Show more

Letting \(P_{n}\) be the \(n\)th Pell number:

\[{P_{n}}^2=\frac{(-1)^{n+1}}{4}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{3^n\cdot8^k}{4\cdot9^k}\Bigg)\]\[{P_{n}}^4=\frac{3}{32}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{17^n\cdot288^k}{32\cdot289^k}-\frac{(-3)^n\cdot8^k}{8\cdot9^k}\Bigg)\]

Show thread

Today I've been playing around with maths looking for patterns and formulae for even powers of Pell numbers.

Dani boosted

Do SQL injections cause autism?

Last sequences I published in the OEIS are oeis.org/A282389 and oeis.org/A282390. I made a GIF that illustrates sequence A282389 so well. โ˜บ mathstodon.xyz/media/8Cd3kfHzP

Dani boosted

Life is made up of meetings and partings. That is the way of it.

Dani boosted

that \(\sqrt{2}\) is irrational:
Suppose not, i.e. \( \sqrt{2} = \frac{a}{b} \), \( a,b \in \mathbb{Z} \) and \(a,b\) are coprime.
Then \( \left(\frac{a}{b}\right)^2 = 2\). Then \(a^2 = 2b^2\), so \(a^2\) is even. Let \(a = 2c\), so we have \(4c^2 = 2b^2 \implies b^2 = 2c^2\). So \(b\) is even.
Then both \(a\) and \(b\) are divisible by 2, contradicting the earlier proposition.
So \(\sqrt{2}\) can't be expressed as the ratio of two integers, which means it's irrational.

Mathstodon

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!