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The limiting sequence of terms preceding the 0's in A248128.
4

%I #8 Oct 02 2014 05:37:38

%S 0,15,3,9,27,21,6,3,6,18,9,18,24,3,27,24,3,12,6,9,21,6,12,6,9,6,3,18,

%T 27,3,6,18,6,3,18,27,18,9,24,3,21,6,9,18,24,3,18,9,24,3,21,6,24,3,27,

%U 12,6,9,3,6,18,27,24,3,12,6,9,24,3,27,12,6,9,3,6,18,12,6,9,21

%N The limiting sequence of terms preceding the 0's in A248128.

%C It can be shown that the terms in between two 0's of sequence A248128 consist of some additional terms followed by the preceding chunk of terms delimited by two 0's. This means that this sequence has a limit "from right to left", equal to ...,27,3,24,18,9,18,6,3,6,21,27,9,3,15,0. The present sequence lists this limiting sequence, starting with the rightmost term.

%C It seems natural to take the offset equal to 0, cf formula.

%C By construction of A248128, all terms are divisible by 3; A248129(n) = a(n)/3 yields the n-th *digit* preceding a 0 in A248128.

%F a(n) = A248128(m-n) if A248128(m) = 0 and A248128(m-k) > 0 for all 0 < k <= n.

%Y Cf. A248128, A248129.

%K nonn,base

%O 0,2

%A _M. F. Hasler_, Oct 02 2014