User Portlet
| Discussions |
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| Also, Dear David it would be great if you could submit brief comment and your mathematica program to OEIS A223881. Cheers, Alex |
| Dear Mariusz, Is your submission for your formula in A035002 ready to be proposed for review? > |
| Thanks, Dear Gianluca - it was a typo. Yes, I meant Pi ~= Sqrt[4 E - 1] Also there was a typo in the link to the OEIS sequence which I also corrected. What are those two distinct limit points? Thanks, Best Regards, Alex |
| Here is the answer given by GPT-4: If the FullSimplify function in Mathematica simplifies the right-hand side of an equation to look exactly like the left-hand side, it suggests that the identity is mathematically valid. However, whether or not it... |
| Enno resent his proof with more comments and details&[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/52cec594-59b5-47d6-969d-53d4dd1d4237 |
| Besides using AI based ChatGPT as the language interpretation interface to Wolfram Alpha engine, Wolfram Alpha should incorporate elements of AI in its core to assist solving difficult mathematical and logical problems. |
| Thanks for the solution! I wish that WolframAlpha would incorporate/internalize this approach directly into the sum command to make it more straightforward - perhaps via employing elements of AI, such as GPT (Mr. Stephen Wolfram recent post described... |
| Why there's no (in addition to Chat-GPT interface with Wolfram Alpha) Google's Minerva math symbolic input interface to WolframAlpha? |
| 1. a) for the first four (smallest) Heegner numbers {1, 2, 3, 7} I came up with the formula: a(n) = 1+((1 + sqrt(3))^(n-1) - (1 - sqrt(3))^(n-1))/(2*sqrt(3)) for n = 1,2,3,4 b) for the last (largest) four Heegner numbers {19, 43, 67, 163}... |