bibkey: cicuttin2016a272170 authors: Andres Cicuttin year: 2016 title: “OEIS A272170, Second most significant bit of Fibonacci numbers > 1 written in base 2” doi: null url: https://oeis.org/A272170 claim: “%N Second most significant bit of Fibonacci numbers > 1 written in base 2. %C It is conjectured that there are no more than two consecutive "0’s" or "1’s" (tested up to n=10^5). The sequence looks quasiperiodic and its Fourier spectrum seems to have a fractal structure. - Andres Cicuttin, Apr 21 2016” strata_touched:
- D5/S3/Arith/FibonacciSecondBitRunLength license: citation-only triage: anchor
OEIS A272170
The entry lists, for each Fibonacci number exceeding one, the binary digit just
below the leading one. The sequence begins
0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, …, starting at
fib 3 = 2.
Verified locator
- URL: https://oeis.org/A272170
- Locator: COMMENTS, “It is conjectured that there are no more than two consecutive “0’s” or “1’s” (tested up to n=10^5). The sequence looks quasiperiodic and its Fourier spectrum seems to have a fractal structure. - Andres Cicuttin, Apr 21 2016“
- Revision read: #34, Dec 08 2025. The comment is present and carries no answer.
Reading of the statement
Writing size m for the number of binary digits of m, the entry’s term for a
Fibonacci number m > 1 is m / 2 ^ (size m - 2) % 2. The assertion is that no
three consecutive terms agree.
Scope of the recorded answer
The comment holds. Two facts carry it. First, the pair of bounds
8 · fib n ≤ 5 · fib (n+1) and 8 · fib (n+1) ≤ 13 · fib n holds for every
n ≥ 5; the content of that pair is that the interval [8/5, 13/8] is carried
into itself by x ↦ 1 + 1/x, so each half feeds the other across a step, with
both ends attained at n = 5. Second, bracketing a number between two
consecutive powers of two turns the second digit into an arithmetic condition:
if 2 · 2^L ≤ m < 4 · 2^L then the digit is one exactly when 3 · 2^L ≤ m.
Together these give short contradictions. Three ones would force the third
Fibonacci number below 12 · 2^L while its window demands at least that; three
zeros would force the third at least 6 · 2^L inside a window where that already
means a one.
No real analysis enters: the argument needs neither logarithms, nor the
irrationality of log 2 of the golden ratio, nor equidistribution.
Bounded prior-resolution evidence
The entry was read in full at revision #34 and records no proof and no reference to one. A companion comment of the same shape sits at A271591 for the tribonacci numbers, also unanswered. Citation indices were not exhaustively reachable, so this is a bounded negative finding and no worldwide priority claim is made.