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Is there (likely to be) any formula for $$ \sum_{m' \geq m} {m+j \choose m'} {B - m' \choose m+i-m'} $$ ?

I am mainly interested in the case $B=1$ (or a prime power), if that's any easier.

EDIT: as suggested, here are the $B=1$ answers for $m=0\ldots 4$, $i,j< 10$, arranged as matrices (one for each $m$).

$m=0: \begin{pmatrix}{2}& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ {2}& {2}& 1& 1& 1& 1& 1& 1& 1& 1\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& {-1}& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& {-2}& 1& 0& 0& 0& 0& 0\\ 0& 0& {-1}& {3}& {-3}& 1& 0& 0& 0& 0\\ 0& 0& 1& {-4}& {6}& {-4}& 1& 0& 0& 0\\ 0& 0& {-1}& {5}& {-10}& {10}& {-5}& 1& 0& 0\\ 0& 0& 1& {-6}& {15}& {-20}& {15}& {-6}& 1& 0\\ 0& 0& {-1}& {7}& {-21}& {35}& {-35}& {21}& {-7}& 1\\ \end{pmatrix}$

$m=1: \begin{pmatrix}{2}& {2}& {3}& {4}& {5}& {6}& {7}& {8}& {9}& {10}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& {-1}& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& {-2}& 1& 0& 0& 0& 0& 0& 0\\ 0& {-1}& {3}& {-3}& 1& 0& 0& 0& 0& 0\\ 0& 1& {-4}& {6}& {-4}& 1& 0& 0& 0& 0\\ 0& {-1}& {5}& {-10}& {10}& {-5}& 1& 0& 0& 0\\ 0& 1& {-6}& {15}& {-20}& {15}& {-6}& 1& 0& 0\\ 0& {-1}& {7}& {-21}& {35}& {-35}& {21}& {-7}& 1& 0\\ 0& 1& {-8}& {28}& {-56}& {70}& {-56}& {28}& {-8}& 1\\ \end{pmatrix}$

$m=2: \begin{pmatrix}{2}& {3}& {6}& {10}& {15}& {21}& {28}& {36}& {45}& {55}\\ {-2}& {-2}& {-6}& {-10}& {-15}& {-21}& {-28}& {-36}& {-45}& {-55}\\ {2}& 1& {7}& {10}& {15}& {21}& {28}& {36}& {45}& {55}\\ {-2}& 0& {-9}& {-9}& {-15}& {-21}& {-28}& {-36}& {-45}& {-55}\\ {2}& {-1}& {12}& {6}& {16}& {21}& {28}& {36}& {45}& {55}\\ {-2}& {2}& {-16}& 0& {-20}& {-20}& {-28}& {-36}& {-45}& {-55}\\ {2}& {-3}& {21}& {-10}& {30}& {15}& {29}& {36}& {45}& {55}\\ {-2}& {4}& {-27}& {25}& {-50}& 0& {-35}& {-35}& {-45}& {-55}\\ {2}& {-5}& {34}& {-46}& {85}& {-35}& {56}& {28}& {46}& {55}\\ {-2}& {6}& {-42}& {74}& {-141}& {105}& {-112}& 0& {-54}& {-54}\\ \end{pmatrix}$

$m=3: \begin{pmatrix}{2}& {4}& {10}& {20}& {35}& {56}& {84}& {120}& {165}& {220}\\ {-4}& {-7}& {-20}& {-40}& {-70}& {-112}& {-168}& {-240}& {-330}& {-440}\\ {6}& {9}& {31}& {60}& {105}& {168}& {252}& {360}& {495}& {660}\\ {-8}& {-10}& {-44}& {-79}& {-140}& {-224}& {-336}& {-480}& {-660}& {-880}\\ {10}& {10}& {60}& {95}& {176}& {280}& {420}& {600}& {825}& {1100}\\ {-12}& {-9}& {-80}& {-105}& {-216}& {-335}& {-504}& {-720}& {-990}& {-1320}\\ {14}& {7}& {105}& {105}& {266}& {385}& {589}& {840}& {1155}& {1540}\\ {-16}& {-4}& {-136}& {-90}& {-336}& {-420}& {-680}& {-959}& {-1320}& {-1760}\\ {18}& 0& {174}& {54}& {441}& {420}& {792}& {1071}& {1486}& {1980}\\ {-20}& {5}& {-220}& {10}& {-602}& {-350}& {-960}& {-1155}& {-1660}& {-2199}\\ \end{pmatrix}$

$m=4: \begin{pmatrix}{2}& {5}& {15}& {35}& {70}& {126}& {210}& {330}& {495}& {715}\\ {-6}& {-14}& {-45}& {-105}& {-210}& {-378}& {-630}& {-990}& {-1485}& {-2145}\\ {12}& {26}& {91}& {210}& {420}& {756}& {1260}& {1980}& {2970}& {4290}\\ {-20}& {-40}& {-155}& {-349}& {-700}& {-1260}& {-2100}& {-3300}& {-4950}& {-7150}\\ {30}& {55}& {240}& {519}& {1051}& {1890}& {3150}& {4950}& {7425}& {10725}\\ {-42}& {-70}& {-350}& {-714}& {-1477}& {-2645}& {-4410}& {-6930}& {-10395}& {-15015}\\ {56}& {84}& {490}& {924}& {1988}& {3520}& {5881}& {9240}& {13860}& {20020}\\ {-72}& {-96}& {-666}& {-1134}& {-2604}& {-4500}& {-7569}& {-11879}& {-17820}& {-25740}\\ {90}& {105}& {885}& {1323}& {3360}& {5550}& {9495}& {14840}& {22276}& {32175}\\ {-110}& {-110}& {-1155}& {-1463}& {-4312}& {-6600}& {-11715}& {-18095}& {-27236}& {-39324}\\ \end{pmatrix}$

Allen Knutson
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    Have you calculated any examples to see if there are any patterns? – Gerry Myerson May 22 '11 at 23:58
  • Yes, and I did, but not enough to guess an answer. Examples added. – Allen Knutson May 23 '11 at 00:55
  • Although I don't know if your summation is handled by the methods in the following book, you might want to check it out. The methods here are often called "revolutionary" as regards summations of binomial coefficients: http://www.math.upenn.edu/~wilf/AeqB.html – Fixee May 23 '11 at 00:57
  • It seems you are using the notation ${n\choose k}$ outside of the range $0\le k\le n$. What numbers does it denote? – Did May 23 '11 at 03:22
  • ${a\choose k} = a(a-1)\cdots (a-k+1)/k!$ for $k\geq 0$, so ${-n\choose k} = (-1)^k {k+n-1\choose k}$. – Allen Knutson May 23 '11 at 03:45
  • Here is a quite comprehensive source of similar-looking identities: http://www.math.wvu.edu/~gould/Vol.4.PDF – Doc Oct 30 '13 at 13:24

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