[seqfan] Re: Divisible by the digits of a(n-1)

[Allan Wechsler]

Looking at the graphs gives a clue. The graph of A237851 shows a tangled
skein of strands, all increasing but at different rates. I believe each
strand corresponds to a different set of divisibility constraints. For
example, there's a strand in there somewhere for when the previous number
contains only 1's, 2's, 4's, and 5's, all its members being divisible by 20.

How many strands are there? Well, there's one for each possible LCM of
subsets of digits. These numbers are all the divisors of 8*9*5*7 = 2520,
and there are 4*3*2*2 = 48 such divisors, and thus 48 increasing strands.
It's perfectly feasible to keep a separate counter for each divisor.

For example, a(999) = 3936. The LCM of the digits is 18, and the next
unused multiple of 18 turned out to be a(1000) = 2826. The counters can't
be completely independent, though: one would have to worry whether 2826 had
been used on an earlier occasion, as the next multiple of 2, 3, 6, or 9.
But this could be done by checking those counters to see if they were past
2826 yet.

At any rate, unless I've missed some subtlety, the entire state of the
algorithm is kept in 48 counters, eliminating the necessity of recording
every number used so far.




On Sat, Mar 15, 2014 at 8:07 AM, Giovanni Resta <g.resta at iit.cnr.it> wrote:

> On 03/15/2014 11:13 AM, Lars Blomberg wrote:
>
>> I found a way to avoid the explicit "used" list and after 22 days
>> reached 10000 terms for A237860, the largest being 891,356,053,262.
>>
>
> Kudos for the perseverance !
> Can you give a hint about the method you used ?
>
>
> Giovanni
>
>
>
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