Reposted by Vince Vatter
The 254th Carnival of Mathematics is posted! Check out all the cool math we collated from the internet in August.
@CarnivalOfMath @aperiodical.bsky.social #CarnivalOfMath #iTeachMath #MathSky #MathsToday
karendcampe.wordpress.com/2026/09/07/c...
karendcampe.wordpress.com
Carnival Of Mathematics
Welcome to the 254th installment of the Carnival of Mathematics, a roundup of interesting mathematics content from August 2026. The Carnival is a monthly post hosted by a different blog each month.…
![ChatGPT-4o:
"go to OEIS bc i’m tired: [1, 1, 1, 2, 5, 17, 73, …] → OEIS A001339"](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:4wrxafbsjycxbpjo4e42ht6q/bafkreiabbattgt5qjv7mlj5fl4cfg5r4yziyesed5alqery53gzanugcry)
![$2^{2^n}+1$ isn't prime for $5\le n\le 32$. By the prime number theorem, $m$ is prime with probability ${\approx\nicefrac{1}{\ln m}}$. The probability that another number of this form is prime is about
\[
\sum_{n=33}^\infty \frac{1}{\ln (2^{2^n}+1)}
< \sum_{n=33}^\infty \frac{1}{2^n\ln 2}
= \frac{1}{2^{32}\ln 2}
\approx 0.000000000359.
\]](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:4wrxafbsjycxbpjo4e42ht6q/bafkreiaxritc4v2f2eo74jt6ox2ylnzfqqxfaioaq7ja3vhkyiwikej3ny)
![Theorem. If $n$ is an integer and $n^2$ is even, then $n$ is itself even.
Proof. Contrapositives are for cowards, so assume $n$ is an integer and $n^2$ is even. Then $n^2=2k$ for some integer $k$, and thus $n^2-2k=0$. Behold:
\[ n = n + (n^2-2k) = n(n+1)-2k. \]
Both $n(n+1)$ and $2k$ are even, so $n$ is even.
QED.](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:4wrxafbsjycxbpjo4e42ht6q/bafkreihucnue4bdyzulcrzddytan5ttdx56rg5ci6cn4vhxw5lebnbdzma)
