The order of appearance of integers at most one away from Fibonacci numbers (Q2891257)

\{This is a joint review also for Zbl 1268.11025.\}NEWLINENEWLINENEWLINENEWLINEThe papers under review, which we label as P1 (The order of appearance of integers at most one away from Fibonacci numbers) and P2 (The order of appearance of product of consecutive Fibonacci numbers), are concerned with the function \(z(n)\) where \(n\in\mathbb{N},\) (which denotes the positive integers in what follows). These papers continue the author's research reported in earlier papers.NEWLINENEWLINEThe function \(z(n)\) orders the positive integers according to their appearance as divisors of elements of the Fibonacci sequence, \(\{ F_n \}.\) \(z(n) = k\) where \(n\mid F_k\) and \(n \nmid F_m, m<k\). For example \( Z(F_n)=n. \) In the same spirit, in P2, it is shown that NEWLINE\[NEWLINEz(F_nF_{n+1}) = n(n+1), n \geq 3,NEWLINE\]NEWLINE and that NEWLINE\[NEWLINEz(F_nF_{n+1}F_{n+2}) =\begin{cases} n(n+1)(n+2), &n \text{ odd}, \\ \frac{n(n+1)(n+2)}{2}, &n\text{ even}.\end{cases}NEWLINE\]NEWLINE NEWLINE\[NEWLINEz(F_nF_{n+1}F_{n+2}F_{n+3}) = \begin{cases} \frac{n(n+1)(n+2)(n+3)}{2},&n \not\equiv 0\pmod{3}, \\ \frac{n(n+1)(n+2)(n+3)}{3}, &n \equiv 0,9\pmod{12}, \\ \frac{n(n+1)(n+2)(n+3)}{6}, &n \equiv 3, 6\pmod{12}.\end{cases}NEWLINE\]NEWLINE Armed with these formulae, the author is prompted to consider the obvious generalization of these results. In [\textit{N. J. A. Sloane} (ed.), The on-line encyclopedia of integer sequences (2010), \url{http://oeis.org/}], the notion of ``Fibonacci factorial'' was introduced: NEWLINE\[NEWLINEn_F! = F_1F_2 \cdots F_n.NEWLINE\]NEWLINE The conjecture then is NEWLINE\[NEWLINEz(n_F!)\mid n!NEWLINE\]NEWLINE with the consequence: NEWLINE\[NEWLINEF_1 \cdots F_n\mid F_{n!}.NEWLINE\]NEWLINE The conjecture is true for \(n=1,2,\dots,10\). However, \(z(110_F!) \nmid 110!.\) Thus \(n=11\) is a counterexample to the conjecture.NEWLINENEWLINENEWLINEThe second theorem in P2 concerns the related function \(\nu_p(n)\) which gives the highest power of the prime \(p\) that divides \(n\), the so-called \(p\)-adic order of \(n\). The theorem tells us that \(v_p(F_1 \cdots F_n) \leq \nu_p(F_{n!}) \) for \(p \in \{2,3,5,7\}\). It is this theorem which is used to defeat the conjecture.NEWLINENEWLINEThe first of the two papers under review, P1, is concerned with the value of \(z(n)\) for integers which are one away from a Fibonacci number. The first theorem and its corollary are portmanteau results for which we give as a sample parts (i) of each of the Theorem and the Corollary: NEWLINE\[NEWLINEz(F_{4m} \pm 1 ) = 2(4m^2 - 1) \text{ if }m > 1,NEWLINE\]NEWLINE and NEWLINE\[NEWLINEz(F_{12m}^2 - 1) = 2(36m^2 - 1),\text{ if } m\geq 1.NEWLINE\]NEWLINE NEWLINEThe second theorem in the paper (Theorem 1.3) deals with the order of the values of \(z(F_m)\) for three consecutive values of \(F_m\) where \(m\) takes on a certain proper subset of integer values. For example, part (i) of the two part theorem tells us that NEWLINE\[NEWLINEz(F_{12t+7} -1) < z(F_{12t+7}) > z(F_{12t+7} + 1). NEWLINE\]NEWLINE P1 culminates with a proposition that combines both the Lucas numbers and the Fibonacci numbers. Thus: NEWLINE\[NEWLINEz(F_mL_n) = 2mn, \text{ where }m,n \in\mathbb{N}, m \text{ odd}, n > 1, \gcd(m,n)=1.NEWLINE\]NEWLINE And the difficulty in getting divisor theorems of the sort achieved for the Fibonacci numbers is discussed and some conjectures offered.NEWLINENEWLINEThe proofs of the results in these two papers use the standard identities satisfied by the numbers in the Fibonacci sequence, and in the case of the second theorem in P1, well-known identities between the Fibonacci sequence and the Lucas sequence. These identities are listed for the reader. In particular the Binet formula plays an important role in the arguments.NEWLINENEWLINEThe results are a useful extension of our knowledge about, especially, the Fibonacci numbers.