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A374209
Number of terms in Zeckendorf representation needed to write A113177(n), where A113177 is fully additive with a(p) = Fibonacci(p).
2
0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 3, 2, 2, 2, 1, 2, 2, 3, 1, 2, 1, 2, 3, 2, 2, 2, 1, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 3, 1, 2, 1, 2, 3
OFFSET
1,9
COMMENTS
Indices for the first occurrences of k=0..6 are: 1, 2, 9, 63, 693, 7623, 105105.
The claim a(n) <= bigomega(n) is true because A007895(n) is the minimum number of Fibonacci numbers which sum to n, regardless of adjacency or duplication. See Apr 17 2015 comments there.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci Number.
FORMULA
a(n) = A007895(A113177(n)).
a(p) = 1 for all primes p.
a(n) <= A001222(n), see comments.
PROG
(PARI)
A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A113177(n) = if(n<=1, 0, my(f=factor(n)); sum(i=1, #f~, f[i, 2]*fibonacci(f[i, 1])));
A374209(n) = if(isprime(n), 1, A007895(A113177(n)));
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 02 2024
STATUS
approved