# [seqfan] Re: A057652 has no more terms below 10^4

Alonso Del Arte alonso.delarte at gmail.com
Wed Oct 20 02:09:24 CEST 2010

```Very interesting, Hugo. I have access to C but not Perl, though on the other

To David: your "disappointing" conjectures are conjectures which I think
Erdos himself would have made if he had pondered this problem. But I also
think he might've continued to think about it, even if he could only chip
away tiny flecks.

My very tiny fleck for today: I confirm that there are no more terms up to
50000. After I implement a more efficient algorithm, I hope to confirm up to
a million.

Al

P.S. Thanks to Charles and Tony for signing up for upcoming Sequences of the
Day. This month has plenty of open slots, and November is wide open.

On Tue, Oct 19, 2010 at 11:16 AM, <hv at crypt.org> wrote:

> hv at crypt.org wrote:
> :A negative result:
> :
> :I tried constructing the minimal modular pattern after each new lucky
> :number was sieved out, and for mod m, checked how many values mod m were
> :candidates for this sequence taking into account powers up to the first
> :2^k > m expecting that they would fairly rapidly descend to zero, thus
> :proving that the sequence is finite.
>
> So I thought, maybe we should look instead at:
>  a(n) := min(k: lucky(k - 2^i) forall i in {1..n})
>
> I expected this would grow reasonably smoothly, but enough faster than 2^n
> to give supporting evidence that further elements of A057652 are unlikely.
> Instead, I found that the existing terms of A057652 yielded the sequence
> [ 1, 3, 5, 11, 17, 647, 647, 647, 647, 647 ], and I didn't find the next.
>
> Letting f(k) represent the highest power of 2 achieved for a given k, I
> found
> the following for 0 < k < 1e6, f(k) >= 5:
>
> f(k) = 5: 2411 15263 67931 104555 152099 167681 226901 245843 248993 338237
>  354581 403595 434843 567353 594359 627497 645299 701495 762317 880919
>  923597 945983
> f(k) = 6: 7199 16397 66581 157175 303839 328079 348617 350927 513503 628085
>  932669
> f(k) = 7: 6479 290531 323585 472601
> f(k) = 8: 772433
> f(k) = 9: 647
>
> I guess this is pretty close to what you'd expect, though I hadn't thought
> it would be quite so lumpy. I guess a(10) should appear somewhere between
> 1e6 and 3e6. It does suggest though that f(647)=9 was super unlikely.
>
> Given my perl code managed up to 1e6 easily enough, I think rewriting in C
> would let me generate a list of the lucky numbers up to 1e8 fairly easily,
> maybe closer to 1e9. Let me know if that would be useful.
>
> Hugo
>
>
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```