[seqfan] Fwd: Add neighbours, use their digits

[M. F. Hasler]

On Sat, Nov 2, 2013 at 5:22 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> Hello SeqFans,
> this is a nightmare to compute by hand:
> S = 1,10,2,19,73,100,3,20,4,39,54,...

Hello Eric,
for the computer that's very easy:

u=0;a=1;for(n=1,99,print1(a",");u+=1<<a;for(t=1,9e9,bittest(u,t)&&next;#setminus(Set(digits(a+t)),Set(concat(digits(a),digits(t))))&&next;a=t;break))

1,10,2,19,72,5,20,3,29,63,18,13,68,100,4,30,6,40,7,50,8,60,9,12,79,108,23,69,27,14,28,54,17,24,58,117,124,47,25,126,35,16,45,38,15,
36,26,37,81,32,91,21,80,31,70,41,73,104,39,56,105,46,71,43,101,65,51,64,102,42,149,62,129,67,103,52,139,82,106,34,59,132,86,120,57,
114,127,44,200,11,90,107,92,109,78,110,89,122,97,...

Note that I do not consider multiplicity : "to write 11", it is
sufficient that there is one digit "1" in a(n),a(n+1). Hope that's OK.

Have a nice week-end,

Maximilian

>
> Let R be a(n)+a(n+1)
> We want R to be written with some (or
> all) the digits belonging to a(n) or a(n+1)
> or both.
> S is extended using the smallest
> integer not yet present in S and
> not leading to a contradiction.