A374924 Zero-avoiding Fibonacci sequence: a(n) is the largest zeroless number that can be written as a(i) + a(j) where 1 ≤ i < j < n with a(1) = a(2) = 1.

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 521, 898, 1419, 2317, 3736, 5155, 8891, 12627, 21518, 34145, 55663, 77181, 132844, 166989, 299833, 466822, 766655, 1233477, 1366321, 2599798, 3966119, 6565917, 9165715, 15731632, 24897347, 31463264, 47194896, 62926528, 94389792, 141584688, 188779584

Offset: 1

Author

Bryle Morga, Jul 24 2024

Keywords

• nonn
• base

Comments

Matches the Fibonacci sequence for the first 14 terms. This breaks after the 15th term because the 15th term of the Fibonacci sequence contains a 0.

Empirically, the ratio between consecutive term approaches 1. Is this sequence eventually constant?

Examples

			a(15) = 521 because:
a(13) + a(14) = 233 + 377 = 610. (contains a 0.)
a(12) + a(14) = 144 + 377 = 521.
		

Links

• Bryle Morga, Table of n, a(n) for n = 1..10000
• Bryle Morga, Log plot of the first 100,000 terms.
• Bryle Morga, Scatter plot of ratios between consecutive terms for the first 100,000 terms.

Crossrefs

Cf. A000045, A004719.

Programs

Python

from itertools import islice
def z(n): return int(str(n).replace("0", ""))
def agen(): # generator of terms
    yield 1
    alst = [1, 1]
    an = 1
    while True:
        yield an
        an = max(max(z(ai+an) for ai in alst[:-1]), an)
        alst.append(an)
print(list(islice(agen(), 45))) # Michael S. Branicky, Jul 24 2024

Formula

a(n+1) = max{a(n), max{A004719(a(i)+a(n)) for 1 <= i < n}}. - Michael S. Branicky, Jul 24 2024