Search Results

This can be explained as follows. Assume that $(a_1,\ldots,a_{n})$ is a suitable sequence. For every $j\in \{0,\ldots,k-1\}$ denote $A(j)=\{i:a(i)=j\}$ and denote by $m(i)$ and $M(i)$ the minimal and …

I think, yes, simply by averaging. Take a very long initial segment of your sequence, which contains $N_i$ items equal to $i$ for $i=1,\ldots,B$. Assume that the sum between consecutive $j$'s is at le …

or another cheating example: positive integers and remainders of positive integers modulo 100000000.

It is a known example: sequence 1,2,3,5,7,11,13,$\dots$ of non-composite numbers coincides with the sequence of orders of finite simple groups until 60 appears (in this last sequence).

If the limit exists, it of course equals 1. Indeed, $A(m):=a(m)a(m+1)\dots a(2m-1)=ds_{10}(3^m)/ds_{10}(3^{2m})\in [\frac1{9m},9m]$. But if $\lim a(m)\ne 1$, then $A(m)$ is either exponentially large …

Conjecture 1. Assume that $p$ is a Wiefrich prime, that is, $p^2$ divides $2^{p-1}-1$. Denote $n=p^km$ where $p$ does not divide $m$. By lifting the exponent lemma, $p^{1+k}$ divides $2^{(p-1)p^{k-1}\ …

There are no other solutions than $n=3$ and those from Claim: $n=2^{p-1}$ such that $2^p-1$ is prime. Consider several cases. $n=2^ks$ is even (here $k\geqslant 1$ and $s$ is odd). Then $$\frac{\p …

For $S\subset \mathbb{Z}$, we have $a\in \mathfrak{c}(S)$ if and only if the following condition $C(S,a)$ holds: $a\in \mathbb{Z}$ and for every prime power $p^m$ there exists an element $x\in S$ cong …

The set $\{2,3,\ldots,2n-1\}$ is mapped to itself, and if we use the new variable $y=x-1$, then it is doubled modulo $2n-1$. Thus the minimal $k$ is the multiplicative order of 2 modulo $2n-1$.

WhatsUp's generating function allows to get an asymptotics: we have $$\sum S_n t^n=\frac{t}{e^{t/e}-t},\sum S_n e^nt^n=\frac{t}{e^{t-1}-t}.$$ Denote $t-1=x$, then $$\frac{t}{e^{t-1}-t}=\frac{1+x}{e^x- …

Q3: certainly yes. Take $n=2\cdot 3^{k}$, then $2^n-1$ is divisible by $3^{k+1}$,thus $3^k$ divides $a(n)$. As for Q1 and Q2, I do not see how to avoid the scenario when $2^n-1$ is always divisible by …

Question 2. Since $P_n(2\cos t)=\sin(nt)/\sin t$, we get that the roots of $P_n$ are $2\cos(\pi k/n)$, $k=1,\dots,n-1$. That is, a number $\kappa=2\cos(\pi a/b)$, $0<a<b$, gcd$(a,b)=1$, is a root of …

Is $g(a)=1\iff a=b^2+1$ (corresponding to $x=0$ and $a-x=b^2+1$ or $x=1$ and $a-x=b^2$) No. It may happen, that, say, $x,a-x$ are perfect squares and $x+1$, $a-x-1$ are twice perfect squares. These P …

This is true, if you replace $n-1$ in "the exponent in the highest power of 2 dividing $n−1$" to $n$. First of all, we study the coefficients of the series $(1-4t)^{-1/2}=\sum C_nt^n$ modulo 4. We hav …