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%I #10 Jan 02 2023 12:30:50
%S 50,15,0,3,15,0,9,3,15,0,27,9,3,15,0,6,21,27,9,3,15,0,18,6,3,6,21,27,
%T 9,3,15,0,3,24,18,9,18,6,3,6,21,27,9,3,15,0,9,6,12,3,24,27,3,24,18,9,
%U 18,6,3,6,21,27,9,3,15,0,27,18,3,6,9,6,12,6,21,9,6,12,3,24
%N Start with a(0)=50, then a(n) = three times the n-th digit of the sequence, for all n > 0.
%C The terms in between 0's in the sequence converge "from right to left" to a limiting sequence ...,18,6,3,6,21,27,9,3,15,0. This sequence is listed in A248129. Sequence A248130 lists the individual digits, starting from some 0 and going to the left (until another 0 would be reached); they are equal to A248129/3.
%C It seems natural to use offset 0 to have the initial term equal to a(0) and a(n) directly related to the n-th digit of the sequence.
%C All terms a(n) with index n>0 are divisible by 3, the sequence a(n)/3 is nothing else than the individual digits of this sequence.
%H E. Angelini, <a href="http://list.seqfan.eu/oldermail/seqfan/2014-October/013722.html">Brute force density: triples and cubes</a>, SeqFan list, Oct 01 2014
%e a(0)=50 by definition, a(1) = 15 = 3*5 (= 3 x the 1st digit of "50"), a(2) = 0 = 3*0 (3 x the 2nd digit of "50,15"), a(3) = 3 = 3*1 (= 3 x the 3rd digit of the sequence which is the 1st digit of a(1) and equals 1).
%o (PARI) a(n,s=50,d=[])={for(i=1,n,print1(s",");d=concat(d,if(s,digits(s)));s=3*d[1];d=vecextract(d,"^1"));s}
%K nonn,base
%O 0,1
%A _Eric Angelini_ and _M. F. Hasler_, Oct 02 2014
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