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Problem 167

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Investigating Ulam sequences

For two positive integers a and b, the Ulam sequence U(a,b) is defined by U(a,b)₁ = a, U(a,b)₂ = b and for k > 2,
U(a,b)_k is the smallest integer greater than U(a,b)_(k-1) which can be written in exactly one way as the sum of two distinct previous members of U(a,b).

For example, the sequence U(1,2) begins with
1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;
5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.

Find ∑U(2,2n+1)_k for 2 ≤ n ≤10, where k = 10¹¹.

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乌拉姆序列研究

任取两个正整数a和b，乌拉姆序列U(a,b)按如下方式定义：U(a,b)₁ = a，U(a,b)₂ = b，对于k > 2，U(a,b)_k是比U(a,b)_(k-1)更大，且存在用U(a,b)之前的这些项中的不同两项之和唯一表示的最小整数。

例如，序列U(1,2)的开头部分如下所示
1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;
5不在这个序列是，因为5 = 1 + 4 = 2 + 3，有两种表示方法，同样地7也是如此因为7 = 1 + 6 = 3 + 4。

对于2 ≤ n ≤10，求∑U(2,2n+1)_k，其中k = 10¹¹.

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