Li Shanlan identity

The Li Shanlan identity states that

${\displaystyle \sum _{k=0}^{p}{p \choose k}^{2}{{n+2p-k} \choose {2p}}={{n+p} \choose p}^{2}}$.

Li Shanlan did not present the identity in this way. He presented it in the traditional Chinese algorithmic and rhetorical way.[5]

Li Shanlan had not given a proof of the identity in Duoji bilei. The first proof using differential equations and Legendre polynomials, concepts foreign to Li, was published by Pál Turán in 1936, and the proof appeared in Chinese in Yung Chang's paper published in 1939.[2] Since then at least fifteen different proofs have been found.[2] The following is one of the simplest proofs.[6]

The proof begins by expressing ${\displaystyle n \choose q}$ as Vandermonde's convolution:

${\displaystyle {n \choose q}=\sum _{k=0}^{q}{{n-p} \choose k}{p \choose {q-k}}}$

Pre-multiplying both sides by ${\displaystyle n \choose p}$,

${\displaystyle {n \choose p}{n \choose q}=\sum _{k=0}^{q}{n \choose p}{{n-p} \choose k}{p \choose {q-k}}}$.

Using the following relation

${\displaystyle {n \choose p}{{n-p} \choose k}={{p+k} \choose k}{n \choose {p+k}}}$

the above relation can be transformed to

${\displaystyle {n \choose p}{n \choose q}=\sum _{k=0}^{q}{p \choose {q-k}}{{p+k} \choose k}{n \choose {p+k}}}$.

Next the relation

${\displaystyle {p \choose {q-k}}{{p+k} \choose {k}}={q \choose k}{{p+k} \choose {q}}}$

is used to get

${\displaystyle {n \choose p}{n \choose q}=\sum _{k=0}^{q}{q \choose k}{n \choose {p+k}}{{p+k} \choose q}}$.

Another application of Vandermonde's convolution yields

${\displaystyle {{p+k} \choose q}=\sum _{j=0}^{q}{p \choose j}{k \choose {q-j}}}$

and hence

${\displaystyle {n \choose p}{n \choose q}=\sum _{k=0}^{q}{q \choose k}{n \choose {p+k}}\sum _{j=0}^{q}{p \choose j}{k \choose {q-j}}}$

Since ${\displaystyle p \choose j}$ is independent of k, this can be put in the form

${\displaystyle {n \choose p}{n \choose q}=\sum _{j=0}^{q}{p \choose j}\sum _{k=0}^{q}{q \choose k}{n \choose {p+k}}{k \choose {q-j}}}$

Next, the result

${\displaystyle {q \choose k}{k \choose {q-j}}={q \choose j}{j \choose {q-k}}}$

gives

${\displaystyle {n \choose p}{n \choose q}=\sum _{j=0}^{q}{p \choose j}\sum _{k=0}^{q}{q \choose j}{j \choose {q-k}}{n \choose {p+k}}}$

${\displaystyle =\sum _{j=0}^{q}{p \choose j}{q \choose j}\sum _{k=0}^{q}{j \choose {q-k}}{n \choose {p+k}}}$

${\displaystyle =\sum _{j=0}^{q}{p \choose j}{q \choose j}{{n+j} \choose {p+q}}}$

Setting p = q and replacing j by k,

${\displaystyle {n \choose p}^{2}=\sum _{k=0}^{p}{p \choose k}^{2}{{n+k} \choose {2p}}}$

Li's identity follows from this by replacing n by n + p and doing some rearrangement of terms in the resulting expression:

${\displaystyle {{n+p} \choose p}^{2}=\sum _{k=0}^{p}{p \choose k}^{2}{{n+2p-k} \choose {2p}}}$

The term duoji denotes a certain traditional Chinese method of computing sums of piles. Most of the mathematics that was developed in China since the sixteenth century is related to the duoji method. Li Shanlan was one of the greatest exponents of this method and Duoji bilei is an exposition of his work related to this method. Duoji bilei consists of four chapters: Chapter 1 deals with triangular piles, Chapter 2 with finite power series, Chapter 3 with triangular self-multiplying piles and Chapter 4 with modified triangular piles.[7]