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Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.

Write $[n{]}^{(r)}$ for the rank-$r$ uniform layer of the Boolean cube. For a uniform set family $\mathcal{F}\subseteq [n{]}^{(r)}$, with $1\le r\le n$, let $\partial \mathcal{F}$ be its lower shadow. The Local LYM inequality isFor its proof, count incidences $(B,A)$ with $A\in \mathcal{F}$, $B\subset A$, and $|B|=r-1$. Each member of $\mathcal{F}$ contributes $r$ incidences, while each member of the lower shadow contributes at most $n-r+1$. Thus $r|\mathcal{F}|\le (n-r+1)|\partial \mathcal{F}|$, which is the displayed bound by the ratio of consecutive binomial coefficients. Complementing every set gives the corresponding upper shadow bound, $|\nabla \mathcal{F}|/\left(\genfrac{}{}{0ex}{}{n}{r+1}\right)\ge |\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$ for $r<n$.

For an antichain $\mathcal{A}$ in the Boolean lattice, the LYM inequality isHere is the proof using the Local LYM inequality. Denote the rank-$r$ part by ${\mathcal{A}}_r$, and let ${\mathcal{D}}_r$ consist of the $r$-sets containing some member of $\mathcal{A}$ of size at most $r$. These families satisfyThe union is disjoint because a member of the antichain cannot contain a strictly smaller member. Apply the upper shadow form of the Local LYM inequality to ${\mathcal{D}}_{r-1}$:Iterating gives a bound for the entire sum by $|{\mathcal{D}}_n|/(\genfrac{}{}{0ex}{}{n}{n})\le 1$. This also covers the empty antichain and an antichain containing the empty set.

For the second proof, each permutation of $[n]$ specifies a maximal chain in a Boolean lattice by taking its successive prefixes. There are $n!$ such maximal chains in a Boolean lattice, and a fixed $r$-set belongs to $r!(n-r)!$ of them. A maximal chain in a Boolean lattice meets an antichain at most once. Double counting these incidences therefore gives ${\sum }_r|{\mathcal{A}}_r|r!(n-r)!\le n!$, exactly the LYM inequality.

The Sperner theorem says that an antichain in $\mathcal{P}([n])$ has at most $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$ members; a full middle rank attains this bound. Indeed, every binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ is at most the central one, so the LYM inequality givesNow let $\mathcal{F}$ be an intersection-free uniform set family. If it is empty, there is nothing to prove. Fix $X\in \mathcal{F}$ and consider the tracesIf $A,B\ne X$ are distinct and $X\cap A\subseteq X\cap B$, then $X\cap A\subseteq B$. This contradicts the defining restriction on the three distinct members $X,A,B$. The intersection has size less than $r$, so it is a proper subset of $B$ even if the printed subset symbol is interpreted strictly. In particular, equal traces are impossible, and the traces form an antichain. Thus $|\mathcal{T}|=|\mathcal{F}|-1$. Applying the Sperner theorem to the $r$-element ground set $X$ proves the antichain trace bound for intersection-free familiesThe argument also handles $r=0$: a uniform set family then has at most one member.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

Use the Chebyshev estimate from central binomial coefficients. DefineThe first is the Chebyshev theta function; the second counts prime powers with the same logarithmic prime weight. For a positive integer $m$, every prime $m<p\le 2m$ divides $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$, henceSumming over dyadic intervals yields $\vartheta (2^j)\le 2^{j+1}\log 2$. By monotonicity and rounding $x$ upward to a power of two,

For the lower bound, the central binomial coefficient is the largest of the $2m+1$ coefficients whose sum is $4^m$, soThe exponent of a prime in the central coefficient isEach summand is zero or one. Therefore $\log \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le \psi (2m)$. Higher prime powers contribute onlyCombining the lower bound with $2m$ just below $x$ shows $\vartheta (x)\ge cx$ for sufficiently large $x$, with, for example, $c=(\log 2)/2$.

Let $\pi (x)$ denote the number of primes at most $x$. Since every prime weight is at most $\log x$,For the upper bound, separate the primes at $\sqrt{x}$. The small ones number at most $\sqrt{x}$, and every larger one has weight at least $(\log x)/2$, givingSince $\sqrt{x}=o(x/\log x)$, this provesfor all sufficiently large $n$, for instance with $A=(\log 2)/2$ and $B=8\log 2+1$. This elementary Chebyshev estimate does not assume the Prime number theorem.

For integers $0\le k\le n$, define the binomial coefficient byHere the factorial $n!$ is the product of the positive integers up to $n$. This also counts the $k$-element subsets of an $n$-element set: an ordered choice has $n!/(n-k)!$ possibilities, and each subset is ordered in $k!$ ways.

Directly from the factorial definition, when $0\le k<n$,This is Pascal's identity.

The required hockey-stick identity follows by mathematical induction on $m$, with $n\ge 0$ fixed. For $m=0$, both sides are $1$. If it holds at $m$, adding the next binomial coefficient and using Pascal's identity givesTherefore for every $n,m\ge 0$,

For $1\le j\le p-1$, the identity for binomial coefficientsshows that $p$ divides $j\left(\genfrac{}{}{0ex}{}{p}{j}\right)$. Since $p$ is a prime number and $j$ is not divisible by $p$, cancellation modulo $p$ givesThe binomial theorem consequently gives $(1+x{)}^p\equiv 1+x^p$ as a coefficientwise polynomial modular congruence. Apply the same identity with $x$ replaced by $x^{p^r}$ and use mathematical induction to obtainThis is an iteration of the Frobenius endomorphism in characteristic $p$. Comparing the intermediate coefficients gives the prime-power binomial divisibility result

The Kruskal-Katona theorem. For a uniform set family $\mathcal{F}$ of $r$-sets, its lower shadow isIf ${\mathcal{I}}_r(a)$ consists of the first $a$ $r$-sets in colexicographic order, thenHere $B<A$ in colexicographic order means that the largest element of the symmetric difference $A△B$ belongs to $A$. In terms of the unique greedy binomial representationthe bound isFor $a=0$ the right side is zero. We prove both the minimizing assertion and this formula.

First, the lower shadow of a colexicographic initial segment is a colexicographic initial segment. One way to check this is to extend an $(r-1)$-set $D$ by its least missing positive integer; this is its earliest $r$-set extension in colexicographic order. If $E<D$, the earliest extension of $E$ is no later than the earliest extension of $D$. To see the latter assertion, let $j=\max (D△E)\in D$. If $E$ misses an integer below $j$, its least missing integer is below $j$, so its earliest extension is still earlier than the earliest extension of $D$. Otherwise $E$ contains every integer below $j$. Above $j$ the two sets agree, and their equal sizes force $D$ to contain $j$ and all but one integer below $j$. Adding that missing integer to $D$, or adding $j$ to $E$, gives the same earliest extension. Thus every set earlier than a member of the lower shadow is also in that lower shadow.

We now prove the minimizing assertion by induction on the size $N$ of a finite ground set. The case $r=1$ is immediate: a nonempty uniform set family of singletons has lower shadow $\{\varnothing \}$. The empty case and $r=N$ are also immediate. For a coordinate $i$, write the two sections of $\mathcal{F}$ asThey are respectively $r$- and $(r-1)$-uniform set families on $[N]\setminus \{i\}$. Perform a colexicographic section compression: replace each section by the colexicographic initial segment of the same size. The two sections of the lower shadow before this operation areBy the inductive hypothesis the individual lower shadows do not grow. After colexicographic section compression, both families in the displayed union are colexicographic initial segments, so their union has the larger of their two sizes. This is no greater than the original union. Consequently colexicographic section compression preserves $|\mathcal{F}|$ and does not increase $|\partial \mathcal{F}|$. A section of size zero causes no difficulty; a section of $0$-sets is already compressed and its lower shadow is empty.

Repeat any colexicographic section compression that changes the family. The sum of the positions of its members in colexicographic order strictly decreases, since restriction to either fixed-coordinate section preserves that order. Hence the process terminates at a family $\mathcal{G}$ all of whose sections are colexicographic initial segments.

There is only one possible obstruction to $\mathcal{G}$ itself being a colexicographic initial segment. Suppose $B<A$, $A\in \mathcal{G}$ and $B\notin \mathcal{G}$. No coordinate can belong to both $A,B$, or to neither: either possibility would contradict the compressed section at that coordinate. Thus $A,B$ partition $[N]$, and $N=2r$. Moreover, every such inversion must use this same complementary pair: any omitted set below $A$ equals $A^c=B$, and any included set above $B$ equals $B^c=A$. These observations also rule out an inversion completely before or after this pair. A set strictly between $B$ and $A$ would have to be both present, by comparison with $A$, and absent, by comparison with $B$. Therefore $B,A$ are consecutive in colexicographic order. Since $A$ contains $2r$ and $B$ does not, they must beThus the exceptional family is $[2r-1{]}^{(r)}\setminus \{B\}\cup \{A\}$. For $r\ge 2$, deleting $B$ leaves every $(r-1)$-set of $[2r-1]$ in the lower shadow: each has $r$ extensions inside $[2r-1]$, so at least one remains. Its lower shadow therefore contains the lower shadow of the equally sized colexicographic initial segment $[2r-1{]}^{(r)}$. The case $r=1$ was already handled. This proves the minimizing assertion in all cases.

Finally the greedy binomial representation describes the successive blocks of a colexicographic initial segment. Its first block is $[k_r{]}^{(r)}$, and the remaining sets are obtained by adding $k_r+1$ to an initial segment of $(r-1)$-sets inside $[k_r]$. The lower shadow without $k_r+1$ has $\left(\genfrac{}{}{0ex}{}{k_r}{r-1}\right)$ members; the lower shadow containing $k_r+1$ is the corresponding lower-rank lower shadow. Repeating this decomposition proves the displayed binomial coefficient formula and completes the Kruskal-Katona theorem proof.

The specified initial segment covers every $r$-set of $[m]$. Assume $m\ge r+1$ and $r\ge 1$, the nontrivial range. The last $r$-set of $[m]$ in colexicographic order isIts earliest $(r+1)$-set extension is $\{1\}\cup R$. Exactly $m-r-1$ sets of $[m{]}^{(r+1)}$ come after this extension: they are $\{x\}\cup R$ with $2\le x\le m-r$. Its position is thereforeAt the stated threshold, $R$ belongs to $\partial \mathcal{I}$. Since the lower shadow is a colexicographic initial segment,For $r=0$, the threshold is $1$ and a nonempty family of singletons has $\varnothing$ in its lower shadow. If $m=r\ge 1$, the threshold is again $1$; the first $(r+1)$-set is $[r+1]$, which contains $[r{]}^{(r)}$. If $m<r$, the desired inclusion is empty and automatic.

The nonuniform bound. Suppose first that $1\le r\le m$ and, contrapositively, that $|\mathcal{B}|\le \left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$. Let $\mathcal{J}$ be the colexicographic initial segment of $r$-sets of size $\left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$; this consists of $[m{]}^{(r)}$ with its last member $R$ removed. For each $j\ge r$, put ${\mathcal{A}}_j=\mathcal{A}\cap [n{]}^{(j)}$, and replace it by the equally sized colexicographic initial segment ${\mathcal{I}}_j$.

Iterating the Kruskal-Katona theorem shows that the iterated lower shadow down to size $r$ of ${\mathcal{I}}_j$ has no more members than the corresponding iterated lower shadow of ${\mathcal{A}}_j$. Indeed, at each step initial segments minimize the next lower shadow, and its minimum size is increasing in the input size. The latter iterated lower shadow lies in $\mathcal{B}$. Therefore every $r$-subset of every member of ${\mathcal{I}}_j$ lies in $\mathcal{J}$.

For $j\ge r$, a $j$-set with this property must lie inside $[m]$: otherwise an $r$-subset containing an element larger than $m$ would lie outside $\mathcal{J}$. Among $j$-sets inside $[m]$, exactly those containing $R$ are forbidden. HenceAdding these bounds proves the contrapositive. Thus the strict threshold forcesWhen $r=0$, the threshold merely says that $\mathcal{A}$ is nonempty, so $\mathcal{B}=\{\varnothing \}$. When $m<r$, the claimed lower bound is zero. The strict inequality is sharp: the family of all subsets of $[m]$ of size at least $r$ that do not contain $R$ attains the displayed sum and has exactly $\left(\genfrac{}{}{0ex}{}{m}{r}\right)-1$ members in its size-$r$ iterated lower shadow.

We use the standard convention that a two-intersecting family satisfies $|A\cap B|\ge 2$ for every pair of members, including $A=B$. In particular every member has at least two elements.

An intersecting shadow lemma. If $\mathcal{F}$ is an intersecting family of $r$-sets, $r\ge 1$, thenHere is a proof, so no additional shadow theorem is being assumed. For $i<j$, an elementary set shift replaces a member $A$ containing $j$ but not $i$ by $(A\setminus \{j\})\cup \{i\}$ exactly when that replacement is not already in the family. This preserves cardinality and the intersecting family property. The only potentially troublesome pair consists of a moved member and an unmoved member $B$ containing $j$ but not $i$. But then $B$'s replacement was already present, and the two original members obtained by using that replacement would have been disjoint if the new pair were disjoint.

Alsoso an elementary set shift cannot increase the lower shadow. For completeness, a shadow set containing neither $i,j$ or both has an unchanged witness in the original lower shadow. A shadow set containing $i$ but not $j$ is either originally present or the replacement of an originally present shadow set. If a shadow set contains $j$ but not $i$, its witnessing member survived the elementary set shift; either that member contains both coordinates, or its replacement was already present. In either case both the shadow set and its replacement were originally present, so the former survives in the shifted lower shadow.

Repeated elementary set shifts terminate because the sum of the elements of all members decreases whenever the family changes. In the resulting shifted set family $\mathcal{H}$, if $a_1<\cdots <a_r$ is a member and $b_1<\cdots <b_r$ satisfies $b_i\le a_i$ for every $i$, then $\{b_1,\dots ,b_r\}$ is also a member. Every $A\in \mathcal{H}$ has a prefix $[2q-1]$ containing $q$ of its elements. Otherwise its sorted elements satisfy $a_i\ge 2i$, and the first $r$ elements outside $A$ form a disjoint member componentwise no larger than $A$, contradicting the intersecting family property.

Choose the first such prefix and reflect membership inside it:This has size $r-1$. The set $D\cup \{2q-1\}$ is componentwise no larger than $A$: before the first excess, the prefix counts of $A$ are at most half the prefix length, and reflection only moves selected positions to the left. Thus $D$ belongs to $\partial \mathcal{H}$. This ballot reflection injection is injective: in $D$, the reflected prefix ends at the first position where unselected positions outnumber selected ones, and reflecting it again recovers $A$. Consequently $|\partial \mathcal{H}|\ge |\mathcal{H}|=|\mathcal{F}|$. Since elementary set shifts did not increase the lower shadow, this proves the intersecting shadow lemma.

The nonuniform two-intersecting family bound. Write ${\mathcal{A}}_i=\mathcal{A}\cap [n{]}^{(i)}$. No member of $\partial {\mathcal{A}}_i$ can be the complement of a member of ${\mathcal{A}}_{n-i+1}$: such an inclusion would make the intersection of the two original members have at most one element. The intersecting shadow lemma therefore givesFor $n=2h$, sum this over $1\le i\le h$. These pairs cover all nonempty levels, soEquality is attained by all subsets of size at least $h+1$: any two intersect in at least two elements.

For $n=2h+1$, the same pairs cover every level except $h+1$. Complements of the members of ${\mathcal{A}}_{h+1}$ form an intersecting family of $h$-sets, sinceFor $h\ge 1$, the permitted Erdős-Ko-Rado theorem bounds this middle level by $\left(\genfrac{}{}{0ex}{}{2h}{h-1}\right)=\left(\genfrac{}{}{0ex}{}{2h}{h+1}\right)$. HenceEquality is attained by all $(h+1)$-sets inside $[2h]$, together with all subsets of $[2h+1]$ of size at least $h+2$. Two members of the first group intersect in at least two elements; a member from each group intersects in at least two; two members of the second group intersect in at least three. For $n=1$ the only two-intersecting family is empty, which also gives equality. The empty-ground-set case is likewise immediate.

The pair-cover bound for a two-intersecting uniform family. If $\mathcal{A}$ is nonempty, fix $A_0\in \mathcal{A}$, with $|A_0|=r$. Every member contains one of the $\left(\genfrac{}{}{0ex}{}{r}{2}\right)$ pairs inside $A_0$. Each fixed pair belongs to exactly $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets, so counting their union givesThe empty family is immediate; a nonempty two-intersecting family requires $r\ge 2$.

Witness-pair concentration for an intersecting uniform family. Let $\mathcal{A}$ now be an intersecting family of $r$-sets. If it is two-intersecting, the preceding bound, together with $\left(\genfrac{}{}{0ex}{}{r}{2}\right)\le (r-1{)}^2$ for $r\ge 2$, shows that any chosen $x\in [n]$ has the required property. Otherwise there are members $A,B$ with $A\cap B=\{x\}$. Every member $C$ not containing $x$ must contain some $a\in A\setminus \{x\}$ and some $b\in B\setminus \{x\}$. These two sets of possible witnesses are disjoint. There are $(r-1{)}^2$ witness pairs, and each is contained in $\left(\genfrac{}{}{0ex}{}{n-2}{r-2}\right)$ $r$-sets. ThusFor $r=1$, a nonempty intersecting family consists of one singleton; choose its element and the number avoiding it is zero. This is the natural separate interpretation of the degenerate case, where the printed binomial coefficient has lower index $-1$. For an empty family any $x\in [n]$ works, provided $n\ge 1$ as usual.

Write $p=p(n)$ and let $T$ be the triangle count in a binomial random graph. For every three-element set $S$ of vertices, let $I_S$ be the indicator random variable that its three edges are present. ThenFor two distinct triples $S,R$, their indicator random variables are independent random variables unless they share an edge. Sharing just one vertex still leaves the sets of relevant edges disjoint. If they share an edge, their union has five edges, and thereforeThere are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)$ unordered pairs of distinct triangles in a graph sharing an edge: choose that edge, and then choose their two additional vertices. The variance of a sum consequently gives the exact formulaFor sufficiently large $n$, the hypothesis $np\to \infty$ ensures $p>0$, so $\mu >0$. Dropping negative terms and simplifying the binomial coefficients,Here $n^{2}p=n(np)\to \infty$ as well. The event $T\ge 2\mu$ implies $|T-\mu |\ge \mu$. Applying the Chebyshev inequality from part (i) gives the requested conclusion:Thus the triangle count is concentrated around its expectation throughout this range of the binomial random graph.

First establish the binomial theorem needed throughout this question. For integers $0\le m\le n$, define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ to be the number of $m$-element subsets of an $n$-element set. Counting ordered selections and then forgetting their order giveswith the factorial convention $0!=1$. In the product of $n$ factors $(1+z)$, choose $z$ from exactly $m$ factors and $1$ from the others. Each choice contributes $z^m$, and there are $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ choices. The distributive law therefore provesfor every complex number $z$, including $n=0$ by the empty-product convention.

To evaluate the even-indexed sum, use the even part of a polynomial: averaging the binomial theorem at $z$ and $-z$ cancels the odd powers. For $z=i\sqrt{3}$,The two complex numbers satisfy $1\pm i\sqrt{3}=2e^{\pm i\pi /3}$. Their $6n$th powers are both $2^{6n}$, since $e^{\pm 2\pi in}=1$. Thus the sum is

Think of a vertex of this hypergraph as an edge of the complete graph on $[N]$. A hyperedge is the edge set of one $t$-vertex clique. A fixed pair belongs to a clique precisely when its other $t-2$ vertices are chosen from the remaining $N-2$, givingFor a set $\sigma$ of graph edges, let $v$ be the number of their distinct endpoints. If $v>t$, no $t$-clique contains them and the codegree is zero. Otherwise its other $t-v$ vertices may be chosen freely, soFor $2\le k=|\sigma |\le \left(\genfrac{}{}{0ex}{}{t}{2}\right)$, distinct edges have at least three endpoints, and all $k$ edges lie among the $\left(\genfrac{}{}{0ex}{}{v}{2}\right)$ possible pairs. Thus $3\le v\le t$ and $k\le \left(\genfrac{}{}{0ex}{}{v}{2}\right)$. In particular,This also identifies why the relevant exponent is $2/(t+1)$.

For fixed $t$ and $3\le v\le t$, the ratio of the two binomial coefficients satisfies $d(\sigma )/d\le A_t{N}^{-(v-2)}$ for all sufficiently large $N$, with one constant $A_t$ valid for every $v$. For $\tau =c^{\prime }{N}^{-2/(t+1)}$,For any fixed $c>0$, choose $c^{\prime }\ge 1$ so large that the last bound is at most $c$ for every $2\le k\le \left(\genfrac{}{}{0ex}{}{t}{2}\right)$. Zero codegrees satisfy the bound automatically. The asserted small-codegree condition therefore holds uniformly. Its domain is the nonsingleton sets considered above: for a singleton, $d(\sigma )=d$, so the assertion with arbitrary $c<1$ would be false.

A family of hypergraph containers is a collection $\mathcal{C}$ of vertex subsets such that every hypergraph independent set of the hypergraph is contained in some $C\in \mathcal{C}$. Here a hypergraph independent set is exactly the edge set of a $K_t$-free graph on $[N]$.

We use the following edge-sparse form of the hypergraph container theorem. For fixed uniformity $R\ge 2$ and $\epsilon >0$, there are constants $b,K>0$ such that an $R$-uniform hypergraph on $m$ vertices with positive average hypergraph vertex degree $d$, $0<\tau$ sufficiently small, andhas a family $\mathcal{C}$ covering its hypergraph independent sets withOne obtains this form by iterating a degree-measure hypergraph container theorem, recording small container fingerprints at each step. To ensure the requested strict edge inequality, apply it with $\epsilon /2$.

For our clique hypergraph, $m=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$, $R=\left(\genfrac{}{}{0ex}{}{t}{2}\right)$ and $e(G)=\left(\genfrac{}{}{0ex}{}{N}{t}\right)$. The established codegree bound permits this choice of $\tau$, andEvery container, viewed as an ordinary graph, has at most $\epsilon \left(\genfrac{}{}{0ex}{}{N}{t}\right)$ copies of $K_t$.

We also use clique supersaturation: for each fixed $\gamma >0$, graphs with at least $(1-1/(t-1)+\gamma )\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ edges have at least $b_{\gamma }{N}^t$ copies of $K_t$, for some $b_{\gamma }>0$ and all sufficiently large $N$. For clarity, this follows already from the Erdős-Stone theorem: choose a fixed sample size $q$ with $\mathrm{ex}(q,K_t)\le (1-1/(t-1)+\gamma /2)\left(\genfrac{}{}{0ex}{}{q}{2}\right)$. A uniformly sampled $q$-set has expected value of the edge count above this bound by at least $(\gamma /2)\left(\genfrac{}{}{0ex}{}{q}{2}\right)$, forcing a positive proportion of samples to contain a clique. Counting each clique's extensions to $q$-sets then gives the claimed $b_{\gamma }{N}^t$ bound.

Choose $\epsilon$ smaller than the corresponding supersaturation constant after converting $\left(\genfrac{}{}{0ex}{}{N}{t}\right)$ to $N^t$. Each container has fewer than $(1-1/(t-1)+\gamma )\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ ordinary edges. Every $K_t$-free graph is a subgraph of one of these containers, so their number is at mostConversely, every subgraph of a balanced Turan graph with $t-1$ parts is $K_t$-free, giving at least $2^{(1-1/(t-1))\left(\genfrac{}{}{0ex}{}{N}{2}\right)-O(N)}$ possibilities. Letting $\gamma$ be arbitrarily small proves the enumeration formula:

Only routes with $n_r>0$ contribute to proportional fairness or to the weighted logarithmic utility; omit inactive routes from expressions with a rate in the denominator. Active routes have positive rates whenever a positive feasible allocation exists. For a feasible competitor $y$, the elementary logarithm inequality $\log u\le u-1$ givesA zero competitor rate on an active route has utility $-\infty$ and also cannot improve the objective. Therefore a proportionally fair allocation maximizes the logarithmic objective. Conversely, the feasible set is a convex set, and differentiating the objective along $x+t(y-x)$ at $t=0+$ shows that an optimizer satisfies the proportional fairness inequality. Strict concavity gives uniqueness on active coordinates; rates assigned to nonexistent flows are immaterial.

In the linear flow network, put $z_0=n_0{x}_0$ and $z_i=n_i{x}_i$ for active routes. A local route $i$ uses only resource $i$. If $n_i>0$ and $z_0+z_i<1$, increasing $x_i$ increases the objective without violating any other constraint. Thus each active local route saturates its resource:Write $L={\sum }_{i=1}^I{n}_i$ and $N=n_0+L$. For $n_0>0$ and $L>0$, substitute $z_i=1-z_0$ into the objective. Apart from constants its dependence on $z_0$ isIts derivative vanishes when $n_0/z_0=L/(1-z_0)$, and its second derivative is negative. Hence the per-flow through-route rate isIf $L=0<n_0$, the optimal through-route aggregate rate is one, so $x_0=1/n_0=1/N$ still holds. If $n_0=0<L$, each active local route has $x_i=1/n_i$, also given by the displayed local formula. When $N=0$ there are no flows and no departures. Inactive coordinates need not be assigned these undefined per-flow formulas.