[seqfan] Re: Self-stuffable numbers

[Ray Chandler]

Using Lars' example of concatenation to arrive at 21021021021021021021 as a self-stuffable primitive root, I have been able to find similar concatenations that show any of the digits 1-9 may be the ending digit of such a number.  

# digits	M	L	m
2	22	4	0
3	126	9	0
20	21	3	3
52	23	5	5
59	32004	9	3
94	1035	9	5
114	100503	9	6
137	12026	11	6
202	2115	9	11
204	111035	11	9
266	132004	10	13
355	308	11	16
436	2205	9	24
490	220115	11	22
525	21205	10	26
576	4009	13	22
603	303	6	50
605	11211	6	50
648	1303	7	46
704	1213	7	50
816	104013	9	45
867	219	12	36
888	302013	9	49
1072	123007	13	41

The first two lines represent the familiar 22 and 126 original examples.  The third line represents the one Lars found.  To my knowledge the rest are new.

The corresponding self-stuffable primitive root can be computed from the parameters in the table using the following formula:

Sspr_M_m =  M* Sum(10^(k*L),{k,0,2m}) where L is the sum of digits of M.  

These primitive solutions may be extended by similarly concatenating any odd number of copies of a primitive solution.  

Ray


> 
> Thanks Ray
> 
> Ottieni Outlook per Android<https://aka.ms/ghei36>
> 
> ________________________________
> From: SeqFan <seqfan-bounces at list.seqfan.eu> on behalf of M. F. Hasler
> <seqfan at hasler.fr>
> Sent: Wednesday, January 2, 2019 10:11:21 PM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: Self-stuffable numbers
> 
> On 2 Jan 2019 12:15, "Hans Havermann" <gladhobo at bell.net> wrote:
> 
> MFH: "The main challenge is to find larger (primitive) roots, i.e., terms of
> A322323 (without repetitions and) with trailing zeros removed.
> Arbitrarily many larger terms of A322323 are then computed
> straightforwardly."
> 
> Lars Blomberg has found 210210210210210210210 to be self-stuffable. While
> it is trivially true that it could have been computed from the primitive root
> 21021021021021021021, I don't believe that a b-file for A322002 with terms
> that large is in the offing.
> 
> 
> Such "sporadic" solutions are unavoidable.
> For  n = 210  with  S(n) = 221100 we have S(n) = -30 (mod 210) and more
> generally for n_m = n * R_m(10^3),  R_m = (X^m - 1)/(X - 1) we have S(n_m)
> = R_m(10^6) * S(n) = - 30 m  (mod 210) i.e., S( R_2(10^3) * 210 ) = S( 210 210 )
> = 150  (mod 210) S( R_3(10^3) * 210 ) = S( 210 210 210  ) = 120  (mod 210) ...
> and finally  S( R_7(10^3)*210 ) = 0 (mod 210) and it turns out that also
>  R_7(10^3)*210  |  R_7(10^6)*221100
> so that  210_7 = 210 210 210 210 210 210 210 is also self-stuffable (and of
> course this is also the case for 7 replaced by any odd multiple of 7).
> I try to formulate a general sufficient condition for some (probably always
> odd) number of concatenations of n to be eventually self-stuffable (and thus
> yield a primitive root).
> 
> - Maximilian
> PS: I use the occasion to acknowledge improvements due to Ray, in my
> comment in A322323 (clarification M | S(M)) and example in A322002.
> 
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