Three Cousins of Recaman's Sequence
MA Alekseyev, JS Myers, R Schroeppel… - arXiv preprint arXiv …, 2020 - arxiv.org
MA Alekseyev, JS Myers, R Schroeppel, SR Shannon, NJA Sloane, P Zimmermann
arXiv preprint arXiv:2004.14000, 2020•arxiv.orgAlthough 10^ 230 terms of Recaman'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>= 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)<= n^ 2-2n-1.(ii){B (n), n>= 1} is a multiplicative
analog of {A (n)}. Start with n, and successively multiply by n+ 1, n+ 2,..., stopping after …
Here three distant cousins of that sequence are described, one of which is also
mysterious.(i){A (n), n>= 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)<= n^ 2-2n-1.(ii){B (n), n>= 1} is a multiplicative
analog of {A (n)}. Start with n, and successively multiply by n+ 1, n+ 2,..., stopping after …
Although 10^230 terms of Recaman'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 >= 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) <= n^2 - 2n - 1. (ii) {B(n), n >= 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) = (1/2 + o(1)) log n loglog n. (iii) The third sequence, {C(n), n >= 1}, is the most interesting, because the most mysterious. Concatenate the decimal digits of n, n+1, n+2, ... until the concatenation n||n+1||...||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 <= 1000 except for two cases. Some of the numbers involved are quite large. For example, C(92) = 218128159460, and the concatenation 92||93||...||(92+C(92)) is a number with about 2*10^12 digits. We have only a probabilistic argument that such a k exists for all n.
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