[seqfan] Re: Fractions with no repeated digits.

[Éric Angelini]

Waow, good job, Maximilian --
and fun serendipity seqs appearing 
(from the OEIS) indeed!
Thanks and
Best,
É.



> Le 10 oct. 2019 à 23:41, M. F. Hasler <seqfan at hasler.fr> a écrit :
> 
>> On Thu, Oct 10, 2019 at 11:09 AM M. F. Hasler wrote:
>> 
>> In the spirit of Eric's idea, it would rather be :
>> T(m,k) = #{ t  s.t. concat(t*m, t*k) has no digit twice (or more)}
>> Then it's always finite since we must have concat(t*m, t*k) < 9876543210.
>> and  T(m,k) = T(k,m)  and  T(m,m) = 0  so one could/should impose k < m
>> and also T(m,k) = 0 whenever 10 | m and 10 | k, or m = k*10^z.
>> 
> 
> I started to compute this.
> My simple brute-force program stops searching further when there is no new
> member among all integers of a given length.
> (Would have to double-check that this yields the correct result.
> At least the values T(2,1) = 304 and T(3,1) = 153 correspond to Eric's
> count of "fractions n/2n and n/3n".
> But for T(20,19) it finds 14 -- one more than Eric's list... in which 720
> (i.e., t = 36) was missing.)
> 
> For this table, with 1 <= n <= m, I find
> 304, (m=2)
> 153, 157,
> 197, 124, 97,
> 221, 156, 69, 171,
> 73, 88, 142, 68, 69,
> 129, 73, 81, 86, 62, 46,
> 189, 88, 40, 67, 48, 51, 24,
> 89, 80, 77, 31, 63, 68, 41, 20,  (m=9)
> 0, 132, 80, 90, 58, 32, 63, 99, 37, (m=10)
> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,  (m=11)
> 76, 0, 96, 31, 62, 54, 27, 31, 49, 41, 0,  (m=12)
> 84, 72, 0, 31, 58, 47, 26, 23, 34, 43, 0, 20,
> 100, 64, 52, 0, 51, 44, 51, 42, 22, 38, 0, 18, 20,   (m=14)
> ...
> In theory we could also consider column k=0.
> However, this is harder to compute. It means to find all the multiples of m
> which don't have a digit 0 and all digits distinct (which ensures the count
> is finite, as before).
> I found that T(1,0) = oeis.org/A285268(9,9) = 986409.
> For T(2,0) I don't have the result yet.
> The count of t's with d=1,2,3... digits are
> (8,56,364,2016,9240,33600,90720,161280, ...)
> = 4 * (2, 14, 91, 504, 2310, 8400, 22680, 40320, ...)
> (sequences not in OEIS -- any inspiration ?)
> 
> I noticed a fun fact when I tried to compute T(2,0) and forgot the "unique
> digits" condition, i.e., I computed
> Z2(L) :=  #{ k < 10^L | 2k has no digit 0 }
> and I found
> Z2 = ( 8, 76, 688, 6196, 55768, 501916, 4517248, 40655236, ...)
> = 4 * ( 2, 19, 172, 1549, 13942, 125479, 1129312, 10163809, ...)
> where seems to appear oeis.org/A224753, Eric's (a priori totally unrelated)
> "magic sequence" s.th. a(n) + a(n+1) = concat(a(n), 1st digit of a(n+1)).
> 
> I did not check whether this remains true (Z2 =?= 4 * A224753), and if so,
> why.
> (It might work only that far because the terms a(2..8) start with digit
> '1', but idk.)
> Maybe someone wants to look at this!
> 
> - Maximilian
> 
> 
>>> On Wed, Oct 9, 2019 at 7:17 PM Éric Angelini  wrote:
>>> As MH wrote me in a private mail, the word "fraction"
>>> is useless and misleading.
>>> What I meant is {n,2n} doesn't show any repeated digit...
>>> Sorry for that.
>>> Best,
>>> É.
>>> 
>>>> Le 8 octobre 2019 à 14:24, Éric Angelini a écrit :
>>>> Hello SeqFans,
>>>> Here are the 304 integers such that the fraction n/2n repeats no digit:
>>>> 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 17, 18, 19, 23, 27, 28, 29,
>> (...)
>>>> 46185, 46851, 48135, 48351, 48513, 48516, 48531, 48615, 48651.
>>>> 
>>>> And here are the 153 integers such that the fraction n/3n repeats
>> no digit either:
>>>> 1, 2, 3, 4, 6, 7, 8, 9, 12, 16, 18, 19, 21, 23, 26, 27, 29, 32, 34, 36,
>>>> (...), 9136, 9168, 16794, 17694, 20583, 23058, 30582, 32058.
>>>> 
>>>> Those examples suggest more of such sequences. For instance 20/19 would
>>> produce the finite (and quite uninteresting, I have to admit) seq 20, 40,
>>> 60, 80, 620, 920, 4160, 7320, 8360, 51360, 52380, 72540, 91320. We see
>>> indeed that 20/19 = 91320/86754, both frations repeating no digits.
>> 
> 
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