Observation on Erdős–Moser equation.

457 Views Asked by At

Define: $ S(n,m)= \sum_{i=1}^{n}i^m$ where $n,m\in \mathbb{Z}_+$

Define: $F_m$ is function as, there exist smallest integer $k$ with respect to $m$ such that, $$k^m\le S(k-1,m)$$ so $F_m=k$.

Example: $F_2=5$.

Let $S(F_m-1,m)$ convert in base $F_m$ so we can represent as $$S(F_m-1,m)= (\alpha_m,\alpha_{m-1},...,\alpha_1,\alpha_0)_{F_m}$$

Where $\alpha_i$ are digits of $S(F_m-1,m)$ in base $F_m$.

Example: $m=2\rightarrow S(F_2-1,2)=S(4,2)=30=(1,1,0)_5$.


Table

$$\begin{split} S(F_m-1,m) &= (\alpha_m,\alpha_{m-1},...,\alpha_1,\alpha_0)_k \\ S(2,1) &= (1,0)_3 \\S(4,2) &=(1,1,0)_5 \\S(5,3) &=(1,0,1,3)_6 \\S(7,4) &=(1,1,1,0,4)_8 \\S(8,5) &=(1,0,3,6,6,0)_9 \\S(10,6) &=(1,1,3,1,4,5,0)_{11}\\S(11,7) &=(1,0,6,11,8,6,1,0)_{12}\\S(13,8) &=(1,1,6,2,9,0,7,6,7)_{14}\\S(14,9) &=(1,0,11,3,0,11,11,3,9,0)_{15}\\S(15,10) &=(1,0,1,10,2,7,12,9,3,2,8)_{16}\\S(17,11) &=(1,0,16,7,11,6,5,16,11,4,16,9)_{18}\\S(18,12) &=(1,0,5,0,6,18,13,14,3,5,11,17,0)_{19}\\ \vdots &= \vdots \end{split}$$

Claim

1) For all $m$, $\alpha_m\in\{1\}$

2) For all $m$, $\alpha_{m-1}\in\{0,1\}$

3) For all $m>2$, $\alpha_{m-2}\notin\{0\}$

4) $F_{m+1}\ge F_m$

The Erdős–Moser equation is ${\displaystyle 1^{m}+2^{m}+\cdots +(n-1)^{m}=(n)^{m}}$ where $n$ and $m$ are positive integers. The only known solution is $1^1 + 2^1 = 3^1$.

Third claim is consequence to Erdős–Moser equation conjecture.

The first two claims are posted here in a different mathematical format with specificity check here

I was not worked hard on this observation. may be you can disprove by counter example.thank you.


Edit 1: Answer for first claim check here

Edit 2: find $x,y$ s.t.$S(x,m)=y^m$ This post is consequence to Erdős–Moser equation.

1

There are 1 best solutions below

4
On

Later edit - everything below is a consequence of the fact that $F_m$ is $2+$ the nearest integer to $m/\ln2$, see comments to the OP. I am still leaving this - for amusement?

Instead of answer - just some curious facts about the numbers $F_m$. If one believes in 4), there is a quick Mathematica code to play with them:

iF[m_]:=iF[m]=If[m==0,2,Module[{k=iF[m-1]},While[HarmonicNumber[k-1,-m]<k^m,k++];k]]

They go like $$ 2,3,5,6,8,9,11,12,14,15,16,18,19,21,22,24,25,27,28,29,31,32,... $$ The sequence ${F'}_m:=F_{m+1}-F_m$ goes like $$ 1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,2,1,... $$ Let $\{{}^2F_1,{}^2F_2,{}^2F_3,...\}$ be the sequence $\{m\mid{F'}_m={F'}_{m+1}\}$; it goes like $$ 9,18,27,34,43,52,61,70,79,88,95,104,113,122,131,140,149,158,165,174,183,... $$ and the sequence of its differences ${}^2{F'}_m:={}^2F_{m+1}-{}^2F_m$ goes like $$ 9,9,7,9,9,9,9,9,9,7,9,9,9,9,9,9,9,7,9,9,9,9,9,9,7,9,9,9,9,9,9,7,9,9,9,9,9,9,9,7,... $$ Let further $\{{}^3F_1,{}^3F_2,{}^3F_3,...\}$ be the sequence $\{m\mid{}^2{F'}_m=7\}$; it goes like $$ 3,10,18,25,32,40,47,54,61,69,76,83,90,98,... $$ and ${}^3{F'}_m:={}^3F_{m+1}-{}^3F_m$ goes like $$ 7,8,7,7,8,7,7,7,8,7,7,7,8,7,7,8,7,7,7,8,7,7,8,7,7,7,8,7,7,8,... $$ Next, let $\{{}^4F_1,{}^4F_2,{}^4F_3,...\}$ be $\{m\mid{}^3{F'}_m=8\}$; it is $$ 2,5,9,13,16,20,23,27,30,34,38,41,45,... $$ and ${}^4{F'}_m:={}^4F_{m+1}-{}^4F_m$ is $$ 3,4,4,3,4,3,4,3,4,4,3,4,3,4,3,4,4,3,4,3,4,4,3,4,3,4,3,4,4,3,4,3,4,4,3,4,3,4,... $$ Continuing in the same way, with $\{m\mid{}^4{F'}_m=3\}$ I get ${}^5{F'}_m$ $$ 3,2,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,2,3,2,3,2,2,3,2,... $$ then ${}^6{F'}_m$ $$ 3,3,2,3,2,3,3,2,3,2,3,3,2,3,2,3,3,2,3,3,2,3,2,3,3,2,3,2,3,3,2,3,2,3,3,2,3,3,... $$ then ${}^7{F'}_m$ $$ 2,3,2,3,2,3,3,2,3,2,3,2,3,3,2,3,2,3,2,3,3,2,3,2,3,2,3,3,2,3,2,3,... $$

I wonder if this goes forever...