做题了做题了

[YWY]

参见 viewtopic.php?t=88734

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

我感觉下面的命题成立，大家可试着证一下

命题：一个正整数n不在集合S里当且仅当n是2的方幂（a natural number n is not in S if and only if n = 2^i for some non-negative integer i）。

[verdelite]

参见 viewtopic.php?t=88734

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

我感觉下面的命题成立，大家可试着证一下

命题：一个正整数n不在集合S里当且仅当n是2的方幂（a natural number n is not in S if and only if n = 2^i for some non-negative integer i）。

根据这个部分解 viewtopic.php?t=88734#p545031，只需要证明2的幂不能被表示为连续正整数的和。从m开始连续n个正整数的和可以表示为mn+ n(n-1)/2；先证明n不能为奇数；假定n为奇数，则这个和有公约数n，根据题意n=2的幂，矛盾；再假定n为偶数2k，则前式为2mk+k(2k-1)=k(2m+2k-1)，即奇数2m+2k+1为原数的约数，这和题意矛盾。证毕

[YWY]

根据这个部分解 viewtopic.php?t=88734#p545031，只需要证明2的幂不能被表示为连续正整数的和。从m开始连续n个正整数的和可以表示为mn+ n(n-1)/2；先证明n不能为奇数；假定n为奇数，则这个和有公约数n，根据题意n=2的幂，矛盾；再假定n为偶数2k，则前式为2mk+k(2k-1)=k(2m+2k-1)，即奇数2m+2k+1为原数的约数，这和题意矛盾。证毕

赞！

[（ヅ）]

赞！

这不还差一半呢？

另外这个结果怎么会跟2有关系呢？有什么表面下的原因？

[（ヅ）]

这不还差一半呢？

另外这个结果怎么会跟2有关系呢？有什么表面下的原因？

想了个构造性证明，z作为一个正整数，如果不是2^i，那么一定有一个奇数质数因子，记为p，记q = z / p. 那么z就可以分解成p个以q为中心的连续整数相加。比如3*4 = (4-1) + 4 +（4+1)

如果这个p/2比q小，那么就不会在左边越界超过0，自然成立
如果p/2比q大，那会是这么一种情况,比如4*11，分解成
-1 + 0 +1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

这时候左边超出小于0的部分，跟其对称的大于0 部分(这个例子里面-1 + 0 + 1）可以消掉，从而留下剩下的2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

这也解释了为什么只要不是2^i就可以(有奇数因数)

再补充一下两个猜想，

• 对于不同的基数因子q，这样的拆解是唯一的。举例:9拆成q = 9，那么得到的是p=1,求和是-3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 = 4 + 5.如果取q = 3, p = 3，得到的求和是2 + 3 + 4. 不同的q(3和9)，造成的sequence一定不同(4 + 5和2+3+4)

• 所有的"大于1的奇数因数"构造了所有的拆解。举例:9就只有4 + 5和2 + 3 + 4两种

[YWY]

想了个构造性证明，z作为一个正整数，如果不是2^i，那么一定有一个奇数质数因子，记为p，记q = z / p. 那么z就可以分解成p个以q为中心的连续整数相加。比如3*4 = (4-1) + 4 +（4+1)

如果这个p/2比q小，那么就不会在左边越界超过0，自然成立
如果p/2比q大，那会是这么一种情况,比如4*11，分解成
-1 + 0 +1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

这时候左边超出小于0的部分，跟其对称的大于0 部分(这个例子里面-1 + 0 + 1）可以消掉，从而留下剩下的2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

这也解释了为什么只要不是2^i就可以(有奇数因数)

赞！

[（ヅ）]

赞！

想了两个延伸问题，大家继续

[verdelite]

谁有时间可以看看openAI证明得对不对？

me: Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Prove that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.

chatGPT: First, let's prove that if n is in S, then n cannot be equal to 2^i for any non-negative integer i. Suppose for the sake of contradiction that n is in S and n = 2^i for some non-negative integer i. This means that n can be expressed as the sum of two or more consecutive positive integers. Let's consider the smallest such sum, which must have at least two terms.

Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:

x + (x+1) + (x+2) + ... + (x+k) = n

If we simplify this equation, we get:

kx + (1+2+...+k) = n

kx + k(k+1)/2 = n

If we solve for k in this equation, we get:

k = [2(n - (k(k+1)/2))]/k

Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2. However, since n is equal to 2^i for some non-negative integer i, n is an even number, so n - (k(k+1)/2) must also be an even number. Therefore, k/2 must be an integer. But since k is a positive integer, it follows that k must be even.

Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:

(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))

This simplifies to:

(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)

If we simplify this further, we get:

(k+1)x + k(k+1)/2 = n

But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is in S, then n cannot be equal to 2^i for any non-negative integer i.

Now, let's prove the converse: if n is not equal to 2^i for any non-negative integer i, then n is not in S. Suppose for the sake of contradiction that n is not equal to 2^i for any non-negative integer i, but n is in S. Then n can be expressed as the sum of two or more consecutive positive integers.

Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:

x + (x+1) + (x+2) + ... + (x+k) = n

If we simplify this equation, we get:

kx + (1+2+...+

me: it is not done yet. please continue.

chatGPT: Sure, I apologize for the incomplete proof. Here is the rest of the proof:

kx + (1+2+...+k) = n

kx + k(k+1)/2 = n

If we solve for k in this equation, we get:

k = [2(n - (k(k+1)/2))]/k

Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2.

We know that n is not equal to 2^i for any non-negative integer i, so n cannot be an even number. Therefore, n - (k(k+1)/2) must also be an odd number. However, since k/2 is an integer, it follows that k must be even. This means that k/2 is also an integer.

Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:

(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))

This simplifies to:

(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)

If we simplify this further, we get:

(k+1)x + k(k+1)/2 = n

But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is not equal to 2^i for any non-negative integer i, then n is not in S.

This completes the proof that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.

[YWY]

谁有时间可以看看openAI证明得对不对？

me: Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Prove that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.

chatGPT: First, let's prove that if n is in S, then n cannot be equal to 2^i for any non-negative integer i. Suppose for the sake of contradiction that n is in S and n = 2^i for some non-negative integer i. This means that n can be expressed as the sum of two or more consecutive positive integers. Let's consider the smallest such sum, which must have at least two terms.

Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:

x + (x+1) + (x+2) + ... + (x+k) = n

If we simplify this equation, we get:

kx + (1+2+...+k) = n

kx + k(k+1)/2 = n

If we solve for k in this equation, we get:

k = [2(n - (k(k+1)/2))]/k

Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2. However, since n is equal to 2^i for some non-negative integer i, n is an even number, so n - (k(k+1)/2) must also be an even number. Therefore, k/2 must be an integer. But since k is a positive integer, it follows that k must be even.

Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:

(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))

This simplifies to:

(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)

If we simplify this further, we get:

(k+1)x + k(k+1)/2 = n

But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is in S, then n cannot be equal to 2^i for any non-negative integer i.

Now, let's prove the converse: if n is not equal to 2^i for any non-negative integer i, then n is not in S. Suppose for the sake of contradiction that n is not equal to 2^i for any non-negative integer i, but n is in S. Then n can be expressed as the sum of two or more consecutive positive integers.

Without loss of generality, let the smallest term in the sum be x, and let the second smallest term be x+1. Then the sum of the consecutive integers from x to x+k for some positive integer k is equal to n. This can be expressed as:

x + (x+1) + (x+2) + ... + (x+k) = n

If we simplify this equation, we get:

kx + (1+2+...+

me: it is not done yet. please continue.

chatGPT: Sure, I apologize for the incomplete proof. Here is the rest of the proof:

kx + (1+2+...+k) = n

kx + k(k+1)/2 = n

If we solve for k in this equation, we get:

k = [2(n - (k(k+1)/2))]/k

Since k is a positive integer, it follows that 2(n - (k(k+1)/2)) is a multiple of k. In other words, n - (k(k+1)/2) is a multiple of k/2.

We know that n is not equal to 2^i for any non-negative integer i, so n cannot be an even number. Therefore, n - (k(k+1)/2) must also be an odd number. However, since k/2 is an integer, it follows that k must be even. This means that k/2 is also an integer.

Now, let's consider the sum x + (x+1) + ... + (x+k). This sum can be expressed as:

(x + (x+k)) + (x+1 + (x+k-1)) + ... + (x+k/2 + (x+k/2))

This simplifies to:

(2x + k) + (2(x+1) + k-2) + ... + (2(x+k/2-1) + 2) + (2(x+k/2) + 0)

If we simplify this further, we get:

(k+1)x + k(k+1)/2 = n

But we already know that n = kx + k(k+1)/2, so we have arrived at a contradiction. Therefore, if n is not equal to 2^i for any non-negative integer i, then n is not in S.

This completes the proof that a natural number n is not in S if and only if n = 2^i for some non-negative integer i.

how does chatgpt get kx + (1+2+...+k) = n???

[verdelite]

how does chatgpt get kx + (1+2+...+k) = n???

你的惊讶不亚于柯洁吧，哈哈

下面不知道证对了没有。

[YWY]

再补充一下两个猜想，

• 对于不同的基数因子q，这样的拆解是唯一的。举例:9拆成q = 9，那么得到的是p=1,求和是-3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 = 4 + 5.如果取q = 3, p = 3，得到的求和是2 + 3 + 4

• 所有的"大于1的奇数因数"构造了所有的拆解。举例:9就只有4 + 5和2 + 3 + 4两种

想了两个延伸问题，大家继续

关于你的延伸问题，每个合乎条件的序列（2 or more consecutive positive integers），如果数列长度是奇数p，则考虑数列正中间的数q，那么加起来就是n=pq（对应n的奇数因子p）；如果数列长度是偶数2q，考虑正中间的两个数a & a+1，令p = 2a+1，则数列总和是n=pq（对应n的奇数因子p），同时这个数列对应于以q为中心长度为p的数列（然后如你所说划掉正负相约掉的项和零）。

[YWY]

你的惊讶不亚于柯洁吧，哈哈

下面不知道证对了没有。

chapgpt还说We know that n is not equal to 2^i for any non-negative integer i, so n cannot be an even number.，这也不对。整个证明我没仔细看，感觉chatgpt的文科是强项，理科还需多加训练。

[Caravel]

你的惊讶不亚于柯洁吧，哈哈

下面不知道证对了没有。

似乎只证明了一半，就是如果 n = 2^i， n is not in S。我大致follow一下，似乎是对的，当然没有查的那么精细

[（ヅ）]

关于你的延伸问题，每个合乎条件的序列（2 or more consecutive positive integers），如果数列长度是奇数p，则考虑数列正中间的数q，那么加起来就是n=pq（对应n的奇数因子p）；如果数列长度是偶数2q，考虑正中间的两个数a & a+1，令p = 2a+1，则数列总和是n=pq（对应n的奇数因子p），同时这个数列对应于以q为中心长度为p的数列（然后如你所说划掉正负相约掉的项和零）。

我觉得我陈述不清楚，应该是，所有的非1奇数因子跟一个这样的求和sequence有一一对应关系

• 不同的的奇数因子生成的sequence不一样

• 每一个sequence都可以由某一个奇数因子按照这个步骤这样构造(这个是比较显而易见的)

[FGH]

假设n是连续k个数之和。情况1，k是奇数，那么n等于k乘以k个数最中间的那个数，所以是k的倍数。
情况2，k是偶数，那么n等于k乘以k个数中间两个整数的平均值，所以等于k/2乘以那两个整数之和奇数。
两种情况下，n都有一个大于1的奇数因子。

[（ヅ）]

假设n是连续k个数之和。情况1，k是奇数，那么n等于k乘以k个数最中间的那个数，所以是k的倍数。
情况2，k是偶数，那么n等于k乘以k个数中间两个整数的平均值，所以等于k/2乘以那两个整数之和奇数。
两种情况下，n都有一个大于1的奇数因子。

这证明了2^i一定不是S里面的元素。另外一半是需要证明所有的非2^i一定在S里面

[FGH]

如果n不是2的次方，可以写成n=2^i * k，k是大于1的奇数。
或者把n写成连续k个数之和，中间的是2^i。要求(k+1)/2<=2^i。
或者把n写成2^(i+1)个连续数之和，中间两个数之和是k。要求k\le 2^(i+1)+1。
两个条件必定可得其一。

[YWY]

我觉得我陈述不清楚，应该是，所有的非1奇数因子跟一个这样的求和sequence有一一对应关系

• 不同的的奇数因子生成的sequence不一样

• 每一个sequence都可以由某一个奇数因子按照这个步骤这样构造(这个是比较显而易见的)

赞！