Binomial coefficient from previous related coefficient

[jxu]

Are the following useful enough for loops? This is what I did for project euler problems when asked to compute (n choose i) mod large prime over loop index i before I noticed factorials could be precomputed. This is most similar to the "factoring in" rule.

$$\binom{k}{i} = \frac{k-i+1}{i} \binom{k}{i-1}$$

$$\binom{k}{i+1} = \frac{k-i}{i+1} \binom{k}{i}$$

$$\binom{k-i}{i} = \frac{(k-2i+2)(k-2i+1)}{i(k-i+1)} \binom{k-(i-1)}{i-1}$$ WA

etc.

[adamant-pwn]

Useful in what sense? I don't see any immediate usage over e.g. precomputed factorials?

[jxu]

Perhaps useful for some math problems, but I don't know any specific uses.

[jakobkogler]

It might make sense to notice the first formula.
It does indeed can be useful, if you try to find a closed formula for a combinatorics problem on paper.

But the second formula ist just the same formula as the first one (by replacing i-> i+1).
The third one looks interesting, but that one seems to be very rarely useful.

[jxu]

There are more identities listed in Ch 5 of Concrete Mathematics. "Factoring in" is called "Absorption"

[jxu]

Also should "Freshman's Dream" be mentioned for binomial coefficients? https://en.wikipedia.org/wiki/Freshman%27s_dream
$$(x+y)^p = x^p + y^p $$
in field Z/pZ[x], that is polynomials with coefficients mod p.

due to divisibility of $p \choose n$