Let $f(n)$ be A000045(n), i.e., Fibonacci numbers: $f(n)=f(n-1)+f(n-2)$ for $n>1$ with $f(0)=0$ and $f(1)=1$.
Let $g(n)$ be A072649, i.e., $n$ occurs $f(n)$ times. The sequence begins with $$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7$$
Let $$h(n)=f(g(n)), p(n)=n-h(n), q(n)=f(g(n)-1)$$ Let $$a(n)=[p(n)<2q(n)]\cdot([p(n)<h(n)]\cdot q(n) + [p(n)\geqslant h(n)]\cdot(h(n) - a(p(n)-q(n)))), a(0)=0$$ Here square brackets denote Iverson brackets.
Let $$r(n)=f(g(n)+1), s(n)=n-r(n), t(n) = g(s(n)), u(n)=b(s(n)), v(n)=u(n)-a(s(n))$$ Let $$b(n) = r(n) + [s(n)>0]\cdot([g(n)=(t(n)+2)]\cdot v(n)+[g(n)=(t(n)+3)]\cdot(f(t(n)+1)+v(n))+[g(n)>(t(n)+3)]\cdot u(n)), b(0)=0$$ I conjecture that $b(n)$ is a permutation of the nonnegative integers.
Here is the PARI prog to verify this conjecture:
g(n)=if(n>0, my(A=0); until(fibonacci(A)>n, A++); A-2)
a(n)=if(n>0, my(A=g(n), B=fibonacci(A), C=n-B, D=fibonacci(A-1)); if(C<2*D, if(C<B, D, B - a(C-D))))
b(n)=if(n>0, my(A=g(n), B=fibonacci(A+1), C=n-B, D=g(C), E=fibonacci(D+1), F=b(C), G=a(C), H=F-G); B + if(C>0, (A==(D+2))*H + (A==(D+3))*(E+H) + (A>(D+3))*F))
test(n) = my(A=0); while(!(b(A)==n), A++); A
Is there a way to to prove it? Is there at least one another permutation of the nonnegative integers $c(n)$ such that if we change $u(n)=b(s(n))$ to $u(n)=c(s(n))$, then the resulting sequence $b(n)$ is also a permutation of the nonnegative integers?
