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?

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 <n + m $ 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}. $