Timeline for Pascal triangle and prime numbers
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29 at 18:26 | comment | added | Michael Lugo | If you just have the constraint that three consecutive binomial coefficients be in arithmetic progression, then there are more examples, namely ${n \choose r-1}, {n \choose r}, {n \choose r+1}$ where $(n,r) = (7,2), (14,5), (23,9), (34,14), \ldots$ and of course the reflections over the axis of the triangle. The general form is $(n,r) = (k^2-2, (k^2 \pm k - 2)/2)$ - that shift of $\pm k/2$ corresponds to the width of the approximating Gaussian curve. | |
| Oct 12, 2016 at 21:08 | history | edited | darij grinberg | CC BY-SA 3.0 |
latex
|
| Jun 30, 2010 at 5:22 | comment | added | Gerry Myerson | ${n\choose r}=2{n\choose r-1}$ is equivalent to $n=3r-1$, and ${n\choose r+1}=(3/2){n\choose r}$ is equivalent to $2n=5r+3$, so the only way to get $a,2a,3a$ as consecutive entries in a row of Pascal's triangle is the one that delighted you, corresponding to the solution $n=14$, $r=5$ of that pair of linear equations. | |
| May 14, 2010 at 19:26 | comment | added | Will Jagy | I once carefully drew the triangle as far as $ \left( \begin{array}{c} 16 \\\ \ast \end{array} \right) $ in an effort to show visually how many gifts were given by day $n$ in the Twelve Days of Christmas song. I was delighted to find the consecutive values 1001, 2002, 3003 in the $ \left( \begin{array}{c} 14 \\\ \ast \end{array} \right) $ row. I do not think the values $a, 2a, 3a$ can occur all that often in a row next to each other, for one thing $2 a$ must lie very near the inflection point of the approximating multiple of a Gaussian distribution curve. But note primes 7,11,13. | |
| May 14, 2010 at 18:03 | history | answered | HenrikRüping | CC BY-SA 2.5 |