Are there other primes in the sequences: $a_n = \sum\limits_{k=0}^{n-1} k \cdot n^{n-k-1}$, $ b_n = \sum\limits_{k=1}^{n-1} k \cdot n^{k-1}$

Question

Context: I was playing with the numbers 123,456,789 and 987,654,321 which are interesting because they are composed of the digits 1–9 in ascending and descending order, respectively. These numbers are also the 10th terms of the following two sequences, which I investigated further in other bases:

I defined the sequences:

$$ a_n = \sum_{k=0}^{n-1} k \cdot n^{n-k-1}, \quad b_n = \sum_{k=1}^{n-1} k \cdot n^{k-1} $$

For $ n = 1 $ to $ 10 $, I computed:

$$ \begin{aligned} a_n &= 0,\ 1,\ 5,\ 27,\ 194,\ 1865,\ 22875,\ 342391,\ 6053444,\ \mathbf{123456789} \\ b_n &= 0,\ 1,\ 7,\ 57,\ 586,\ 7465,\ 114381,\ 2054353,\ 42374116,\ \mathbf{987654321} \end{aligned} $$

I then checked these sequences for prime values and noticed that only $ a_3 = 5 $ and $ b_3 = 7 $ are prime for $ n < 100 $. I couldn't compute further due to overflow and performance issues, but it appears these might be the only primes in the sequences.

Here's the Python code I used to compute and check primality:

import math
def isprime(n):
    n = abs(int(n))
    if n < 2:
        return False
    if n == 2: 
        return True    
    if not n & 1: 
        return False
    for x in range(3, int(n**0.5)+1, 2):
        if n % x == 0:
            return 0
    return 1
n=100
p1=[]
p2=[]
sum=0
for i in range(1,n+1):
    sum=0
    for j in range(0,i):
        sum=sum+j*pow(i,i-j-1)
    if isprime(sum)==1:
        p1.append((i,sum))
for i in range(1,n+1):
    sum=0
    for j in range(1,i):
        sum=sum+j*pow(i,j-1)
    if isprime(sum)==1:
        p2.append((i,sum))
print(p1)
print(p2)

My question is: Are $a_3 , b_3$ the only primes in their respective sequences?

Another interesting property is

$$\gcd(a_n, b_n)= \begin{cases} n-1, & \text{if $n$ is even} \\ (n-1)/2, & \text{if $n$ is odd} \end{cases}$$ But I couldn't prove/disprove that.

$a_n = (n^n-n^2+n-1)/(n-1)^2$ and $b_n = ((n+1)^{(n+1)}*(n-1)+1)/n^2$ Not sure if it helps — Aurel-BG, May 16 '25 at 19:08
I think your formula for $b_n$ is wrong, @Aurel-BG I haven't checked $a_n.$ — Thomas Andrews, May 16 '25 at 19:51
$b_n=f_n'(n)$ when $f_n(x)=\frac{x^n-1}{x-1}.$ Then $$f_n'(x)=\frac{(n-1)x^n-nx^{n-1}+1}{(x-1)^2}.$$ So $$b_n=\frac{(n-1)n^n-n\cdot n^{n-1}+1}{(n-1)^2}=\frac{n^{n+1}-2n^n+1}{(n-1)^2}$$ — Thomas Andrews, May 16 '25 at 20:01
@pie Do you know how to prove the unjustified claims in the answer you accepted? (cf. my comment on that answer). If not you should ask for elaboration. Note that no ad hoc guessing or ingenuity is needed if we use the Euclidean algorithm, and the arithmetic is simpler (and this way answers both of your questions), as I explain in my answer and comments on it. $\ \ $ — Bill Dubuque, May 19 '25 at 02:27

user1172706 · Accepted Answer · 2025-05-17T09:23:08.153

6

Modulo $n-1$ both series are simply $1+2+...+ (n-1)$ and are therefore divisible by $$\begin{cases} n-1, & \text{if $n$ is even} \\ (n-1)/2, & \text{if $n$ is odd} \end{cases}$$ This gives you the required result about primality.

Now let $S$ be the integer, coprime to $n$, such that$$1+n+...+n^{n-2}=(n-1)S.$$ Then $$b_n=1+2n +...(n-1)n^{n-2}=n^{n-1}-S$$ and is coprime to $n$ and $S$.

Since $a_n+b_n=n(n-1)S$ the GCD is a divisor of $n(n-1)S$ and is therefore a divisor of $n-1$. From the primality argument your second conjecture is therefore also true.

edited May 17 '25 at 09:23

answered May 16 '25 at 20:01

user1172706

2,791

1

Nice and simple, but it only shows the gcd is divisible by $n-1$ or $(n-1)/2,$ not whether they are equal to the gcd. – Thomas Andrews May 16 '25 at 20:06
No, that's why I didn't make that claim! – user1172706 May 16 '25 at 20:07
I've now added a proof of the second conjecture. – user1172706 May 17 '25 at 09:25
Please explain (in the answer) how you prove your claims that $,b_n = n^n - S,,$ and $,b_n$ is coprime to $S,,$ and how you infer the gcd is $(n-1)/2$ when $n$ is odd, since these may not be obvious to many readers.$\ \ $ – Bill Dubuque May 17 '25 at 10:56

Thomas Andrews · Answer 2 · 2025-05-17T02:22:19.203

3

Just to answer the GCD part of the question..

You can get the closed forms for $a_n,b_n:$

$$ \begin{align} a_n&=\frac{n^n-n^2+n-1}{(n-1)^2}\\ b_n&=\frac{n^{n+1}-2n^n+1}{(n-1)^2} \end{align}$$

Then you can show $$b_n-(n-2)a_n=n-1$$

That means the GCD is a divisor of $n-1.$ But the other answer shows that$$a_n,b_n\equiv 0\text{ or }\frac{n-1}2\pmod{n-1}$$ depending on whether $n$ is even or odd, thus validating your conjecture.

edited May 17 '25 at 02:22

answered May 16 '25 at 22:56

Thomas Andrews

186,215

My answer explains how to mechanically derive the key equation $,b_n-(n!-!2)a_n=n-1,,$ by using the $\rm\color{#0a0}{polynomial}$ Euclidean algorithm. Although the $\rm\color{#c00}{exponential}$ expressions (in $,n^n,$ etc) are simple enough here that one might be able to derive the equation in an ad-hoc way without too much difficulty, that will prove much more difficult for larger such expressions. But the Euclidean algorithm always works - efficiently and uniformly (no ingenuity is needed), no matter how complex the $\rm\color{#0a0}{polynomial}$ expressions. $\ \ $ – Bill Dubuque May 17 '25 at 06:45

score 2 · Answer 3 · edited May 19 '25 at 03:11

The gcd is a mental mechanical calculation by the Euclidean $\rm\color{#90f}{division}$ algorithm. But the key idea is that we must do this for the generating polynomials in $\:\!x,\,$ i.e. before evaluating them at $\,x=n\,$ (which maps polynomials like $\,x^n\,$ into exponentials $\,n^n\,$ so Euclid's algorithm no longer applies).

We seek the gcd $\:\!(a_n,b_n)\:\!$ for $\:\!b_x = [\:\!(x^n\!-\!1)/(x\!-\!1)]'\:\!$ and $\:\!a_x\:\!$ is its reverse polynomial i.e.

$$ b_x = \dfrac{(\color{#0a0}{(n\!-\!1)x\!-\!n})x^{n-1}+1}{(x\!-\!1)^2} =:\dfrac{\!\!B}{(x\!-\!1)^2}\qquad$$

$$ a_x = \dfrac{x^n-nx+n\!-\!1}{(x\!-\!1)^2} =: \dfrac{\!\!A}{(x\!-\!1)^2}$$

so $\ \ xB-(\color{#0a0}{(n\!-\!1)x\!-\!n})A = n(n\!-\!1)(x\!-\!1)^2\,$ by $\rm\color{#90f}{division}$ $\,xB\div A$

so $\ \ \,\color{#0af}{b_n}-(n\!-\!2)a_n = \color{#0af}{n\!-\!1}\ $ by eval at $\:\!x=n\:\!$ then cancel $\:\!n(n\!-\!1)^2$

so $\,\ \ (a_n,\color{#0af}{b_n}) = (a_n,\color{#0af}{n\!-\!1})\ $ by $\:\!\rm\color{brown}{Euclid}\:\!$ & $\,\color{#0af}{b_n\equiv n\!-\!1}\pmod{\!a_n}$

so $\,\ \bbox[6px,border:2px solid #0a0]{\! (\color{#f40}{a_n},b_n) = (\color{#f40}{n(n\!-\!1)/2},\:\!n\!-\!1) = n\!-\!1\,\ {\rm if}\ \,2\mid n,\ {\rm else}\ \,(n\!-\!1)/2}$

by $\ \ \ \:\!\color{#f40}{a_n}\!\equiv a_1\! \equiv\sum_{k=1}^{n-1} k\equiv \color{#f40}{\frac{n(n-1)}2}\!\pmod{\!n\!-\!1}\,$ and $\rm\color{brown}{Euclid}$.

By above $,a_n,b_n$ are both divisible by $,d = n!-!1$ or $,(n!-!1)/2,$ for $:!n:!$ even or odd resp. For $,n>3,$ we have $, 1 < d < a_n,b_n,$ so $,d,$ is a proper factor, thus you have found all of the primes in $,a_n,b_n.,$Therefore the above gcd calculation answers both of your questions (without pulling anything out of a hat like magic, i.e. discovery of the proper factor and division reduction step occur naturally during the Euclidean algorithm). $\ \ $ — Bill Dubuque, May 17 '25 at 08:07
${\rm\color{brown}{\text{Euclid:}}}\ \ A\equiv a\pmod{! m},\Rightarrow\ (A,m) = (a,m),,$ i.e. gcd mod reduction, is what we used above to compute the gcd (it's a generalization of $, (A,m) = (A\bmod m,,m)$ = descent step in the Euclidean algorithm). — Bill Dubuque, May 19 '25 at 01:59

Are there other primes in the sequences: $a_n = \sum\limits_{k=0}^{n-1} k \cdot n^{n-k-1}$, $ b_n = \sum\limits_{k=1}^{n-1} k \cdot n^{k-1}$

3 Answers3