Results 111 to 120 of about 33,775 (162)

The fitness landscape of a form II rubisco in a photosynthetic bacterium guides engineering of oxygen tolerance. [PDF]

open access: yesSci Adv
Hoffmann UA   +14 more
europepmc   +1 more source

Digital infrastructure to support and share daily high-throughput experimental materials research.

open access: yesFaraday Discuss
Mittmann LA   +11 more
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Combinatorial DNA-PAINT

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Steen PR   +10 more
europepmc   +1 more source
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A Combinatorial Identity and the World Series

SIAM Review, 1993
In this note the author gives a simple probabilistic proof of a combinatorial identity by calculating the winning probability in the World Series. The winning probabilities and the expected length of the championship series are given by the applications of the identity and its generalization.
exaly   +2 more sources

A Family of Combinatorial Identities

Canadian Mathematical Bulletin, 1972
In a recent paper, Murray Eden [5] generalized the simple identity for the Eulerian product,1.1and obtained the following infinite family of identities:For A= 1,2, 3,…, let1.2where we assume throughout that |x| < 1, empty products equal unity and empty sums equal zero; then1.3As Eden noted, Fh(b;x) is the generating function of ph(m, n) which ...
Andrews, G. E.   +2 more
openaire   +1 more source

COMBINATORIAL IDENTITIES AND HYPERGEOMETRIC FUNCTIONS

Rocky Mountain Journal of Mathematics, 2022
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Alzer, Horst, Richards, Kendall C.
openaire   +1 more source

On two combinatorial identities

Journal of Computational Methods in Sciences and Engineering, 2017
Two combinatorial identities are proved. The first one allows easy to establish connections between Bernoulli and Stirling numbers and values of zeta function of integer arguments and to find numerous identities, including the new ones, for these quantities.
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On a combinatorial identity

Publicationes Mathematicae Debrecen, 2022
The author proves that the only solution of the difference equation \[ \Phi (2(v+1))+\sum^{v}_{k=1}\binom{2v}{2k-1} \Phi (2k) \Phi (2(v-k+1))=0 \] subject to the initial condition \(\Phi (2)=1/4\) is \(\Phi (2k)=((2^{2k}-1)/2k) B_{2k}\) where the \(B_{2k}\) are the even-indexed Bernoulli numbers. Using this result, he then proves that (1/k)\(| B_{2k}|^{
openaire   +2 more sources

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