题目167：研究Ulam数列
Investigating Ulam sequencesFor 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。
题目：
对任意两整数 a，b，Ulam 数列 U(a,b) 定义如下：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) 大的，并且只能唯一表示为数列中以前的两项之和的最小的整数。
比如，U(1,2) 数列前几项如下：
1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8；
5 不属于该序列，因为 5 = 1 + 4 = 2 + 3，有两种不同的两个前项和的表达方式， 7 = 1 + 6 = 3 + 4情况也是一样的。
请给出 ∑U(2,2n+1)k 的结果，其中 2 ≤ n ≤10， k = 1011。
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