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

[Lars Blomberg]

Here, Giovanni, is a sketch of the method I used.

A237851 a(1)=1; a(n) is the smallest integer not yet in the sequence 
divisible by all non-zero digits of a(n-1).
= 1, 2, 4, 8, 16, 6, 12, 10, 3, 9, 18, 24, 20, 14, 28, 32, 30, 15, 5, ...

Combine the divisibility requirements for the digits of a(n-1) into one 
number.
E g 13->3, 28->8, 64->12 and 123456789->2520.
There are only 48 such numbers, described by A165412 - Divisors of 2520.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 
40, 42, 45, 56, 60, 63, 70, 72, 84, 90, 105,
120, 126, 140, 168, 180, 210, 252, 280, 315, 360, 420, 504, 630, 840, 1260, 
2520.

Each a(n) must be a multiple of one of these numbers.
Make a table that for each number contains the latest used multiplier for 
that number,
and update the table as new terms are added.

Denote the latest used multiple for each number by (number, multiplier).
Show only the ones where the multipler is > 0.
After a(1)=1 has been placed we have (1,1).
After a(2)=2 we have (1,2).
Now we want divisibility by 2 so (2,1) could be a candidate,
but 2*1 = 1*2 which is already taken, we must choose (2,2) making a(2)=4.
We want divisibility by 4 so (4,1) could be a candidate,
but 4*1 = 2*2 which is already taken, we must choose (4,2) making a(3)=8.

and a little further along when the sequence is 1, 2, 4, 8, 16, 6, 12, 10, 3
The state is: (1,3) (2,5) (4,2) (6,2) (8,2) and we need divisibility by 3.
Multiplier 1 can not be used because 1 * 3 = 3 which is <= (1,3) = 3.
Multiplier 2 can not be used because 2 * 3 = 6 which is <= (2,5) = 10.
Accepted multiplier 3 which makes a(n) = 9.

/Lars B

-----Ursprungligt meddelande----- 
From: Giovanni Resta
Sent: Saturday, March 15, 2014 1:07 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Divisible by the digits of a(n-1)

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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