Algorithm Analysis – How long does it take to generate all $ k $ combinations of $ n $ elements $ binom {n} {k} $?

I only solved the following problem for reference (441 – Lotto). It basically requires the generation everyone $ k $Combinations of $ n $ items

void backtrack(std::vector& a,
               int index,
               std::vector& sel,
               int selections) {
    if (selections == 6) { // k is always 6 for 441 - lotto
        //print combination
        return;
    }
    if (index >= a.size()) { return; } // no more elements to choose from
    // two choices
    // (1) select a(index)
    sel(index) = true;
    backtrack(a, index+1, sel, selections+1);

    // (2) don't select a(index)
    sel(index) = false;
    backtrack(a, index+1, sel, selections);

}

I wanted to analyze my own code. I know that I am making a call at the top level (level = 0). On the next level (level = 1) of the recursion I have to trace two calls back. At the next level, I have $ 2 ^ 2 $ Calls. The last level would have been $ 2 ^ n $ Sub-problems. We do for every call $ O (1) $ Work of selecting or not selecting the item. So the total time would be $ 1 + 2 + 2 ^ 2 + 2 ^ 3 + … + 2 ^ n = 2 ^ {n + 1} – 1 = O (2 ^ {n}) $

I've been thinking since we started generating $ binom {n} {k} $ Combinations that there could be a better algorithm with a better runtime since then $ binom {n} {k} = O (n ^ 2) $ or maybe my algorithm is wasteful and there is a better way? or is my analysis really incorrect? Which is it?

Calculate – Find $ lim borders_ {n to + infty} np ^ n Sum borders_ {k = n} ^ { infty} binom {k-1} {n-1} (1-p) ^ {kn} $.

This is most likely a probabilistic problem

$$ E (1 / X) = sum_ {k = n} ^ { infty} frac {1} {k} binom {k-1} {n-1} p ^ n (1-p) ^ {kn}, $$
Where $ X $ is the random variable of Negative binomial distribution. So the current limit is very
$$ lim_ {n to infty} nE (1 / X). $$
What's next? Can it only be solved by calculus? without probability?

Asymptotic $ binom {2n} {n} approximately frac {2 ^ {2n}} { sqrt { pi n}} $ or $ binom {2n} {n} approximately2 ^ {2n} $?

The Wikipedia entry on the binomial coefficient has the following (asymptotic) estimate for $ n $ and $ k $ that grow at the same rate (under "Limits and Asymptotic Formulas")

$$ log binom {n} {k} approximately nH_ {2} left ( frac {k} {n} right) $$

Where $ H_ {2} left ( alpha right) = – alpha log left ( alpha right) – left (1- alpha right) log left (1- alpha right) $ is the binary entropy function

Which would imply
$$ binom {2n} {n} approximately2 ^ {2nH_ {2} left ( frac {1} {2} right)} = 2 ^ {2n} $$

However, the same paragraph contains the following asymptotic estimate of the central binomial coefficient

$$ binom {2n} {n} approximately frac {2 ^ {2n}} { sqrt { pi n}} $$

Aren't these two contradictory? If not, why not and if so, which is correct?

Combinatorics – Prove $ binom {N + K-1} {K} = sum_ {i = 1} ^ {N-1} (-1) ^ {i + 1} binom {N} {i} binom {Ni + K-1} {K} $ using a polynomial

I find the identity, $$ binom {N + K-1} {K} = sum_ {i = 1} ^ {N-1} (-1) ^ {i + 1} binom {N} {i} binom {N -i + K-1} {K} $$

by counting the number of distribution channels $ K $ Sweets too $ N $ People ($ N> K) $,

For the right equation, I count the number of ways to distribute candy, if there is at least $ i $ People who can not get sweets.

Can we prove this identity with polynomials?

For the left equation this corresponds to the coefficient of $ x ^ K $ from $ frac {1} {(1-x) ^ N} $, Can we have the right Eq. To get? by manipulating this polynomial?

Show $ p_ {n + 1} (x) = (x-2n-1) p_n (x) -n ^ 2p_ {n-1} (x) $ for $ p_n (x) = (- 1) ^ nn! sum_ {k = 0} ^ {n} binom {n} {k} frac {(- x) ^ k} {k!} $

To let $ n in mathbb N_0 $ and

$$ p_n (x) = (- 1) ^ nn! sum_ {k = 0} ^ {n} binom {n} {k} frac {(- x) ^ k} {k!} $$

I have to show that $ p_0 (x) = 1, p_1 (x) = x-1 $ and $ p_ {n + 1} (x) = (x-2n-1) p_n (x) -n ^ 2p_ {n-1} (x) $,

$ p_0 (x) = 1 $ and $ p_1 (x) = x-1 $ are easy to show, but I have trouble showing $ p_ {n + 1} (x) = (x-2n-1) p_n (x) -n ^ 2p_ {n-1} (x) $

elementary number theory – Show that $ sum_ {k = 0} ^ {n + 1} left ( binom {n} {k} – binom {n} {k-1} right) ^ 2 = frac {2} {n + 1} binom {2n} {n} $

Show that $$ displaystyle sum_ {k = 0} ^ {n + 1} left ( dbinom {n} {k} – dbinom {n} {k-1} right) ^ 2 = dfrac {2} {n + 1} dbinom {2n} {n} $$ Where $ n in mathbb {N} $

Consider $ dbinom {n} {r} = 0 $ to the $ r <0 $ and $ r> n $, Use $ displaystyle sum_ {k = 0} ^ {n} dbinom {n} {k} ^ 2 = dbinom {2n} {n} $, we get $ displaystyle sum_ {k = 0} ^ {n + 1} left ( dbinom {n} {k} – dbinom {n} {k-1} right) ^ 2 = displaystyle sum_ {k = 0} ^ {n + 1} dbinom {n} {k} ^ 2 + displaystyle sum_ {k = 0} ^ {n + 1} dbinom {n} {k} ^ 2 – 2 displaystyle sum_ {k = 0} ^ {n + 1} dbinom {n} {k} dbinom {n} {k-1} \ hspace {47mm} = 2 dbinom {2n} {n} -2 Display style sum_ {k = 1} ^ {n} dbinom {n} {k} dbinom {n} {k-1} \ hDistance {47mm} = 2 dbinom {2n} {n} -2 Display style sum_ {k = 1} ^ {n} dfrac {1} {k (nk + 1)} dbinom {n} {k-1} ^ 2 \ hspace {47mm} = 2 dbinom {2n } {n} – dfrac {2} {n + 1} displaystyle sum_ {k = 1} ^ {n} left ( dfrac {1} {k} + dfrac {1} {n- k + 1} right) dbinom {n} {k-1} ^ 2 $

Now I'm stuck. How do I proceed with this proof? Is there a better way to solve this problem?

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?