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Introduction

The number $33333331$ is a prime (or $1$), regardless of how many $3$s are removed.

That is, $31,331,3331,33331,333331,3333331,33333331$ are all prime numbers.

I was wondering if we can find a longer sequence of such numbers in another number base.


Problem

If we in general allow any prime digit $\beta$ (as well as the original $1$) as the unit digit, we can ask:

Does there exist $b\ge 2$ and $\alpha,\beta\lt b$ where $\beta$ is prime or $1$, such that for some $L\gt 7$, the numbers

$$ n_b(l)=\alpha \sum_{k=1}^{l} b^k+\beta=\alpha\left(b\frac{ b^l -1}{b-1}\right)+\beta,\quad l=1,2,\dots,L\tag{$\star$} $$

are all prime numbers ?

In the introduction, we have $b=10,\alpha=3,\beta=1$ and $L=7$.

It turns out that $L=8$ appears for the first time in $b=774$, for $\alpha,\beta=(495, 137)$.

$$ 63840216500331090348696887=(495, 495, 495, 495, 495, 495, 495, 495, 137)_{774} $$

But, does such an example exist for $\beta = 1$?

Similarly as in the introduction example, but considering any number base $b$, we ask:

Question $1.$ What is the smallest $L\gt 7$ example of all $(\star)$ being prime, if $\beta=1$?

More generally, I'm wondering if it is possible to provide any arguments for the following:

Question $2.$ Does for every $L\in\mathbb N$, there exists $b\ge 2$, in which all $(\star)$ are primes?

E.g. Are there results about "prime sequences with differences $O(b^1),O(b^2),\dots,O(b^L)$" ?


Computations

When searching for $\beta = 1$ in number bases $b=2,3,\dots,10^4$, we find the following records:

b    L [α, α, ..., α, 1]
2    2 [1, 1, 1]
4    2 [3, 3, 1]
5    3 [2, 2, 2, 1]
6    3 [5, 5, 5, 1]
10   7 [3, 3, 3, 3, 3, 3, 3, 1]
3675 7 [1650, 1650, 1650, 1650, 1650, 1650, 1650, 1]
3789 7 [1218, 1218, 1218, 1218, 1218, 1218, 1218, 1]
4784 7 [462, 462, 462, 462, 462, 462, 462, 1]
4943 7 [2660, 2660, 2660, 2660, 2660, 2660, 2660, 1]

What is the smallest $L=8$ example (for $\beta=1$)?

For comparison, a simple search up to $b=10^3$ for any $\beta$ finds the following records:

b   L [α, α, ..., α, β]
2   2 [1, 1, 1]
4   2 [3, 3, 1]
4   2 [1, 1, 3]
4   2 [2, 2, 3]
5   3 [2, 2, 2, 1]
6   3 [5, 5, 5, 1]
6   4 [1, 1, 1, 1, 5]
10  7 [3, 3, 3, 3, 3, 3, 3, 1]
38  7 [30, 30, 30, 30, 30, 30, 30, 31]
56  7 [15, 15, 15, 15, 15, 15, 15, 23]
69  7 [10, 10, 10, 10, 10, 10, 10, 37]
195 7 [156, 156, 156, 156, 156, 156, 156, 11]
290 7 [42, 42, 42, 42, 42, 42, 42, 277]
310 7 [249, 249, 249, 249, 249, 249, 249, 47]
314 7 [1, 1, 1, 1, 1, 1, 1, 209]
320 7 [91, 91, 91, 91, 91, 91, 91, 281]
340 7 [111, 111, 111, 111, 111, 111, 111, 113]
342 7 [205, 205, 205, 205, 205, 205, 205, 271]
350 7 [117, 117, 117, 117, 117, 117, 117, 233]
399 7 [232, 232, 232, 232, 232, 232, 232, 391]
410 7 [11, 11, 11, 11, 11, 11, 11, 37]
509 7 [340, 340, 340, 340, 340, 340, 340, 299]
510 7 [434, 434, 434, 434, 434, 434, 434, 157]
542 7 [70, 70, 70, 70, 70, 70, 70, 113]
551 7 [500, 500, 500, 500, 500, 500, 500, 499]
552 7 [380, 380, 380, 380, 380, 380, 380, 517]
687 7 [14, 14, 14, 14, 14, 14, 14, 683]
735 7 [358, 358, 358, 358, 358, 358, 358, 491]
740 7 [552, 552, 552, 552, 552, 552, 552, 641]
752 7 [273, 273, 273, 273, 273, 273, 273, 331]
774 8 [495, 495, 495, 495, 495, 495, 495, 495, 137]

Where the smallest $L=8$ exmaple was found (for $\beta\gt 1$).

Vepir
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  • Maybe, the generalized Bunyakovsky conjecture implies arbitary long such chains, but I have not worked out yet a complete proof. This conjecture is of course open anyway, but this would at least be a good heuristic. – Peter Sep 10 '20 at 15:57
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    A solution with $L=8$ and $\beta=1$ : $[55912, 70164, 8]$ , please check ! – Peter Sep 10 '20 at 16:44
  • @Peter That was fast! Indeed, $b=70164,\alpha=55912$ is a $L=8$ solution! – Vepir Sep 10 '20 at 17:06
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    $[1584, 1724, 1453, 9]$ is a $L=9$-solution – Peter Sep 10 '20 at 17:34
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    $$[6218, 930, 2251, 11]$$ is a $L=11$-solution, please check! – Peter Sep 11 '20 at 10:20

0 Answers0