cp's OEIS Frontend

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.

%I A374924 #16 May 30 2025 19:29:33
%S A374924 1,1,2,3,5,8,13,21,34,55,89,144,233,377,521,898,1419,2317,3736,5155,
%T A374924 8891,12627,21518,34145,55663,77181,132844,166989,299833,466822,
%U A374924 766655,1233477,1366321,2599798,3966119,6565917,9165715,15731632,24897347,31463264,47194896,62926528,94389792,141584688,188779584
%N 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.
%C A374924 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.
%C A374924 Empirically, the ratio between consecutive term approaches 1. Is this sequence eventually constant?
%H A374924 Bryle Morga, <a href="/A374924/b374924.txt">Table of n, a(n) for n = 1..10000</a>
%H A374924 Bryle Morga, <a href="/A374924/a374924.png">Log plot of the first 100,000 terms.</a>
%H A374924 Bryle Morga, <a href="/A374924/a374924_1.png">Scatter plot of ratios between consecutive terms for the first 100,000 terms.</a>
%F A374924 a(n+1) = max{a(n), max{A004719(a(i)+a(n)) for 1 <= i < n}}. - _Michael S. Branicky_, Jul 24 2024
%e A374924 a(15) = 521 because:
%e A374924 a(13) + a(14) = 233 + 377 = 610. (contains a 0.)
%e A374924 a(12) + a(14) = 144 + 377 = 521.
%o A374924 (Python)
%o A374924 from itertools import islice
%o A374924 def z(n): return int(str(n).replace("0", ""))
%o A374924 def agen(): # generator of terms
%o A374924     yield 1
%o A374924     alst = [1, 1]
%o A374924     an = 1
%o A374924     while True:
%o A374924         yield an
%o A374924         an = max(max(z(ai+an) for ai in alst[:-1]), an)
%o A374924         alst.append(an)
%o A374924 print(list(islice(agen(), 45))) # _Michael S. Branicky_, Jul 24 2024
%Y A374924 Cf. A000045, A004719.
%K A374924 nonn,base
%O A374924 1,3
%A A374924 _Bryle Morga_, Jul 24 2024