A214365 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.

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

Offset: 0

Author

Eric Angelini and M. F. Hasler, Feb 16 2013

Keywords

• nonn
• base

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) 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.

From Michael S. Branicky, Dec 14 2020: (Start)

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)

Examples

			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]
		

Links

• Lars Blomberg, Table of n, a(n) for n = 0..10000
• Eric Angelini (and replies from others), Fibonaccit, posts to the SeqFan list, Feb 15 2013.
• Hans Havermann, Chart of the first differences of the indices of 18 in A214365

Crossrefs

Cf. A093086.

Programs

Mathematica

Module[{rd = {0, 1}, a}, Join[rd, Table[rd = Join[Rest[rd], IntegerDigits[a = Total[Take[rd, 2]]]]; a, 100]]] (* Paolo Xausa, Aug 22 2025 *)

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)
print(aupto(89)) # Michael S. Branicky, Dec 14 2020