_\infty&=(1-q)(1-q^2)(1-q^3)(1-q^4)(1-q^5)(1-q^6)\cdots\quad(间隔2项取)\\ &=(q:q^3)_\infty(q^2:q^3)_\infty(q^3:q^3)_\infty\\ &=(q^3:q^3)_\infty(q^2:q^3)_\infty(q:q^3)_\infty\quad(令x=q^2,或者x=q均可)\\ &=\sum_n(-1)^nq^{3\binom n2}q^{2n}\\ &=\sum_n(-1)^nq^{3\binom n2+2n}. \end{aligned} (q:q)∞=(1−q)(1−q2)(1−q3)(1−q4)(1−q5)(1−q6)⋯(间隔2项取)=(q:q3)∞(q2:q3)∞(q3:q3)∞=(q3:q3)∞(q2:q3)∞(q:q3)∞(令x=q2,或者x=q均可)=n∑(−1)nq3(2n)q2n=n∑(−1)nq3(2n)+2n.

定理14 Euler五角形数定理的组合证明

令Qe(n)(Q_e(n)\big(Qe(n)(或Qo(n))Q_o(n)\big)Qo(n))为nnn分成偶数(或奇数)个部分互异的分拆数, 则
Qe(n)−Qo(n)={(−1)k,n=3k2±k20,n≠3k2±k2Q_e(n)-Q_o(n)= \begin{cases} (-1)^k,&n=\dfrac{3k^2\pm k}2\\ 0,&n\ne \dfrac{3k^2\pm k}2 \end{cases} Qe(n)−Qo(n)=⎩⎪⎨⎪⎧(−1)k,0,n=23k2±kn=23k2±k

证明:

对于nnn的部分互异的分拆: λ=(λ1,⋯,λr)>\lambda=(\lambda_1,\cdots,\lambda_r)_{>}λ=(λ1,⋯,λr)>,

定义:

S(λ)S(\lambda)S(λ)表示nnn 的互异分拆λ\lambdaλ的最小部分,(S(λ)=λrS(\lambda)=\lambda_rS(λ)=λr).

σ(λ)\sigma(\lambda)σ(λ)表示从λ1\lambda_1λ1开始, 连续整数的长度.

例子:

当s(λ)≤σ(λ)s(\lambda)\leq\sigma(\lambda)s(λ)≤σ(λ), 对λ=(7,6,4,3,2)\lambda=(7,6,4,3,2)λ=(7,6,4,3,2)作变换:
当s(λ)>σ(λ)s(\lambda)>\sigma(\lambda)s(λ)>σ(λ), 对λ′=(8,7,4,3)\lambda'=(8,7,4,3)λ′=(8,7,4,3)作变换:

在上述两种变换下:
Qe(n)⇄Qo(n)Q_e(n)\rightleftarrows Q_o(n) Qe(n)⇄Qo(n)
但是当出现下述情况时, 无法执行:

对于第一种变换, 若nnn的互异分拆部分数为kkk, 且s(λ)=σ(λ)=ks(\lambda)=\sigma(\lambda)=ks(λ)=σ(λ)=k, 即
n=k+(k+1)+⋯+(k+(k−1))=k2+(k2)=3k2−k2,n=k+(k+1)+\cdots+(k+(k-1))=k^2+\binom k2=\frac{3k^2-k}2, n=k+(k+1)+⋯+(k+(k−1))=k2+(2k)=23k2−k,
于是
Qe(n)=Qo(n)+(−1)k.Q_e(n)=Q_o(n)+(-1)^k. Qe(n)=Qo(n)+(−1)k.
对于第二种变换, 若nnn的互异分拆部分数为kkk, 且σ(λ)=k,s(λ)=k+1\sigma(\lambda)=k,s(\lambda)=k+1σ(λ)=k,s(λ)=k+1, 即
n=(k+1)+⋯+2k=k2+(k+12)=3k2+k2,n=(k+1)+\cdots+2k=k^2+\binom {k+1}2=\frac{3k^2+k}2, n=(k+1)+⋯+2k=k2+(2k+1)=23k2+k,
于是
Qe(n)=Qo(n)+(−1)k.Q_e(n)=Q_o(n)+(-1)^k. Qe(n)=Qo(n)+(−1)k.

综上:

nnn不是五角形数: Qe(n)=Qo(n)Q_e(n)=Q_o(n)Qe(n)=Qo(n).
nnn是五角形数: n=12k(3k±1)n=\dfrac12k(3k\pm1)n=21k(3k±1), Qe(n)=Qo(n)+(−1)kQ_e(n)=Q_o(n)+(-1)^kQe(n)=Qo(n)+(−1)k.

推论: (Euler五角形数定理)
(q:q)∞=∑k(−1)kqk(3k−1)2=∑k(−1)kqk(3k+1)2=1+∑k≥1(−1)k(qk(3k−1)2+qk(3k+1)2)\begin{aligned} (q:q)_\infty&=\sum_k(-1)^kq^{\frac{k(3k-1)}2}=\sum_k(-1)^kq^{\frac{k(3k+1)}2}\\ &=1+\sum_{k\geq1}(-1)^k\left(q^{\frac{k(3k-1)}2}+q^{\frac{k(3k+1)}2}\right) \end{aligned} (q:q)∞=k∑(−1)kq2k(3k−1)=k∑(−1)kq2k(3k+1)=1+k≥1∑(−1)k(q2k(3k−1)+q2k(3k+1))

证明: 根据
1+∑k≥1(−1)k(qk(3k−1)2+qk(3k+1)2)=1+∑n≥1(Qe(n)−Qo(n))qn1+\sum_{k\geq1}(-1)^k\left(q^{\frac{k(3k-1)}2}+q^{\frac{k(3k+1)}2}\right)=1+\sum_{n\geq1}\big(Q_e(n)-Q_o(n)\big)q^n 1+k≥1∑(−1)k(q2k(3k−1)+q2k(3k+1))=1+n≥1∑(Qe(n)−Qo(n))qn
而:
(q:q)∞=∏k≥1(1−qk)=∏k≥1∑ak=0,1(−1)akqkak=∑k≥1a1,a2,...,ak=0,1(−1)∑kakq∑kkak=∑a1=0,1∑a2=0,1⋯(−1)a1+a2+⋯q1⋅a1+2⋅a2+⋯\begin{aligned} (q:q)_\infty&=\prod_{k\geq1}(1-q^k)=\prod_{k\geq1}\sum_{a_k=0,1}(-1)^{a_k}q^{ka_k}\\ &=\sum_{\stackrel{a_1,a_2,...,a_k=0,1}{k\geq1}}(-1)^{\sum_ka_k}q^{\sum_kka_k}\\ &=\sum_{a_1=0,1}\sum_{a_2=0,1}\cdots(-1)^{a_1+a_2+\cdots}q^{1\cdot a_1+2\cdot a_2+\cdots} \end{aligned} (q:q)∞=k≥1∏(1−qk)=k≥1∏ak=0,1∑(−1)akqkak=k≥1a1,a2,...,ak=0,1∑(−1)∑kakq∑kkak=a1=0,1∑a2=0,1∑⋯(−1)a1+a2+⋯q1⋅a1+2⋅a2+⋯
取[qn][q^n][qn], 即
n=1a1+2a2+⋯=1a12a2⋯,ai=0or1.n=1a_1+2a_2+\cdots=1^{a_1}2^{a_2}\cdots,\quad a_i=0\ or\ 1. n=1a1+2a2+⋯=1a12a2⋯,ai=0 or 1.
为一部分互异的分拆, 这里∑iai\sum_ia_i∑iai为nnn的部分互异分拆的部分数.

定理15 (p(n)p(n)p(n)的递推关系式)
p(n)=p(n−1)+p(n−2)−p(n−5)−p(n−7)+⋯(n>0)=∑k≥1(−1)k−1(p(n−k(3k−1)2)+p(n−k(3k+1)2))\begin{aligned} p(n)&=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots\quad(n>0)\\ &=\sum_{k\geq1}(-1)^{k-1}\left(p\left(n-\frac{k(3k-1)}2\right)+p\left(n-\frac{k(3k+1)}2\right)\right) \end{aligned} p(n)=p(n−1)+p(n−2)−p(n−5)−p(n−7)+⋯(n>0)=k≥1∑(−1)k−1(p(n−2k(3k−1))+p(n−2k(3k+1)))

n<0n<0n<0, 规定p(n)=0p(n)=0p(n)=0.

证明:

由分拆数生成函数, 得
∑n≥0P(n)qn=∏k≥111−qk=1(q:q)∞⇒(q:q)∞∑n≥0p(n)qn=1\sum_{n\geq0}P(n)q^n=\prod_{k\geq1}\frac1{1-q^k}=\frac1{(q:q)_{\infty}}\Rightarrow (q:q)_\infty\sum_{n\geq0}p(n)q^n=1 n≥0∑P(n)qn=k≥1∏1−qk1=(q:q)∞1⇒(q:q)∞n≥0∑p(n)qn=1
再由Euler五角数定理, 得到
⇒∑n≥0p(n)qn⋅(1+∑k≥1(−1)k(qk(3k−1)2+qk(3k+1)2))=1,\Rightarrow \sum_{n\geq0}p(n)q^n\cdot\left(1+\sum_{k\geq1}(-1)^k\left(q^{\frac{k(3k-1)}2}+q^{\frac{k(3k+1)}2}\right)\right)=1, ⇒n≥0∑p(n)qn⋅(1+k≥1∑(−1)k(q2k(3k−1)+q2k(3k+1)))=1,
比较两端[qn][q^n][qn], (n>0n>0n>0)
p(n)+∑k≥1(−1)k[p(n−k(3k−1)2)+p(n−k(3k+1)2)]=0,p(n)+\sum_{k\geq1}(-1)^k\left[p\left(n-\frac{k(3k-1)}2\right)+p\left(n-\frac{k(3k+1)}2\right)\right]=0, p(n)+k≥1∑(−1)k[p(n−2k(3k−1))+p(n−2k(3k+1))]=0,
即:
p(n)=∑k≥1(−1)k−1(p(n−k(3k−1)2)+p(n−k(3k+1)2)).p(n)=\sum_{k\geq1}(-1)^{k-1}\left(p\left(n-\frac{k(3k-1)}2\right)+p\left(n-\frac{k(3k+1)}2\right)\right). p(n)=k≥1∑(−1)k−1(p(n−2k(3k−1))+p(n−2k(3k+1))).

分拆函数p(n)p(n)p(n)的同余性质

定理:
p(5n+4)≡0(mod5)p(7n+5)≡0(mod7)p(11n+6)≡0(mod11)\begin{aligned} p(5n+4)&\equiv0\pmod5\\ p(7n+5)&\equiv0\pmod7\\ p(11n+6)&\equiv0\pmod{11}\\ \end{aligned} p(5n+4)p(7n+5)p(11n+6)≡0(mod5)≡0(mod7)≡0(mod11)

证明:
∑n≥0p(n)qn=∏k≥111−qk=1(q:q)∞,\sum_{n\geq0}p(n)q^n=\prod_{k\geq1}\frac1{1-q^k}=\frac1{(q:q)_{\infty}}, n≥0∑p(n)qn=k≥1∏1−qk1=(q:q)∞1,
得到:
∑n≥0p(m)qm+1=q(q:q)∞=q(q:q)∞4(q:q)∞5≡q(q:q)∞4(q5:q5)∞(mod5),∑n≥0p(m)qm+2=q2(q:q)∞=q2(q:q)∞6(q:q)∞7≡q2(q:q)∞6(q7:q7)∞(mod7).\sum_{n\geq0}p(m)q^{m+1}=\frac q{(q:q)_{\infty}}=\frac{q(q:q)_{\infty}^4}{(q:q)_\infty^5}\equiv\frac{q(q:q)_\infty^4}{(q^5:q^5)_\infty}\pmod5,\\ \sum_{n\geq0}p(m)q^{m+2}=\frac {q^2}{(q:q)_{\infty}}=\frac{q^2(q:q)_{\infty}^6}{(q:q)_\infty^7}\equiv\frac{q^2(q:q)_\infty^6}{(q^7:q^7)_\infty}\pmod7. n≥0∑p(m)qm+1=(q:q)∞q=(q:q)∞5q(q:q)∞4≡(q5:q5)∞q(q:q)∞4(mod5),n≥0∑p(m)qm+2=(q:q)∞q2=(q:q)∞7q2(q:q)∞6≡(q7:q7)∞q2(q:q)∞6(mod7).

其中
(q:q)∞5=(1−q)5(1−q2)5(1−q3)5⋯≡(1−q5)(1−q10)(1−q15)⋯(mod5).(q:q)_\infty^5=(1-q)^5(1-q^2)^5(1-q^3)^5\cdots\equiv(1-q^5)(1-q^{10})(1-q^{15})\cdots\pmod5. (q:q)∞5=(1−q)5(1−q2)5(1−q3)5⋯≡(1−q5)(1−q10)(1−q15)⋯(mod5).
由于
(1+x)p≡1+xp(modp),(pk)≡0(modp)(0<k<p).(1+x)^p\equiv1+x^p\pmod p,\quad \binom pk\equiv0\pmod p\quad(0<k<p). (1+x)p≡1+xp(modp),(kp)≡0(modp)(0<k<p).

由Euler五角数定理以及Jacobi恒等式:
(q:q)∞=∑k(−1)kqk(3k−1)2=∑k(−1)kq3(k2)+k,(q:q)∞3=∑n≥0(−1)n(2n+1)q(n+12).(q:q)_\infty=\sum_k(-1)^kq^{\frac{k(3k-1)}2}=\sum_k(-1)^kq^{3\binom k2 +k},\\ (q:q)_\infty^3=\sum_{n\geq0}(-1)^n(2n+1)q^{\binom {n+1}2}. (q:q)∞=k∑(−1)kq2k(3k−1)=k∑(−1)kq3(2k)+k,(q:q)∞3=n≥0∑(−1)n(2n+1)q(2n+1).

q(q:q)∞4=q(q:q)∞(q:q)∞3=q∑k(−1)kq3(k2)+k∑n≥0(−1)n(2n+1)q(n+12)=∑i≥0∑j(−1)i+j(2i+1)q(i+12)+3(j2)+2j+1\begin{aligned} q(q:q)_{\infty}^4&=q(q:q)_\infty(q:q)_\infty^3\\ &=q\sum_k(-1)^kq^{3\binom k2 +k}\sum_{n\geq0}(-1)^n(2n+1)q^{\binom {n+1}2}\\ &=\sum_{i\geq0}\sum_j(-1)^{i+j}(2i+1)q^{\binom{i+1}2+3\binom j2+2j+1} \end{aligned} q(q:q)∞4=q(q:q)∞(q:q)∞3=qk∑(−1)kq3(2k)+kn≥0∑(−1)n(2n+1)q(2n+1)=i≥0∑j∑(−1)i+j(2i+1)q(2i+1)+3(2j)+2j+1

由于(5,8)=1(5,8)=1(5,8)=1, 我们有
(i+12)+3(j2)+2j+1≡8{(i+12)+3(j2)+2j+1}(mod5)≡4i2+4i+12j2−12j+16j+8(mod5)≡4i2+4i+2j2+4j+3(mod5)≡(2i+1)2+2(j+1)2(mod5)\begin{aligned} \binom{i+1}2+3\binom j2+2j+1&\equiv8\left\{\binom{i+1}2+3\binom j2+2j+1\right\}\pmod5\\ &\equiv 4i^2+4i+12j^2-12j+16j+8\pmod5\\ &\equiv 4i^2+4i+2j^2+4j+3\pmod5\\ &\equiv(2i+1)^2+2(j+1)^2\pmod5 \end{aligned} (2i+1)+3(2j)+2j+1≡8{(2i+1)+3(2j)+2j+1}(mod5)≡4i2+4i+12j2−12j+16j+8(mod5)≡4i2+4i+2j2+4j+3(mod5)≡(2i+1)2+2(j+1)2(mod5)
又因为
(2i+1)2≡0,1,4(mod5),2(j+1)2≡0,2,3(mod5),(2i+1)^2\equiv0,1,4\pmod5,\quad2(j+1)^2\equiv0,2,3\pmod5, (2i+1)2≡0,1,4(mod5),2(j+1)2≡0,2,3(mod5),
而
(i+12)+3(j2)+2j+1≡0(mod5)⟺{(2i+1)2≡0(mod5)⟺2i+1≡0(mod5)2(j+1)2≡0(mod5)\binom{i+1}2+3\binom j2+2j+1\equiv0\pmod5\iff\\[5pt] \begin{cases} (2i+1)^2\equiv0\pmod5\iff2i+1\equiv0\pmod5\\ 2(j+1)^2\equiv0\pmod5 \end{cases} (2i+1)+3(2j)+2j+1≡0(mod5)⟺{(2i+1)2≡0(mod5)⟺2i+1≡0(mod5)2(j+1)2≡0(mod5)
⟺p(5n+4)≡(mod5)\iff p(5n+4)\equiv\pmod5⟺p(5n+4)≡(mod5).

对于第二个式子:
由于
q2(q:q)∞6=q2(q:q)∞3(q:q)∞3=∑i,j≥0(−1)i+j(2i+1)(2j+1)q(i+12)+(j+12)+2,q^2(q:q)_\infty^6=q^2(q:q)_\infty^3(q:q)_\infty^3=\sum_{i,j\geq0}(-1)^{i+j}(2i+1)(2j+1)q^{\binom{i+1}2+\binom{j+1}2+2}, q2(q:q)∞6=q2(q:q)∞3(q:q)∞3=i,j≥0∑(−1)i+j(2i+1)(2j+1)q(2i+1)+(2j+1)+2,
而
(i+12)+(j+12)+2≡8[(i+12)+(j+12)+2](mod7)=(2i+1)2+(2j+1)2(mod7),\binom{i+1}2+\binom{j+1}2+2\equiv8\left[\binom{i+1}2+\binom{j+1}2+2\right]\pmod7\\ =(2i+1)^2+(2j+1)^2\pmod7, (2i+1)+(2j+1)+2≡8[(2i+1)+(2j+1)+2](mod7)=(2i+1)2+(2j+1)2(mod7),
又由
(2n+1)2≡0,1,2,4(mod7),(2n+1)^2\equiv0,1,2,4\pmod7, (2n+1)2≡0,1,2,4(mod7),
当
(i+12)+(j+12)+2≡0(mod7)⟺{(2i+1)2≡0(mod7)⟺2i+1≡0(mod7)(2j+1)2≡0(mod7)⟺2j+1≡0(mod7)\binom{i+1}2+\binom{j+1}2+2\equiv0\pmod7\iff\\ \begin{cases} (2i+1)^2\equiv0\pmod7\iff2i+1\equiv 0\pmod7\\ (2j+1)^2\equiv0\pmod7\iff2j+1\equiv 0\pmod7\\ \end{cases} (2i+1)+(2j+1)+2≡0(mod7)⟺{(2i+1)2≡0(mod7)⟺2i+1≡0(mod7)(2j+1)2≡0(mod7)⟺2j+1≡0(mod7)
⟺p(7n+5)≡(mod7)\iff p(7n+5)\equiv\pmod7⟺p(7n+5)≡(mod7).

或者直接从生成函数方面进行考虑:

∑n≥0p(5n+4)qn=5(q5:q5)∞5(q:q)∞6,\sum_{n\geq0}p(5n+4)q^n=5\frac{(q^5:q^5)_\infty^5}{(q:q)_\infty^6}, n≥0∑p(5n+4)qn=5(q:q)∞6(q5:q5)∞5,
∑n≥0p(7n+5)qn=7(q7:q7)∞7(q:q)∞4+49q(q7:q7)∞7(q:q)∞8.\sum_{n\geq0}p(7n+5)q^n=7\frac{(q^7:q^7)_\infty^7}{(q:q)_\infty^4}+49q\frac{(q^7:q^7)_\infty^7}{(q:q)_\infty^8}. n≥0∑p(7n+5)qn=7(q:q)∞4(q7:q7)∞7+49q(q:q)∞8(q7:q7)∞7.