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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, 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
1,3
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 approach 1. Is this sequence eventually constant?
FORMULA
a(n+1) = max{a(n), max{A004719(a(i)+a(n)) for 1 <= i < n}}. - Michael S. Branicky, Jul 24 2024
EXAMPLE
a(15) = 521 because:
a(13) + a(14) = 233 + 377 = 610. (contains a 0.)
a(12) + a(14) = 144 + 377 = 521.
PROG
(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
CROSSREFS
Sequence in context: A107358 A374266 A243063 * A248740 A185357 A132636
KEYWORD
nonn,base
AUTHOR
Bryle Morga, Jul 24 2024
STATUS
approved