## harmonic numbers – generation function of the series $sum limits_ {n = 1} ^ infty H_n ^ {(m)} binom {2n} {n} x ^ n$?

I want to find the generation function of infinite series $$sum limits_ {n = 1} ^ infty H_n ^ {(m)} binom {2n} {n} x ^ n = ?,$$ from where $$H_n ^ {(m)}$$ is a generalized harmonic number, defined by $$H_n ^ {(m)}: = sum limits_ {k = 1} ^ n frac {1} {k ^ m}$$
and $$H_n = H_n ^ {(1)}$$ is the classical harmonic number. It is known that for $$m = 1$$. $$sum limits_ {n = 1} ^ infty H_n binom {2n} {n} x ^ n = frac {2} { sqrt {1-4x}} log left ( frac {1 + sqrt {1-4x}} {2 sqrt {1-4x}} right),$$
See https://cs.uwaterloo.ca/journals/JIS/VOL19/Chen/chen21.pdf. But for every positive integer $$m$$What is the generating function?

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## nt.number theory – Does the sentence ${ binom x3 + binom y3 + binom z3: x, y, z in mathbb Z }$ contain all integers?

Gauss-Legendre's theorem on sums of three squares states that
$${x ^ 2 + y ^ 2 + z ^ 2: x, y, z in mathbb Z } = mathbb N setminus {4 ^ k (8m + 7): k, m in mathbb N },$$ from where $$mathbb N = {0,1,2, ldots }$$,

It's easy to see that set $${x ^ 3 + y ^ 3 + z ^ 3: x, y, z in mathbb Z }$$ does not contain an integer that is congruent $$4$$ or $$-4$$ modulo $$9$$, In 1992, Heath-Brown suspected that every integer $$m not equiv pm4 pmod9$$ can be written as $$x ^ 3 + y ^ 3 + z ^ 3$$ With $$x, y, z in mathbb Z$$, Recently, A.R. Booker arXiv: 1903.04284 integers $$x, y, z$$ With $$x ^ 3 + y ^ 3 + z ^ 3 = 33$$,

It is well known that
$$left { binom x2 + binom y2 + binom z2: x, y, z in mathbb Z right } = mathbb N,$$ which was claimed by Fermat and proved by Gauss.

Here I ask a similar question.

question: Has the set $${ binom x3 + binom y3 + binom z3: x, y, z in mathbb Z }$$ contain all integers?

Clear,
$$binom {-x} 3 = – binom {x + 2} 3.$$ About Mathematica I found that the only integers under $$0, ldots, 2000$$ not in the set
$$left { binom x3 + binom y3 + binom z3: x, y, z in {- 600, ldots, 600 } right }$$
are
$$522, , 523, , 622, , 633, 642, 843, 863, 918, , 1013, , 1458, 1523, , 1878.$$
For example,
$$183 = binom {549} 3+ binom {-525} 3+ binom {-266} 3$$
and
$$423 = binom {426} 3+ binom {-416} 3+ binom {-161} 3.$$

In my opinion, the question has a positive answer. Your comments are welcome!

## Why is $binom {n} {f} ^ g = O (n ^ {fg})$ true?

Why is that right? I understand why $$n ^ g$$ but how does it work? $$f$$ come to power ??
I believe from the context that it's not just that $$binom {n} {f} ^ g$$ is strictly smaller than $$n ^ {f g}$$but rather that it really belongs to this class, which means that it is also heavily constrained by this class or very narrow. But why is it?

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## Combinatorics – writing $binom {m} {n} binom {m-a} {n-b}$ as a binomial coefficient

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## Proving $binom {n + m} {r} = sum_ {i = 0} ^ {r} binom {n} {i} binom {m} {r – i}$

To prove $$binom {n + m} {r} = sum_ {i = 0} ^ {r} binom {n} {i} binom {m} {r – i},$$

I have shown that equality applies to all $$n,$$ $$m = 0, 1,$$ and all $$r simply fix it $$n$$ and $$r$$ and insert $$0.1$$ to the $$m.$$ Then I go on $$m$$ (and further $$m$$ just).

But I'm not completely confident because I see two placeholders, $$n$$ and $$m.$$ Is this the case when a double induction is required (first)? $$m$$ and then on $$n$$)

Consider all fixated $$n, r geq 0$$ and the following two cases (I know that only one case is needed to complete this inductive proof).

CASE 1

begin {align} binom {n + 0} {r} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {0} {r – i} \ & = binom {n} {0} binom {0} {r} + binom {n} {1} binom {0} {r-1} + cdots + binom {n} {r} binom {0} {0} \ & = 0 + 0 + cdots + binom {n} {r} \ & = binom {n} {r} end

CASE 2

begin {align} binom {n + 1} {r} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {1} {r – i} \ & = binom {n} {0} binom {0} {r} + binom {n} {1} binom {0} {r-1} + cdots + binom {n} {r-1} binom {1} { r – (r-1)} + binom {n} {r} binom {1} {r – r} \ & = 0 + 0 + cdots + binom {n} {r-1} + binom {n} {r} \ & = binom {n} {r-1} + binom {n} {r} end

INDUCTION

Suppose it is true $$m leq k.$$ Now think $$binom {n + (k + 1)} {r}.$$ Pascal's identity follows

$$binom {n + (k + 1)} {r} = binom {n + k} {r} + binom {n + k} {r-1}$$

And,

begin {align} binom {n + k} {r} + binom {n + k} {r-1} & = sum_ {i = 0} ^ {r} binom {n} {i} binom {k} { r – i} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k} {r – 1 – i} \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k} {r – i} + sum_ {i = 0} ^ {r-1} binom {n } {i} binom {k} {r – 1 – i} \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} bigg[binom{k}{r – i} + binom{k}{r – 1 – i}bigg] \ & = binom {n} {r} + sum_ {i = 0} ^ {r-1} binom {n} {i} binom {k + 1} {ri} \ & = sum_ {i = 0} ^ {r} binom {n} {i} binom {k + 1} {ri} end

Therefore, equality applies to $$m = k + 1$$ Since equality is for $$m = 0, 1,$$ and that if equality holds $$m = k,$$ then it applies to $$m = k + 1,$$ It follows that equality holds $$forall m in mathbb {N}.$$