Timeline for Are there an infinite number of integers $n$ such that $n, n+1$, and $n+2$ have the same number of divisors?
Current License: CC BY-SA 4.0
15 events
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| Oct 24, 2020 at 5:47 | history | edited | Tony Huynh | CC BY-SA 4.0 |
deleted 5 characters in body; edited tags; edited title
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| Jul 24, 2016 at 3:02 | review | First posts | |||
| Jul 24, 2016 at 4:05 | |||||
| Jul 16, 2016 at 12:22 | answer | added | Tony Huynh | timeline score: 8 | |
| Jul 15, 2016 at 7:12 | comment | added | Fedor Petrov | @MathivananPalraj of course there are many other examples. For example, triples $(17p,2q,3r)$. | |
| Jul 15, 2016 at 5:50 | comment | added | Mathivanan Palraj | @Fedor Petrov- I think (3p, 2q, 5r) is not a unique set. (33, 34, 35), (85, 86, 87), and (93, 94, 95) conform to the above set. However, set (141, 142, 143) does not conform. | |
| Jul 14, 2016 at 11:51 | comment | added | Fedor Petrov | It is a very special case of widely believed (I think, cause of probabilistic heuristics) en.m.wikipedia.org/wiki/Dickson%27s_conjecture | |
| Jul 14, 2016 at 10:35 | comment | added | Tony Huynh | Heath-Brown proved that there are infinitely many $n$ for which $n$ and $n+1$ have the same number of divisors. It is conjectured that your statement is true. See Richard Guy's Unsolved Problems in Number Theory Problem B18. Indeed, there it is conjectured that there are infinitely many $n$ such that $n, n+1, n+2$ are all the product of two primes (this is stronger than your conjecture and weaker than Fedor's). | |
| Jul 14, 2016 at 10:30 | comment | added | user35593 | @Fedor: why are you sure? | |
| S Jul 14, 2016 at 9:47 | history | suggested | The Thin Whistler | CC BY-SA 3.0 |
"mathematicised" the question
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| Jul 14, 2016 at 9:16 | review | Suggested edits | |||
| S Jul 14, 2016 at 9:47 | |||||
| Jul 14, 2016 at 8:01 | review | Close votes | |||
| Jul 14, 2016 at 12:46 | |||||
| Jul 14, 2016 at 7:55 | comment | added | Todd Trimble | Please use LaTeX for mathematical notation. | |
| Jul 14, 2016 at 7:37 | history | edited | Mathivanan Palraj | CC BY-SA 3.0 |
added 26 characters in body
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| Jul 14, 2016 at 6:01 | comment | added | Fedor Petrov | The answer is surely yes, even for triples $(3p,2q,5r)$ with prime $p,q,r$. Another question is how to prove. | |
| Jul 14, 2016 at 4:25 | history | asked | Mathivanan Palraj | CC BY-SA 3.0 |