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Prove that if $\sigma(F_a)<\sigma(F_b)$, where $1 \leq b < a$, then $\sigma(a)<\sigma(b)$ also. Here $\sigma(x)$ denotes the sum of the divisors of $x$ and $F_x$ is the $x$th Fibonacci number. Additionally, a highly abundant number is a number $k$ for which $\sigma(k) > \sigma(m) \hspace{3 mm} \forall m < k$.

As a special case of the above, if we define $a(n)=\sigma(F_n)$, then whenever $a(n)$ is not a record value of the sequence, n is not highly abundant.


The (now deleted) second part of this question has been moved here so that this one, which has been disproved by @Greg Martin, may be closed.

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  • $\begingroup$ Is your question the statement in the title, or proving that if $F_n$ is highly abundant, then $n$ is highly abundant, or whether the only highly abundant Fibonacci numbers are the ones you know? It might help if you include the definitions of the terms you use (practicality, highly abundant). $\endgroup$ – Gerry Myerson Aug 6 '13 at 9:25
  • $\begingroup$ I find that if I want people to answer my questions, the easier I make it for them, the better I do. So I give them the definitions, instead of making them chase them down. But, suit yourself. Also, it's best if a question is in the body, and not just in the title. $\endgroup$ – Gerry Myerson Aug 6 '13 at 10:29
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The assertion is false for $a=82$ and $b=81$. The 81st and 82nd Fibonacci numbers are $37889062373143906 = 2\times17\times53\times109\times2269\times4373\times19441$ and $61305790721611591 = 2789\times59369\times370248451$ respectively. So \begin{align*} \sigma(37889062373143906) &= 3\times18\times54\times110\times2270\times4374\times19442 \\ &= 61919263145361600, \text{ which is larger than} \\ \sigma(61305790721611591) &= 2790\times59370\times370248452 \\ &= 61328805160719600; \end{align*} but $\sigma(81) = 121$ while $\sigma(82)=126$.

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    $\begingroup$ Great, thanks! Tomorrow I'll edit the post to contain only the first assertion and close this question. I'll also include a link to a new post containing the second assertion about highly abundant Fibonacci numbers in case you'd like to attempt a proof or search for counterexamples. $\endgroup$ – Jaycob Coleman Aug 12 '13 at 10:34

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