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Define $$f(n):=\sum_{j=1}^n j^{n+1-j}=1+2^{n-1}+3^{n-2}+\cdots+(n-1)^2+n$$ for a positive integer $n$.

With PARI/GP, this function can be calculated with the self-defined function

f(n)=sum(j=1,n,j^(n+1-j))

Question $1$ : Is $f(n)$ only a perfect power for $n=3$ ?

I checked upto $n=10^4$ and only found one perfect power, namely for $n=3$. I do not even have an idea for the perfect squares. The PARI/GP-code for those who want to extend the search limit :

gp > for(m=1,10^4,if(ispower(f(m))>0,print1(m," ")))
3
gp >

Question $2$ : If $f(n)$ only prime for $n=2$ and $n=34$ ?

Upto $n=7\ 400$ , there are no other primes $f(n)$.

It seems that $f(n)$ cannot be easily calculated with a formula. So, I think I cannot use fast primality testing tools like PFGW, but someone might have a faster software for testing general numbers.

The first such number from which I do not know a prime factor, is $f(138)$. So, small factors are not forced, hence I guess we can only search for further primes.

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    $\begingroup$ No primes to at least 7800 $\endgroup$
    – Saša
    Commented Nov 27, 2020 at 13:04
  • $\begingroup$ @Oldboy Thank you. By the way, I also continued my search. But of course, I invite you to get as far as you want. $\endgroup$
    – Peter
    Commented Nov 27, 2020 at 13:06
  • $\begingroup$ Just an additional observation: for differences I found for $d(n)=f(n)-f(n-1)$ : $d(8)=2^{11}$ and $d(6)=12^2$ $\endgroup$ Commented Feb 9, 2021 at 8:23

1 Answer 1

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To get a better intuition about the possible primality of $f(n)$ I've looked at periodicities in its primefactorizations.

Some short heuristics.


  • Primefactor $2$:

    • for $n_k=3+4k$ we have $2 \mid f(n)$
    • for $n_k=4+4k$ we have $2 \mid f(n)$

    We analyze the sequences of p-adic valuations for that primefactor below.

  • Primefactor $p=3$:
    with some vector $R$ of length $(p-1) \cdot p$ and residues $\pmod {(p-1) \cdot p^2}$ having
    $\qquad R=[2,7,10,15,17,18]$
    for $\qquad n_{j,k}=r_j+18k \qquad $ we have $ p \mid f(n)$

  • Primefactor $p=5$:
    with some vector $R$ of length $p \cdot (p-1)$ and residues $\pmod {(p-1) \cdot p^2}$ having $\qquad R=[5, 8, 13, 14, 16, 21, 44, 47, 49, 55, 62, 63, 70, 71, 78, 86, 92, 97, 99, 100]$
    for $\qquad n_{j,k}=r_j+100k \qquad$ we have $ p \mid f(n)$

  • Primefactor $p=7,11,13,...$:
    This is heuristically and in all tested cases analogue for the other small primefactors


We can even assume, that the higher p-adic valuations are periodic with $(p-1) \cdot p^A$ with stepwidth of $n$ like $(p-1) \cdot p^A$ . There is something more to say, but I can transfer the empirical tabular results to precise formulae only later.

Possibly we can say something about the covering of $f(n)$ by this periodicity of the primefactors, and how many holes per primefactor can occur allowing $f(n)$ to be composed by one prime factor alone.


Some initial impression:

 n  primefactorization of f(n)
 --|--------------------------------------
 1   1
 2   3
 3       2^3
 4       2  .11
 5              5 .13
 6           11      .19
 7   3.  2^2            .61
 8       2^2   .5          .139
 9                           31.367
10   3      .11                      .1511
11       2^6                               .3637
12       2                                      .575957
13              5                                      .1203757
14              5                                              .41.103.1567
15   3  .2^2                                                              .7.29.78317
16       2^5   .5.13                                                                 .227.2437
17   3                                                                                       .37.65240639
18   3                                                                                               .73.216688337
19       2^3                                                                      .40394337023
20   3^2.2                                               .126821110583
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