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Journal Articles The Fibonacci Quarterly Year : 2021

Three Cousins of Recamán's Sequence

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Richard Schroeppel
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Scott R Shannon
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  • PersonId : 1078136
Neil James Alexander Sloane
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  • PersonId : 1078137

Abstract

Although $10^{230}$ terms of Recamán's sequence have been computed, it remains a mystery. Here three distant cousins of that sequence are described, one of which is also mysterious. (i) ${A(n), n\geq 3}$ is defined as follows. Start with $n$, and add $n+1, n+2, n+3$, . . ., stopping after adding $n + k$ if the sum $n + (n + 1) + . . . + (n + k)$ is divisible by $n + k + 1$. Then $A(n) = k$. We determine $A(n)$ and show that $A(n) \leq n2 − 2n − 1$. (ii) ${B(n), n\geq 1}$ is a multiplicative analog of ${A(n)}$. Start with $n$, and successively multiply by $n + 1, n + 2, . . .,$ stopping after multiplying by $n + k$ if the product $n(n + 1)⋯(n + k)$ is divisible by $n + k + 1$. Then $B(n) = k$. We conjecture that $log^2 B(n) = (\frac{1}{2} + o(1)) \log {n} \log \log n$. (iii) The third sequence, ${C(n), n\geq 1}$, is the most interesting, because the most mysterious. Concatenate the decimal digits of $n, n + 1, n + 2, . . .$ until the concatenation $n\Vert n + 1 . . . \Vert n + k$ is divisible by $n + k + 1$. Then $C(n) = k$. If no such $k$ exists we set $C(n) = −1$. We have found $k$ for all $n \leq 1000$ except for two cases. Some of the numbers involved are quite large. For example, $C(92) = 218128159460$, and the concatenation $92\Vert 93\Vert . . . \Vert(92+C(92))$ is a number with about $2 \cdot {10}^{12}$ digits. We have only a probabilistic argument that such a $k$ exists for all $n$.
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Dates and versions

hal-02951011 , version 1 (28-09-2020)
hal-02951011 , version 2 (02-06-2021)

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Max Alekseyev, Joseph Samuel Myers, Richard Schroeppel, Scott R Shannon, Neil James Alexander Sloane, et al.. Three Cousins of Recamán's Sequence. The Fibonacci Quarterly, In press, ⟨10.48550/arXiv.2004.14000⟩. ⟨hal-02951011v2⟩
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