Oh, it was my mistake. I read it as "n, n+1 and n+2 have no common digits", which is absolutely not what you wrote. My apologies. I have no insight to offer into the problem you were actually posing. On Thu, Oct 10, 2019 at 1:05 PM Éric Angelini <eric.angelini@skynet.be> wrote:
Hello Allan, once again, my question was vague :-(( Let me rephrase that properly: How many terms does the finite lexicographically earliest seq of distinct terms contain such that a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0.
My first terms are:
S = 1,2,3,4,5,6,7,8,9,10,23,45,16,...
We see that no three successive terms of S share any digit so far. Best, É.
à+ É. Catapulté de mon aPhone
Le 10 oct. 2019 à 17:34, Allan Wechsler <acwacw@gmail.com> a écrit :
Coincidentally, Éric's earlier posts made me think of this question, so I am ready with the answer: the only terms are 0,1,2,3,4,5,6,7,8. For all higher n, either n and n+1 will have the same first digit, or n+1 and n+2 will have the same first digit.
On Thu, Oct 10, 2019 at 7:38 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.
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