[seqfan] Re: Fractions with no repeated digits.

[M. F. Hasler]

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.
>