cp's OEIS Frontend

A248128 Start with a(0)=50, then a(n) = three times the n-th digit of the sequence, for all n > 0.

%I A248128 #11 Dec 23 2024 14:53:44
%S A248128 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 A248128 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 A248128 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 A248128 Start with a(0)=50, then a(n) = three times the n-th digit of the sequence, for all n > 0.
%C A248128 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 A248128 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 A248128 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 A248128 E. Angelini, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2014-October/013722.html">Brute force density: triples and cubes</a>, SeqFan list, Oct 01 2014
%e A248128 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 A248128 (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 A248128 nonn,base
%O A248128 0,1
%A A248128 _Eric Angelini_ and _M. F. Hasler_, Oct 02 2014