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%I #6 Oct 28 2019 20:01:37
%S 0,0,1,1,2,1,2,1,3,3,2,1,3,1,4,3,3,1,4,1,3,4,4,1,3,3,3,4,4,1,3,1,4,3,
%T 5,3,3,1,5,4,3,1,4,1,3,3,4,1,3,4,3,3,6,1,5,3,4,5,4,1,3,1,5,3,4,4,2,1,
%U 5,6,2,1,3,1,4,3,5,5,5,1,3,5,5,1,3,7,5,4,3,1,3,5,5,5,4,6,4,1,3,4,5,1,7,1,6,5
%N Number of terms in Zeckendorf expansion needed to write the first Fibonacci based variant of arithmetic derivative of n.
%H Antti Karttunen, <a href="/A328847/b328847.txt">Table of n, a(n) for n = 0..20000</a>
%F a(n) = A007895(A328845(n)).
%F a(p) = 1 for all primes p.
%o (PARI)
%o A328845(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]*fibonacci(f[i,1])/f[i, 1]));
%o A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
%o A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
%o A328847(n) = A007895(A328845(n));
%Y Cf. A000045, A007895, A324907, A328845.
%K nonn
%O 0,5
%A _Antti Karttunen_, Oct 28 2019
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