

A214365


Digitwise Fibonacci: Start with 0,1; then the next term is always the sum of the earliest two consecutive digits not yet summed so far.


3



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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OFFSET

0,4


COMMENTS

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) 2k1 consecutive 9's somewhere later, which in turn will yield (at least) 2k2 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) k1 consecutive 12's, then 2k3 3's, then 2k4 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 k1 16's, 2k3 7's, 2k4 14's, 4k9 5's, 4k10 10's, 8k21 1's, 8k22 2's, 8k23 2's, then 8k24 8's, leading to infinite growth for k > 3. Four consecutive 8's first occur at a(9606). (End)


LINKS

Eric Angelini (and replies from others), Fibonaccit, posts to the SeqFan list, Feb 15 2013.


EXAMPLE

The sequence starts in the same way as the Fibonacci sequence A000045. But after 5+8=13 follows the digitwise continuation, viz: 8+1=9, 1+3=4, 3+9=12, ... (Due to the presence of 2digit 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]


PROG

(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)


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



