[seqfan] Re: 3,29,289,321,...

[M. F. Hasler]

Eric,

your term a(n+1) which follows a(n) is of the form
a(n+1) = a(n)*(a(n+1) mod 10) + floor[a(n+1) / 10].
Writing a(n+1) = 10x+y, 1 <= y <= 9, we must have :
10x + y = a(n)*y + x  <=>  9 x = (a(n)-1)*y

So (a(n)-1)*y must be divisible by 9, and then the quotient yields x.
This is always possible by choosing y=9 (the case of your first 2 terms),
but if a(n)-1 is divisible by 3 we can also choose y=3 for a smaller
solution,
and if a(n)-1 is divisible by 9 then we can choose any y,
in particular y=1, the case of your 3rd term.

You may to want to exclude a(0)=1 because if the ambiguity and leading zero
(the a(n+1) is < 10 and so "the string a(n+1) without the last digit" is
empty = 0):
it could yield as lex.first the constant solution
1, 1*1+0 = 1, 1,1,1...
but it also admits the non-constant solution (choosing the first y=2)
1, 1*2+0 = 2, 2*9+1 = 19, ...
and (y=3)
1, 1*3+0 = 3, 3*9+2 = 29,... (your solution)
or
1, 1*4+0 = 4, 4*3+1 = 13, 43 = 13*3+4,
etc.

So your sequence(s) result from iterations of the map

S(n) /* least possible successor of the term "n" with that property */
= if( n%9==1, n\9*10+1, if( n%3==1, n\3*10+3, n*10-1))
= (1, 19, 29, 13, 49, 59, 23, 79, 89, 11, 109, 119, 43, 139, 149, 53, 169,
179, 21, 199, 209, 73, 229, 239, 83, 259, 269, 31, 289, 299, 103, 319, 329,
113, 349, 359, 41, 379, 389, 133, 409, 419, 143, 439, 449, 51, 469, 479,
163, 499, 509, 173, 529, 539, 61, ...)

So, unless I err, the lexicographic first nontrivial variant is the orbit
of 2:
2, 2*9+1 = 19, 19*1+2 = 21, 21*9+20 = 209, 209*9+208 = 2089, 2089*1+ 232 =
2321, ...
i.e., (2, 19, 21, 209, 2089, 2321, 23209, 77363, 773629, ....)
next comes yours = orbit of 3:
3, 29, 289, 321, 3209, 32089, 106963, 356543, ...
and others.
(PARI: vector(20,n,t=if(n>1,S(t),3)))

Maximilian

On Sat, Feb 13, 2016 at 3:00 PM, Eric Angelini <Eric.Angelini at kntv.be>
wrote:

>
> Hello SeqFans,
> S= 3,29,289,321,3209,...
>
> Look at 3 and 29:
> we do 3*9 = 27
> and 27+ 2 = 29
>
> Look at 29 and 289:
> we do 29*9 = 261
> and 261 + 28 = 289
>
> Look at 289 and 321:
> we do 289*1 = 289
> and 289 + 32 = 321
>
> Look at 321  and 3209
> we do 321*9 = 2889
> and 2889 + 320 = 3209
>
> ...
> The task is:
> a(n)*[the last digit of a(n+1)]
> + [the string a(n+1) without the said digit]
> = a(n+1)
>
> Is this the start of the lexico-first such seq?
> Best,
> É.
> <http://list.seqfan.eu/>

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