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All odd numbers in A003592 are powers of $5$, so this is equivalent to finding all $n \ge 0$ such that $5^n = 2^m+1$ or $2^m - 1$ for some $m \ge 0$.

By quick brute force computation I cannot seem to find any consecutive pairs other than $(1,2)$ and $(4,5)$ for items under $2^{32}$. Furthermore, $(125,128)$ is the only pair that differs by $3$, but $(25,32)$ is the only pair that differs by $7$. I know these are more or less pretty "random" (in the sense that they don't have enough pattern to be worth attention), but I would like to know more about this distribution.

In particular, I would like a mathematical proof that $(1,2)$ and $(4,5)$ are the only pair that differs by one, and in general, I wonder if there is a function for an (approximate?) lower bound on the difference between consecutive terms of A003592.

For quick reference, here is a sorted list of differences below $2^{32}$: https://gist.github.com/SOF3/5c62561770b55fccc261beedd41c9e1f

Btw. A corollary of this conjecture states that $(n,n,2n)$ and $(n,4n,5n)$ are the only tuples taking only (duplicate) terms from A003592 that satisfy an $(a,b,a+b)$ condition. Otherwise, for $a, b \ne 1$, $\gcd(a,b,a+b) = 1 \implies \gcd(a,b) = 1 \implies \text{(wlog) } 5 \mid a \wedge 2 \mid b \implies a + b \not\equiv 5 \pmod{10} \implies 2 \mid a + b \implies \gcd(b, a+b) = 2 \implies 2 \mid a \implies \mathbb F$.

Context why I am interested in this question: I have some OCD where I try to make numbers in my life, e.g. number of steps I walk, number of syllables I talk, etc. be some number in A003592. When I am uncertain where I should count a certain step/syllable, I want the number to be in A003592 no matter it is counted or not.

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    $\begingroup$ Catalan's conjecture, now Mihăilescu's theorem, tells us that the only consecutive perfect powers are $8,9$. $\endgroup$
    – lulu
    Sep 8, 2020 at 15:35
  • $\begingroup$ Great, that answers the specific case for $(1,2)$ and $(4,5)$. But is there any relevant theorem regarding the lower bound of consecutive differences? $\endgroup$
    – SOFe
    Sep 9, 2020 at 4:15
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    $\begingroup$ A lower bound can be obtained from "linear forms in logarithms": see the first answer to this question, where they speak of powers of $2$ and $3$ but the method works equally well for powers of $2$ and $5$. $\endgroup$ Sep 9, 2020 at 4:53

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