Math-5998 · mathematics · Mathematics

Math-5998

Erdős Problem #927

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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolved

Statement

Statement and result.
Let $g(N)$ be the maximum number of different orders of maximal cliques in a graph on $N$ vertices. Is $g(N)=N-\log_2N-\log_*N+O(1)$? Here "clique" must mean maximal complete subgraph, as in the original papers; counting arbitrary complete subgraphs would instead give $g(N)=N$. Spencer disproved the proposed formula by proving \[ g(N)=N-\log_2N+O(1). \] We reconstruct his construction with a flexible final block, obtaining the slightly unoptimized bounds \[ N-\log_2N-5<g(N)\leq N-\lfloor\log_2N\rfloor \quad\text{for all sufficiently large }N. \tag{1} \] All logarithms below have base two.
Upper bound.
In any graph realizing $g=g(N)$ distinct maximal-clique orders, a largest clique $C$ has at least $g$ vertices. A maximal clique $K$ is determined by $S=K\setminus C$: its intersection with $C$ consists of every vertex of $C$ adjacent to all of $S$. Indeed, any omitted such vertex could be added to $K$. Thus there are at most $2^{N-|C|}\leq2^{N-g}$ maximal cliques, so $g\leq2^{N-g}$. Put $l=\lfloor\log N\rfloor$. For $N\geq2$, if $g\geq N-l+1$, then \[ g\geq2^l-l+1>2^{l-1}\geq2^{N-g}, \] a contradiction. The strict middle inequality holds for every integer $l\geq1$. This proves the upper bound.
Small auxiliary parameters.
Fix an integer $r\geq32$. Put $r_0=r$, and recursively let $r_i\geq2$ be the least integer for which \[ 2^{r_i}+r_i-2\geq r_{i-1}, \] stopping at $r_s=2$. Set \[ A_0(r)=1+\sum_{i=1}^s(2^{r_i}+r_i-1),\qquad l=\lfloor r/2\rfloor. \] The elementary bounds needed below are \[ 1+\sum r_i\leq r-l,\qquad A_0(r)\leq4r, \qquad 5r\leq2^l. \tag{2} \] Here are details to avoid relying on asymptotic notation for these choices. If $m=\lceil\log r\rceil$, then $r_1\leq m$, $s\leq m-1$, and $\sum r_i\leq2m$. The last assertion follows inductively from the fact that the next term after an integer $x\geq6$ is at most $\lfloor x/2\rfloor$, with $x=2,3,4,5$ checked directly. Minimality of $r_i$ gives $2^{r_i}+r_i-1<2r_{i-1}-r_i+5$, whence \[ A_0(r)<2r+\sum r_i+5s-3\leq2r+7m-8\leq4r. \] Finally $2m+1\leq r-l$ and $5r\leq2^l$ for $r\geq32$, by their values at the first dyadic intervals and monotonicity on subsequent intervals. These prove $2$.
Graph construction.
Choose integers $A,w$ satisfying \[ A_0(r)\leq A\leq5r,\qquad 1\leq w\leq2^{r-1}. \] Let $d_i=2^{i-1}$ for $1\leq i<r$, and $d_r=w$, and put $T=\sum d_i$. Take disjoint core blocks $C_i$ of size $d_i+1$, another core block $C_*$ of size $A$, selectors $y_1,\ldots,y_r,y_*$, and a vertex $z$. Make all core vertices mutually adjacent and all selectors mutually adjacent. A vertex of $C_i$ is adjacent to every selector except $y_i$. No vertex of $C_*$ is adjacent to $y_*$.

By $2$, among $y_{l+1},\ldots,y_r$ we can label distinct selectors $w_{ij}$, $1\leq i\leq s,1\leq j\leq r_i$, and $w_*$. Partition $A_0(r)$ vertices of $C_*$ into blocks $V_{ij}$ of size $2^{j-1}+1$ and a singleton $\{v_*\}$. Call these the original vertices of $C_*$; the other $A-A_0(r)$ vertices are extra vertices.

Every selector other than these labelled $w$'s and $y_*$ is adjacent to all of $C_*$. Among labelled selectors and original vertices, add exactly the edges from $w_{ij}$ to $V_{ij'}$ with $j'\ne j$, and the edge $w_*v_*$. Every extra vertex is adjacent to all $y_i$, including the labelled selectors. Finally, make $z$ adjacent exactly to the labelled selectors and the original vertices of $C_*$. This explicitly includes the triangle $z,w_*,v_*$. No other edges are added.

There are \[ B=T+r+A\quad\text{core vertices},\qquad N=T+2r+A+2\quad\text{vertices in total}. \tag{3} \] We next exhibit maximal cliques of every order from 3 through $B$.

Middle and large orders.
For any subset $S\subseteq\{1,\ldots,r\}$, take \[ \{y_*\}\cup\{y_i:i\in S\}\cup\bigcup_{i\notin S}C_i. \] This is maximal: the opposite choice in every pair $(y_i,C_i)$ is blocked, and $y_*$ blocks both $C_*$ and $z$. Its order is $T+r+1-\sum_{i\in S}d_i$. The first $r-1$ weights represent all integers from 0 to $2^{r-1}-1$; adjoining $w\leq2^{r-1}$ gives all subset sums from 0 to $T$. Thus these cliques have every order in $[r+1,B-A+1]$.

For $S\subseteq\{1,\ldots,l\}$, instead take \[ C_*\cup\{y_i:i\in S\}\cup\bigcup_{i\notin S}C_i. \] All selected selectors are unlabelled and so adjacent to $C_*$. The opposite choices are again blocked, $y_*$ is blocked by $C_*$, and $z$ is blocked by either $y_1$ or $C_1$. These maximal cliques have orders $B-\sum_{i\in S}2^{i-1}$. Since $A\leq2^l$, they include every order in $[B-A+1,B]$.

Small orders.
For a fixed $i$, choose a nonempty proper subset $S\subset\{1,\ldots,r_i\}$ and take \[ \{z\}\cup\{w_{ij}:j\in S\}\cup\bigcup_{j\notin S}V_{ij}. \] It contains both a labelled selector and an original core vertex from group $i$, so any further neighbor of $z$ outside this group is blocked. Within the group every unchosen selector is blocked by its own $V$-block and every omitted $V$-block by its own selector. It is therefore maximal. Its order is \[ 2^{r_i}+r_i-\sum_{j\in S}2^{j-1}. \] Consequently every order in $[r_i+2,2^{r_i}+r_i-1]$ occurs. The recursion defining $r_i$ makes these intervals overlap or meet consecutively, reaching from 4 through at least $r+1$. The explicitly specified triangle $z,w_*,v_*$ is maximal and supplies order 3. We have proved \[ g(N)\geq B-2=N-r-4. \tag{4} \]
Fitting every sufficiently large number of vertices.
Given $N$, let $q=\lceil\log N\rceil$, with $q\geq33$. If \[ N\geq2^{q-1}+2q+A_0(q)+2, \] take $r=q$, $A=A_0(r)$, and $w=N-2^{r-1}-2r-A-1$. These choices satisfy $1\leq w\leq2^{r-1}$, since $N\leq2^q$, and $3$ holds.

Otherwise take $r=q-1$. In this case $N>2^r$, and $w_0=N-2^{r-1}-2r-A_0(r)-1$ is positive. If $w_0\leq2^{r-1}$, take $w=w_0,A=A_0(r)$. If $w_0>2^{r-1}$, take \[ w=2^{r-1},\qquad A=N-2^r-2r-1. \] Then $A>A_0(r)$, while the defining upper bound on $N$ gives $A<A_0(r+1)+3\leq4r+7\leq5r$. Thus every case meets the construction's requirements and has exactly $N$ vertices. Since $r\leq\lceil\log N\rceil<\log N+1$, $4$ proves the lower bound in $1$.

Scope and attribution.
This is an exposition adapting J. H. Spencer, On cliques in graphs, Israel Journal of Mathematics 9 $1971$, 419--421, https://doi.org/10.1007/BF02771457. The original paper proves a slightly sharper constant. The unbounded $\log_*N$ term in the proposed asymptotic is therefore false. Source statement: https://www.erdosproblems.com/927, checked 24 September 2026. The proof above includes the construction and its interpolation to all sufficiently large $N$; no new result or local Lean verification is claimed.

Restatement from Math-5998-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-5998-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5998

Erdős Problem 927

Statement and result.
Let $g(N)$ be the maximum number of different orders of maximal cliques in a graph on $N$ vertices. Is $g(N)=N-\log_2N-\log_*N+O(1)$? Here "clique" must mean maximal complete subgraph, as in the original papers; counting arbitrary complete subgraphs would instead give $g(N)=N$. Spencer disproved the proposed formula by proving \[ g(N)=N-\log_2N+O(1). \] We reconstruct his construction with a flexible final block, obtaining the slightly unoptimized bounds \[ N-\log_2N-5<g(N)\leq N-\lfloor\log_2N\rfloor \quad\text{for all sufficiently large }N. \tag{1} \] All logarithms below have base two.
Upper bound.
In any graph realizing $g=g(N)$ distinct maximal-clique orders, a largest clique $C$ has at least $g$ vertices. A maximal clique $K$ is determined by $S=K\setminus C$: its intersection with $C$ consists of every vertex of $C$ adjacent to all of $S$. Indeed, any omitted such vertex could be added to $K$. Thus there are at most $2^{N-|C|}\leq2^{N-g}$ maximal cliques, so $g\leq2^{N-g}$. Put $l=\lfloor\log N\rfloor$. For $N\geq2$, if $g\geq N-l+1$, then \[ g\geq2^l-l+1>2^{l-1}\geq2^{N-g}, \] a contradiction. The strict middle inequality holds for every integer $l\geq1$. This proves the upper bound.
Small auxiliary parameters.
Fix an integer $r\geq32$. Put $r_0=r$, and recursively let $r_i\geq2$ be the least integer for which \[ 2^{r_i}+r_i-2\geq r_{i-1}, \] stopping at $r_s=2$. Set \[ A_0(r)=1+\sum_{i=1}^s(2^{r_i}+r_i-1),\qquad l=\lfloor r/2\rfloor. \] The elementary bounds needed below are \[ 1+\sum r_i\leq r-l,\qquad A_0(r)\leq4r, \qquad 5r\leq2^l. \tag{2} \] Here are details to avoid relying on asymptotic notation for these choices. If $m=\lceil\log r\rceil$, then $r_1\leq m$, $s\leq m-1$, and $\sum r_i\leq2m$. The last assertion follows inductively from the fact that the next term after an integer $x\geq6$ is at most $\lfloor x/2\rfloor$, with $x=2,3,4,5$ checked directly. Minimality of $r_i$ gives $2^{r_i}+r_i-1<2r_{i-1}-r_i+5$, whence \[ A_0(r)<2r+\sum r_i+5s-3\leq2r+7m-8\leq4r. \] Finally $2m+1\leq r-l$ and $5r\leq2^l$ for $r\geq32$, by their values at the first dyadic intervals and monotonicity on subsequent intervals. These prove $2$.
Graph construction.
Choose integers $A,w$ satisfying \[ A_0(r)\leq A\leq5r,\qquad 1\leq w\leq2^{r-1}. \] Let $d_i=2^{i-1}$ for $1\leq i<r$, and $d_r=w$, and put $T=\sum d_i$. Take disjoint core blocks $C_i$ of size $d_i+1$, another core block $C_*$ of size $A$, selectors $y_1,\ldots,y_r,y_*$, and a vertex $z$. Make all core vertices mutually adjacent and all selectors mutually adjacent. A vertex of $C_i$ is adjacent to every selector except $y_i$. No vertex of $C_*$ is adjacent to $y_*$.

By $2$, among $y_{l+1},\ldots,y_r$ we can label distinct selectors $w_{ij}$, $1\leq i\leq s,1\leq j\leq r_i$, and $w_*$. Partition $A_0(r)$ vertices of $C_*$ into blocks $V_{ij}$ of size $2^{j-1}+1$ and a singleton $\{v_*\}$. Call these the original vertices of $C_*$; the other $A-A_0(r)$ vertices are extra vertices.

Every selector other than these labelled $w$'s and $y_*$ is adjacent to all of $C_*$. Among labelled selectors and original vertices, add exactly the edges from $w_{ij}$ to $V_{ij'}$ with $j'\ne j$, and the edge $w_*v_*$. Every extra vertex is adjacent to all $y_i$, including the labelled selectors. Finally, make $z$ adjacent exactly to the labelled selectors and the original vertices of $C_*$. This explicitly includes the triangle $z,w_*,v_*$. No other edges are added.

There are \[ B=T+r+A\quad\text{core vertices},\qquad N=T+2r+A+2\quad\text{vertices in total}. \tag{3} \] We next exhibit maximal cliques of every order from 3 through $B$.

Middle and large orders.
For any subset $S\subseteq\{1,\ldots,r\}$, take \[ \{y_*\}\cup\{y_i:i\in S\}\cup\bigcup_{i\notin S}C_i. \] This is maximal: the opposite choice in every pair $(y_i,C_i)$ is blocked, and $y_*$ blocks both $C_*$ and $z$. Its order is $T+r+1-\sum_{i\in S}d_i$. The first $r-1$ weights represent all integers from 0 to $2^{r-1}-1$; adjoining $w\leq2^{r-1}$ gives all subset sums from 0 to $T$. Thus these cliques have every order in $[r+1,B-A+1]$.

For $S\subseteq\{1,\ldots,l\}$, instead take \[ C_*\cup\{y_i:i\in S\}\cup\bigcup_{i\notin S}C_i. \] All selected selectors are unlabelled and so adjacent to $C_*$. The opposite choices are again blocked, $y_*$ is blocked by $C_*$, and $z$ is blocked by either $y_1$ or $C_1$. These maximal cliques have orders $B-\sum_{i\in S}2^{i-1}$. Since $A\leq2^l$, they include every order in $[B-A+1,B]$.

Small orders.
For a fixed $i$, choose a nonempty proper subset $S\subset\{1,\ldots,r_i\}$ and take \[ \{z\}\cup\{w_{ij}:j\in S\}\cup\bigcup_{j\notin S}V_{ij}. \] It contains both a labelled selector and an original core vertex from group $i$, so any further neighbor of $z$ outside this group is blocked. Within the group every unchosen selector is blocked by its own $V$-block and every omitted $V$-block by its own selector. It is therefore maximal. Its order is \[ 2^{r_i}+r_i-\sum_{j\in S}2^{j-1}. \] Consequently every order in $[r_i+2,2^{r_i}+r_i-1]$ occurs. The recursion defining $r_i$ makes these intervals overlap or meet consecutively, reaching from 4 through at least $r+1$. The explicitly specified triangle $z,w_*,v_*$ is maximal and supplies order 3. We have proved \[ g(N)\geq B-2=N-r-4. \tag{4} \]
Fitting every sufficiently large number of vertices.
Given $N$, let $q=\lceil\log N\rceil$, with $q\geq33$. If \[ N\geq2^{q-1}+2q+A_0(q)+2, \] take $r=q$, $A=A_0(r)$, and $w=N-2^{r-1}-2r-A-1$. These choices satisfy $1\leq w\leq2^{r-1}$, since $N\leq2^q$, and $3$ holds.

Otherwise take $r=q-1$. In this case $N>2^r$, and $w_0=N-2^{r-1}-2r-A_0(r)-1$ is positive. If $w_0\leq2^{r-1}$, take $w=w_0,A=A_0(r)$. If $w_0>2^{r-1}$, take \[ w=2^{r-1},\qquad A=N-2^r-2r-1. \] Then $A>A_0(r)$, while the defining upper bound on $N$ gives $A<A_0(r+1)+3\leq4r+7\leq5r$. Thus every case meets the construction's requirements and has exactly $N$ vertices. Since $r\leq\lceil\log N\rceil<\log N+1$, $4$ proves the lower bound in $1$.

Scope and attribution.
This is an exposition adapting J. H. Spencer, On cliques in graphs, Israel Journal of Mathematics 9 $1971$, 419--421, https://doi.org/10.1007/BF02771457. The original paper proves a slightly sharper constant. The unbounded $\log_*N$ term in the proposed asymptotic is therefore false. Source statement: https://www.erdosproblems.com/927, checked 24 September 2026. The proof above includes the construction and its interpolation to all sufficiently large $N$; no new result or local Lean verification is claimed.

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