[seqfan] Re: Concatenate, divide

[Allan Wechsler]

a(4) is 33, 1133%33 = 11. OEIS still has 15 matches.

The next entry can't have just one digit, since the remainder has to be 33.
Can it have two digits? If it does, 3300 + a(5) = 33 mod a(5). That means
that 3300-33 + a(5) is divisible by a(5), and so is 3267. Now, 3267 is
3^3*11^2, and we can list its divisors easily. The first few divisors are
too small; the first divisor that is big enough is 99, so a(5) = 99. OEIS
still has 8 matches.

a(6) cannot have just two digits, because it must be at least 99 to leave a
remainder of 99. Can it have three digits? If it does, it must divide
99000-99 exactly. This number is 99*999, = 3^5*11*37. The smallest divisor
over 99 is 111, so that is a(6). At this point OEIS stops giving any
matches. I am guessing that the sequence continues 1, 3, 9, 11, 33, 99,
111, 333, 999, 1111 ...

On Sun, Jun 7, 2015 at 4:08 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:

> Hello SeqFans,
> concatenate a(n) to a(n+1);
> divide by a(n+1);
> remainder is a(n).
>
>  S = 1, 3, 9, 11, ...
>
> Example:
> 13 divided by 3 leaves 1
> 39 divided by 9 leaves 3
> 911 divided by 11 leaves 9
>
> I don't have enough terms to check
> if this is old hat or not, sorry.
>
> Best,
> É.
>
>
> Catapulté de mon aPhone
>
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