[seqfan] Re: A new digital problem

[Vladimir Shevelev]

It is not difficult to prove that, for the first sequence {a(n) is the smallest m  in case r=1, g_1=n, n>=2}, we have a(n)=A026741(n+1). Maybe, more interesting that, for n>=2, this sequence coincides with A145051 (numerator of the first convergent to the sqrt(n) using the recursion x = (n/x+x)/2). And what is about the second sequence?
If anyone can help me to look at  a few terms of a few first such sequences?
 
Regards,
Vladimir



----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Saturday, April 30, 2011 14:14
Subject: [seqfan] A new digital problem
To: seqfan at list.seqfan.eu

>  Dear SeqFans, 
>  
> I am interested in the following problem.
>  
>  Let g_1,g_2,...,g_r be integers more than 1. Denote 
> G=2*Prod_{i=1,...,r}(g_i-1). Let m be positive integer. Denote 
> s_i(m) sum of digits of mG in base g_i. Then numbers s_i/(g_i-1) 
> are integers.
> Does exist, for given integers  
> 2<=g_1<g_2<...<g_r, any number m=m(g_1,...,g_r) such 
> that all numbers s_i/(g_i-1), i=1,...,r, are even?
>    For example, for g_1=2, g_2=4, g_3=5, we have G=24 
> and the minimal requied m is m=5. Indeed, we have 120_2=1111000, 
> 120_4=1320, 120_5=440. So, s_1(5)/1=4, s_2(5)/3=2, s_3(5)/4=2.
>   In particular, sequences of the following type are of interest:
>  
>   a(n) be the smallest m 1) for r=1, g_1=n, n>=2; 2)  
> for r=2, g_1=2, g_2=n;, n>=3; 3) for r=3, g_1=2, g_2=3, g_3=n, 
> n>=4; 4) for r=4, g_1=2, g_2=3, g_3=4, g_4=n, n>=5, etc.
>  
> Regards,
> Vladimir
> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎