"Lebesgue identity"의 두 판 사이의 차이
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introduction==
fermionic form expression==
\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\)
(proof)
We use the q-binomial identity useful techniques in q-series
\((-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\) and \((-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\)
\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}\)
\(=\sum_{k\geq 0}\frac{a^kq^{k(k-1)/2}}{(q)_{k}}\sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\). Put \(j=r\) and \(i=k-j\).
\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\)
\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\) ■
Lebesgue's identity==
continued fraction expression==
comparison with Rogers-Selberg identities==
encyclopedia==
books==
articles==
imported>Pythagoras0 잔글 (찾아 바꾸기 – “</h5>” 문자열을 “==” 문자열로) |
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| − | <h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">introduction | + | <h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">introduction== |
* [http://pythagoras0.springnote.com/pages/7958020 르벡 항등식] | * [http://pythagoras0.springnote.com/pages/7958020 르벡 항등식] | ||
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| − | <h5 style="line-height: 2em; margin: 0px;">fermionic form expression | + | <h5 style="line-height: 2em; margin: 0px;">fermionic form expression== |
<math>f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}</math> | <math>f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}</math> | ||
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| − | <h5 style="line-height: 2em; margin: 0px;">Lebesgue's identity | + | <h5 style="line-height: 2em; margin: 0px;">Lebesgue's identity== |
* Put a=q, c=z. we get the Lebesgue's identity.<br><math>f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})</math><br> | * Put a=q, c=z. we get the Lebesgue's identity.<br><math>f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})</math><br> | ||
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| − | <h5 style="line-height: 2em; margin: 0px;">specializations | + | <h5 style="line-height: 2em; margin: 0px;">specializations== |
(Theorem) | (Theorem) | ||
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| − | <h5 style="line-height: 2em; margin: 0px;">continued fraction expression | + | <h5 style="line-height: 2em; margin: 0px;">continued fraction expression== |
* [[rank 2 continued fraction]]<br> | * [[rank 2 continued fraction]]<br> | ||
| 85번째 줄: | 85번째 줄: | ||
| − | <h5 style="line-height: 2em; margin: 0px;">related dilogarithm identity | + | <h5 style="line-height: 2em; margin: 0px;">related dilogarithm identity== |
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| − | <h5 style="line-height: 2em; margin: 0px;">comparison with Rogers-Selberg identities | + | <h5 style="line-height: 2em; margin: 0px;">comparison with Rogers-Selberg identities== |
* [[Rogers-Selberg identities]]<br><math>AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}</math><br><math>A(q)W(q)=AG_{3,3}(q)</math><br> where<br><math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math><br> | * [[Rogers-Selberg identities]]<br><math>AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}</math><br><math>A(q)W(q)=AG_{3,3}(q)</math><br> where<br><math>W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}</math><br> | ||
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* http://www.google.com/search?hl=en&tbs=tl:1&q= | * http://www.google.com/search?hl=en&tbs=tl:1&q= | ||
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* [[1 Nahm's conjecture|Nahm's conjecture]]<br> | * [[1 Nahm's conjecture|Nahm's conjecture]]<br> | ||
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* http://en.wikipedia.org/wiki/ | * http://en.wikipedia.org/wiki/ | ||
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* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords= | * http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords= | ||
| − | [[4909919|]] | + | [[4909919|4909919]] |
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* New Proofs of Identities of Lebesgue and Göllnitz via Tilings<br> | * New Proofs of Identities of Lebesgue and Göllnitz via Tilings<br> | ||
2012년 10월 28일 (일) 14:31 판
introduction==
- [Alladi&Gordon1993] 278&279p
\(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\)
\(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\)
fermionic form expression==
\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\)
(proof)
We use the q-binomial identity useful techniques in q-series
\((-z;q)_{n}= \sum_{r=0}^{n} \begin{bmatrix} n\\ r\end{bmatrix}_{q}q^{r(r-1)/2}z^r\) and \((-zq;q)_{k}= \sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\)
\(f(a,z)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}\)
\(=\sum_{k\geq 0}\frac{a^kq^{k(k-1)/2}}{(q)_{k}}\sum_{r=0}^{k} \begin{bmatrix} k\\ r\end{bmatrix}_{q}q^{r(r+1)/2}z^r\). Put \(j=r\) and \(i=k-j\).
\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{(i+j)(i+j-1)/2+j(j+1)/2}}{(q)_{i}(q)_{j}}\)
\(=\sum_{i,j\geq 0}\frac{a^{i+j}z^{j}q^{\frac{i^2+2ij+j^2-i}{2}}}{(q)_{i}(q)_{j}}\) ■
- here we get a 2x2 matrix (rank 2 case)
\( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)
\( \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)
Lebesgue's identity==
- Put a=q, c=z. we get the Lebesgue's identity.
\(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})\)
- special case : we get a rank 2 form of the Lebesgue's identity
\(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}\)
\(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{k\geq 0}\frac{q^{k(k+1)/2}(-zq)_{k}}{(q)_{k}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}=\prod_{m=1}^{\infty} (1+zq^{2m})(1+q^{m})\)
\(f(q,z)=\sum_{k\geq 0}\frac{q^{k}q^{k(k-1)/2}(-zq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{z^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-zq^2;q^2)_{\infty}(-q)_{\infty}\)
specializations== (Theorem) \(f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\) \(f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\) (proof) useful techniques in q-series \((-q)_{n}=\frac{(q^2;q^2)_{n}}{(q;q)_{n}}\) \((-q;q^{2})_{n}=\frac{(-q;q)_{n}}{(-q^{2};q^{2})_{n}}=\frac{(q^{2};q^{2})_{n}(q^{2};q^{2})_{n}}{(q^{4};q^{4})_{n}(q;q)_{n}}=\frac{(q^{2};q^{4})_{n}}{(q^{1};q^{4})_{n}(q^{3};q^{4})_{n}}\) . \((-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}\) Therefore \(f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\) \(f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\). ■
continued fraction expression==
- rank 2 continued fraction
- [Alladi&Gordon1993] 277-278p
Let \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\).
\(F(a,c)=\frac{f(a,c)}{f(aq,c)}=1+a+\frac{acq}{1+aq} {\ \atop+} \frac{acq^2}{1+aq^2}{\ \atop+} \frac{acq^3}{1} {\ \atop+\dots}\)
\(R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}\)
where
\(R^{N}(a,b)=f(q,a^{-1}b)-af(aq,a^{-1}b)=f(aq,a^{-1}bq^{-1})=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}b)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}\)
\(R^{D}(a,b)=f(aq,a^{-1}b)=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}bq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}\)
- application
\(R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q^1;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\)
\(R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
- continued fraction
\(R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} } \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
Let \(f(a,c)=\sum_{k\geq 0}\frac{a^{k}q^{k(k-1)/2}(-cq)_{k}}{(q)_{k}}\).
\(F(a,c)=\frac{f(a,c)}{f(aq,c)}=1+a+\frac{acq}{1+aq} {\ \atop+} \frac{acq^2}{1+aq^2}{\ \atop+} \frac{acq^3}{1} {\ \atop+\dots}\)
\(R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}\)
where
\(R^{N}(a,b)=f(q,a^{-1}b)-af(aq,a^{-1}b)=f(aq,a^{-1}bq^{-1})=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}b)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}\)
\(R^{D}(a,b)=f(aq,a^{-1}b)=\sum_{k\geq 0}\frac{a^{k}q^{k(k+1)/2}(-a^{-1}bq)_{k}}{(q)_{k}}=\sum_{i,j\geq 0}\frac{a^{i}b^{j}q^{\frac{i^2+2ij+j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}\)
\(R^N(1,1)=f(q,q^{-1})=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+j^2+i}{2}}}{(q)_{i}(q)_{j}}=(-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{(q^2;q^4)_{\infty}}{(q^1;q^4)_{\infty}^2(q^3;q^4)_{\infty}^2}\)
\(R^{D}(1,1)=f(q,1)=\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2+i+2j}{2}}}{(q)_{i}(q)_{j}}=(-q^2;q^2)_{\infty}(-q)_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{1}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
\(R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} } \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
related dilogarithm identity==
comparison with Rogers-Selberg identities==
- Rogers-Selberg identities
\(AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}\)
\(A(q)W(q)=AG_{3,3}(q)\)
where
\(W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}\)
- Lebesgue's identity
\(\frac{W(q)^2}{W(q^2)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\)
(proof)
Note that from useful techniques in q-series
\((-q;q^{2})_{\infty}=\frac{(-q;q)_{\infty}}{(-q^{2};q^{2})_{\infty}}=\frac{(q^{2};q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(q^{4};q^{4})_{\infty}(q;q)_{\infty}}=\frac{W(q)}{W(q^2)}\)
Therefore
\((-q;q^2)_{\infty}(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{W(q)^2}{W(q^2)}\). ■
\(W(q)=\frac{\eta(2\tau)}{\eta(\tau)}\)
\(W(q^2)=\frac{\eta(4\tau)}{\eta(2\tau)}\)
\(\frac{W(q)^2}{W(q^2)}=\frac{(q^{2};q^{2})_{\infty}^3}{(q;q)_{\infty}^2(q^{4};q^{4})_{\infty}}=\frac{\eta(2\tau)^3}{\eta(\tau)^2\eta(4\tau)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\)
\(W(q^2)W(q)=\frac{\eta(4\tau)}{\eta(\tau)}=q^{1/8}\sum_{i,j\geq 0}\frac{q^{\frac{i^2+2ij+2j^2}{2}+\frac{i+2j}{2}}}{(q)_{i}(q)_{j}}=\frac{q^{1/8}(q^4;q^4)_{\infty}}{(q;q)_{\infty}}=\frac{q^{1/8}}{(q^1;q^4)_{\infty}(q^2;q^4)_{\infty}(q^3;q^4)_{\infty}}\)
eta product and eta quotient
\(AG_{3,3}(q)=\sum_{n_1,n_{2}\geq0}\frac{q^{n_{1}^2+2n_1n_2+2n_{2}^{2}}}{(q)_{n_1}(q)_{n_{2}}}=\prod_{r\neq 0,\pm 3 \pmod {7}}\frac{1}{1-q^r}=\frac{(q^3;q^7)_\infty (q^4; q^7)_\infty(q^7;q^7)_\infty}{(q)_\infty}\)
\(A(q)W(q)=AG_{3,3}(q)\)
where
\(W(q)=(-q)_{\infty}=\frac{(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}\)
\(\frac{W(q)^2}{W(q^2)}=\sum_{i,j\geq 0}\frac{q^{(i^2+2ij+2j^2)/2+i/2}}{(q)_{i}(q)_{j}}\)
history==
related items==
encyclopedia==
- http://en.wikipedia.org/wiki/
- http://www.scholarpedia.org/
- http://www.proofwiki.org/wiki/
- Princeton companion to mathematics(Companion_to_Mathematics.pdf)
books==
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
4909919
articles==
- New Proofs of Identities of Lebesgue and Göllnitz via Tilings
- DP Little, 2007
- Alladi, Krishnaswami, and Basil Gordon. 1993. “Partition identities and a continued fraction of Ramanujan.” Journal of Combinatorial Theory, Series A 63 (2) (July): 275-300. doi:10.1016/0097-3165(93)90061-C.
- [Alladi&Gordon1993]Partition identities and a continued fraction of Ramanujan ,Krishnaswami Alladi and Basil Gordon, 1993
- DP Little, 2007