I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}, x>1$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.

I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}, x>1$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7 for the sum, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests for the sums.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.

I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.

I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}, x>1$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.

I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}.$

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7, please check your precision ;)

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for x less than 10^7 and we're still running tests.

They aren't equal without the floor function, and each equal x + C, where C is a constant less than 1, and C is different for the integral and sum. As x tends to infinity, C tends to 1.

Please don't get angry with me, I just want to share and see if anyone has advice on a proof.

I've found through evidence and have conjectured on a math publication that:

$$\Big\lfloor\int_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$

where $ x \in \mathbb{N}$.

It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7, please check your precision.

Proving this has proved extremely difficult.

My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests.

They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.