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A214365
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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.
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3
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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, 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, 2, 6, 10, 8, 6, 4, 10, 10, 5, 11, 11, 8, 13, 10, 5, 12, 10, 8, 7, 1, 8, 14, 10, 5
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internal format)
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OFFSET
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0,4
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COMMENTS
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Offset chosen in analogy to the classical Fibonacci sequence. But the present sequence has no term larger than 9+9=18.
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.
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.
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).
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)
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LINKS
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Eric Angelini (and replies from others), Fibonaccit, posts to the SeqFan list, Feb 15 2013.
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EXAMPLE
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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.)
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]
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PROG
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(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}
(Python)
def aupto(n):
alst, remaining = [0, 1], [0, 1]
for i in range(2, n+1):
an = remaining.pop(0) + remaining[0]
alst.append(an)
remaining.extend(list(map(int, str(an))))
return alst # use alst[n] for a(n)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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