[seqfan] Re: Bag of digits

[David Seal]

There is an ambiguity in the definition of this sequence: are digits considered left-to-right or right-to-left? It makes no difference to the contents of the bag, but it does to the question of what remains of k and thus what the term of the sequence is.

If my quickly-thrown-together spreadsheet and quick scan through that spreadsheet are correct (*), this is first relevant for k=414. After processing k=413, I think the bag is {4,5,6,7,8,9}, so a 4 is thrown away and the remaining digits of k are a 1 and a 4 - but is the resulting sequence term 14 (as a result of the digits being considered in left-to-right order) or 41 (for right-to-left order)?

(*) Not guaranteed in either case! A check on the spreadsheet is that I have the bag emptying completely at k=19, 39, 59, 79, 98, 139, 159, 179, 219, 239, 259, 279, 319, 359, 379, 419, 439, 459, 479, 519, 539, 579, 619, 639, 659, 679, 719, 739, 759, 819, 839, 859, 879, 1019, ...

David


> On 10 December 2019 at 21:18 jnthn stdhr <jstdhr at gmail.com> wrote:
> 
> 
> Hi folks.
> 
> Start with an empty bag of digits.  For each k =0,1,2,... compare the
> digits of k with the contents of the bag.  If a digit of k matches a digit
> in the bag throw them in the trash (they cancel each other out).  Add the
> concatenation m of what remains of k to the sequence and toss the digits of
> m into the bag.  Note that if m contains leading zeros they are stripped
> away in the seq. but added to the bag.
> 
> Initially no digits are in the bag so the first ten terms are
> [0,1,2,3,4,5,6,7,8,9] and the bag contains (0,1,2,3,4,5,6,7,8,9).
> 
> Now with k=10, we match both digits and nothing remains to add to the
> sequence or the bag,so the sequence remains unchanged and the bag now
> contains (2,3,4,5,6,7,8,9).
> 
> With k=11, we match nothing, so 11 is added to the seq. ->
> [0,1,2,3,4,5,6,7,8,9,11]  and two ones are added to the bag ->
> (1,1,2,3,4,5,6,7,8,9).
> 
> With k=12 and k=13, all digits are matched and the seq. is unchanged, and
> the bag is left with (4,5,6,7,8,9)
> 
> With k=14, the fours cancel and the one remains, so 1 is added to the seq.
> and to the bag.
> 
> Etc.
> 
> The sequence begins:
> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 1, 1, 1, 20, 1, 22, 3, 4, 25, 6, 27, 8,
> 29, 3, 33, 3, 3, 40, 1, 42, 3, 44, 5, 6, 47, 8, 49, 5, 5, 55, 5, 60, 1, 62,
> 3, 64, 5, 66, 7, 8, 69, 7, 7, 7, 77, 80, 1, 82, 3, 84, 5, 86, 7, 88, 9, 9,
> 9, 9, 9, 99, 100, 1, 2, 103, 4, 105, 6, 107, 8, 10, 1, 11, 11, 11, 11, 11,
> 20, 1, 22, 13, 4, 125, 6, 127, 8, 129, 131, 33, 1, 13, 13, 140, 1, 42, 13,
> 44, 15, 6, 147, 8, 149, 151, 5, 1, 55, 1, 15, 160, 1, 62, 13, 64, 15, 66,
> 17, 8, 169, 171, 7, 1, 7, 1, 77, 1, 180, 1, 82, 13, 84, 15, 86, 17, 88, 19,
> 191, 9, 1, 9, 1, 9, 1, 99, 200]
> 
> The bag size always remains small.  For k > 9 record bag sizes occur at
> (k:record size) 808:11, 8008:12, 10008:13, 80008:15, 800008:16,
> 1000008:17.  Is there a simple explanation for why these k's all end with
> an eight?
> 
> The seq. of integers where no digits are canceled out is:
> [0,1,2,3,4,5,6,7,8,9,11,20,22,25,27,29,33,40,42,44,47,49,55,60,...]
> 
> The seq. of integers where all digits are canceled out is:
> [10, 12, 13, 15, 17, 19, 30, 32, 34, 35, 37, 39, 50, 52, 54, 56, 57, 59,
> 70, 72, 74, 76, 78, 79, 90, 92, 94, 96, 98, 112, 114, 116, 118, 130, 132,
> 135, 137, 139, 150, 152, 157, 159, 170, 172, 179, 190, 192, 210, 213, 215,
> 217, 219, 230, 235, 237, 239, 250, 257, 259, 270, 279, 290, 310,...]
> 
> Should I add these?
> 
> -Jonathan
> 
> Python code:
> 
> seq = []
> bag = ""
> #for first 10K term set range(11256)
> for k in range(1000):
>   m = str(k)
>   for digit in m:
>     if digit in bag:
>       #Remove digit from m and bag.
>       mndx = m.index(digit)
>       m = m[:mndx] + m[mndx+1:]
>       bndx = bag.index(digit)
>       bag = bag[:bndx] + bag[bndx+1:]
>   if m:
>     seq.append(int(m))
>     bag = bag + m
> 
> print(seq)
> 
> --
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