论文标题
Rocaman序列的三个堂兄
Three Cousins of Recaman's Sequence
论文作者
论文摘要
尽管已经计算出10^230的术语，但它仍然是一个谜。在这里描述了该顺序的三个遥远的表亲，其中之一也是神秘的。 （i）{a（n），n> = 3}定义如下。从n开始，然后添加n+1，n+2，n+3，...，如果添加n+k后停止，如果总和n+（n+1）+...+（n+k）可以除以n+k+1。然后a（n）= k。我们确定a（n），并表明a（n）<= n^2-2-2n -1。（ii）{b（n），n> = 1}是{a（n）}的乘数类似物。从n开始，然后依次乘以n+1，n+2，...，如果乘积n（n+1）...（n+k）乘以n+k后停止，则可以除以n+k+1。然后b（n）= k。我们猜想log^2 b（n）=（1/2 + o（1））log n loglog n。 （iii）第三个序列{c（n），n> = 1}是最有趣的，因为最神秘。串联n，n+1，n+2的小数位数，...直到连接n || n+1 || ... || n+k被n+k+1分开。然后c（n）= k。如果不存在这样的k，我们将c（n）= -1设置。我们已经找到了所有n <= 1000的k，但两种情况除外。涉及的一些数字很大。例如，C（92）= 218128159460，并串联92 || 93 || ... ||（92+c（92））是一个数字，大约为2*10^12位。我们只有一个概率论点，即所有n都存在这样的k。
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.
