[seqfan] Re: Is there a better way to find the terms of this sequence (other than trial and error)? Correction

[David Seal]

> I guess an obvious question about this sequence is: Can we
> prove that this sequence has values for all natural numbers?

No, we cannot prove that, because for n = 625, your requirement is that the digits comprising 625i+1 include at least one 2, at least one 6 and at least one 7 (the distinct digits of n+1 = 626 and n+2 = 627) and no digits other than 2, 6 and 7. But 625i+1 is 1 if i=0 and 626 if i=1, neither of which contains a 7, and is otherwise a >=4-digit number ending with one of:

0001 0626 1251 1876 2501 3126 3751 4376 5001 5626 6251 6876 7501 8126 8751 9376

none of which contain only 2s, 6s and 7s.

So a(625) doesn't have a value.

David


> On 10 November 2019 at 19:07 Ali Sada via SeqFan <seqfan at list.seqfan.eu> wrote:
> 
> 
>  Thank you very much Dr. Hasler for your response. I really appreciate it. 
> 
> First, I am sorry for the huge blunder at a(9.) Sometimes I automatically exclude n+1. 
> I used the word "all" so that we include the all the distinct digits of n+1 And n+2. Otherwise, we could use only the digits of n+1 (and that would be just the natural numbers.) For example, for n=34, the digits of a(n) should include 3,5, and 6. We can repeat any digit as many times as we like.
> For a(18)=1122229, we can argue that there is a digit 0 at the far left of the number. However, if we insist on having it "inside" the number then a(18)=10122229.
> What I meant in my question was to have an analytical way to find the terms of the sequence. For example, between n=x and n=y the values of a(n) could be found by the function f(x,y.)  
> 
> I guess an obvious question about this sequence is: Can we prove that this sequence has values for all natural numbers?
> Best,
> Ali
> 
> 
> 
>     On Sunday, November 10, 2019, 10:57:45 AM EST, M. F. Hasler <oeis at hasler.fr> wrote:  
>  
>  On Sun, Nov 10, 2019 at 2:54 AM Ali Sada wrote:
> 
> > a(n)=m+1, where m is the least multiple of n such that a(n) digits consist
> > only of all the distinct digits of n+1 and n+2.
> >
> 
> It depends what you call "better" and what you call "trial and error".
> You can indeed write a program the avoids testing "all" multiples.
> For example, if the first digit is not in the allowed set, you can skip
> increasingly many multiples.
> 
> There are certainly "even more intelligent" (lol) approaches.
> But most probably the programs get much longer. I think that testing
> leading digits and skipping contiguous ranges where a given digit excludes
> success will maybe a good length/efficiency compromise.
> 
> For example, 209 is the least multiple of 19 where m (209+1=210) consists
> > only of the distinct digits of 20 and 21.
> 
> 23, 43, 445,65, 76, 787, 988, 1009, 1111111111, 21, 1321, 11413, 4551, 561,
> > 11176, 8817, 9181,1122229, 210, 221, 232, 243, 254, 265, 276, 287, 298,
> > 22093, 1103, 3211, 32, 34433,3334
> 
> 
> I get different values.
> I notice that you say "only all the digits of n+1 and n+2",
> but on one hand, 445 has twice the digit of 4, idem for 1009,
> so I am not sure whether you really don't want to allow a(9) = 10 = 9+1
> which has all and only digits of 10 and 11 -- or maybe that would be OK ?
> Also, your a(18) does not contain the 0 of 20, several other following
> values are smaller than yours:
> 
> apply(
> {a(n,d=eval(Set(Vec(Str(n+1,n+2)))))=forstep(m=n,oo,n,Set(digits(m+1))==d&&return(m+1))},
> [1..20])
> % = [23, 43, 445, 65, 76, 787, 988, 1009, 10, 21, 1123, 11341, 1145, 561,
> 11176, 817, 9181, 10122229, 210, 21]
> 
> For a(9) and a(20), I'm not sure whether you mean "multiple of n larger
> than n":
> If so, "m=n" must be replaced by "m=2*n" in my code which then gives
> a(9) = 100 = 9*11 + 1 (still much smaller than 11...11), and
> a(20) = 121 = 20*6 + 1, having only and all digits of 21 and 22, but
> also smaller than your 221.
> 
> - Maximilian
> 
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