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%I #41 Jan 02 2023 12:30:48
%S 0,1,1,2,3,5,8,13,9,4,12,13,5,3,3,4,8,8,6,7,12,16,14,13,8,3,3,7,7,5,5,
%T 4,11,11,6,10,14,12,10,9,5,2,2,2,7,7,1,1,5,5,3,3,1,9,14,7,4,4,9,14,8,
%U 2,6,10,8,6,4,10,10,5,11,11,8,13,10,5,12,10,8,7,1,8,14,10,5
%N Digit-wise Fibonacci: Start with 0,1; then the next term is always the sum of the earliest two consecutive digits not yet summed so far.
%C Offset chosen in analogy to the classical Fibonacci sequence. But the present sequence has no term larger than 9+9=18.
%C As observed by H. Havermann in reply to the original post (cf. link), a run of k consecutive 18's will yield a run of (at least) 2k-1 consecutive 9's somewhere later, which in turn will yield (at least) 2k-2 consecutive 18's. Since there are such runs of sufficient length (Z. Seidov pointed out that a(n)=9 for 78532 < n < 78598), the sequence cannot become periodic.
%C In what precedes, a value of k >= 3 is sufficient for infinite growth. But a run of only three 9's is also sufficient because the 2 consecutive 18's will be followed by a number >= 10, which then yields four 9's and subsequently infinitely long runs of 9's, cf Example.
%C From _Michael S. Branicky_, Dec 14 2020: (Start)
%C Likewise, a run of k consecutive 6's will yield (at least) k-1 consecutive 12's, then 2k-3 3's, then 2k-4 6's, leading to infinite growth for k > 4. Five consecutive 6's first occur at a(17072).
%C Similarly, a run of k 8's will yield k-1 16's, 2k-3 7's, 2k-4 14's, 4k-9 5's, 4k-10 10's, 8k-21 1's, 8k-22 2's, 8k-23 2's, then 8k-24 8's, leading to infinite growth for k > 3. Four consecutive 8's first occur at a(9606). (End)
%H Lars Blomberg, <a href="/A214365/b214365.txt">Table of n, a(n) for n = 0..10000</a>
%H Eric Angelini (and replies from others), <a href="http://list.seqfan.eu/oldermail/seqfan/2013-February/010810.html">Fibonaccit</a>, posts to the SeqFan list, Feb 15 2013.
%H Hans Havermann, <a href="http://chesswanks.com/num/eighteen.png">Chart of the first differences of the indices of 18 in A214365</a>
%e The sequence starts in the same way as the Fibonacci sequence A000045. But after 5+8=13 follows the digit-wise continuation, viz: 8+1=9, 1+3=4, 3+9=12, ... (Due to the presence of 2-digit terms, the summed digits lag more and more behind the correspondingly computed term.)
%e The first run of 3 consecutive 9's occurs at a(3862)=a(3863)=a(3864)=9, which then yield a(4975)=a(4976)=18, a(4977)=14 and the first run of four 9's at 6392 <= n <= 6395. [_M. F. Hasler_, Feb 17 2013]
%o (PARI) A214365(n,show=0,d=[0,1])={show & print1(d[1]","d[2]); for(i=2,n, n=d[1]+d[2]; show & print1(","n); d=concat(vecextract(d,"^1"),digits(n))); n}
%o (Python)
%o def aupto(n):
%o alst, remaining = [0, 1], [0, 1]
%o for i in range(2, n+1):
%o an = remaining.pop(0) + remaining[0]
%o alst.append(an)
%o remaining.extend(list(map(int, str(an))))
%o return alst # use alst[n] for a(n)
%o print(aupto(89)) # _Michael S. Branicky_, Dec 14 2020
%Y Cf. A093086.
%K nonn,base
%O 0,4
%A _Eric Angelini_ and _M. F. Hasler_, Feb 16 2013
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