|
%I #5 Oct 29 2019 21:06:09
%S 0,0,1,1,1,1,3,1,2,2,3,1,2,1,3,3,3,1,4,1,2,3,3,1,3,3,2,3,5,1,2,1,3,4,
%T 2,4,1,1,3,2,3,1,4,1,5,5,5,1,3,3,3,3,4,1,5,4,6,4,4,1,3,1,4,5,3,3,4,1,
%U 3,4,3,1,5,1,6,4,5,3,4,1,3,3,6,1,4,3,6,6,6,1,5,3,5,5,4,5,3,1,2,5,3,1,4,1,4,3
%N Number of terms in Zeckendorf expansion needed to write the second Fibonacci based variant of arithmetic derivative of n.
%H Antti Karttunen, <a href="/A328848/b328848.txt">Table of n, a(n) for n = 0..20000</a>
%F a(n) = A007895(A328846(n)).
%o (PARI)
%o A328846(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]*fibonacci(2+primepi(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 A328848(n) = A007895(A328846(n));
%Y Cf. A000045, A007895, A324905, A328846, A328847.
%K nonn
%O 0,7
%A _Antti Karttunen_, Oct 29 2019
|
