[seqfan] Re: Scott Shannon's "Grow" sequence A332580 (and A332584)
Sean A. Irvine
sairvin at gmail.com
Wed Feb 19 20:56:15 CET 2020
I'm not sure how far Scott already searched but I have a(92) > 3000000 and
a(98) > 3500000. (I haven't tried 44).
On Wed, 19 Feb 2020 at 04:27, Neil Sloane <njasloane at gmail.com> wrote:
> The definition of A332580 is:
>
> a(n) = minimal positive k such that the concatenation of the decimal digits
> of n,n+1,...,n+k is divisible by n+k+1, or -1 if no such k exists
>
> .and A332584 gives the values of n+k.
>
> Examples:
>
> a(1) = 1 as '1' || '2' = '12', which is divisible by 3 (where || denotes
> decimal concatenation).
>
> a(7) = 13 as '7' || '8' || '9' || '10' || '11' || '12' || ... || '20' =
> 7891011121314151617181920, which is divisible by 21.
>
> a(8) = 2 as '8' || '9' || '10' = 8910, which is divisible by 11.
>
> a(2) = 80: the concatenation 2 || 3 || ... || 82 is
>
> 23456789101112131415161718192021222324252627282930313233343536373839\
>
> 40414243444546474849505152535455565758596061626364656667686970717273747\
>
> 576777879808182, which is divisible by 83.
>
> What makes this interesting is that the chance that a particular k works
> is, naively, 1/(n+k+1). For fixed n, the sum 1/(n+k+1) diverges, so a(n)
> should always exist. However, Scott and I have not been able to find three
> values up to 100, namely a(44), a(92), and a(98).
>
> Possibly there are number-theoretic reasons why 44, 92, 98, ... are
> special, or maybe we did not search far enough.
>
> I ran Maple out to 200000, on 44, but when I tried to go out to a million,
> Maple managed to put my iMac into a coma and I had to turn off the power.
> Scott ran a Java program a lot further, but I always worry when Java is
> doing number theory.
>
> So we need help! Find a(44), a(92), a(98) or show they do not exist.
>
> --
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>
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