Prime Sequence

Consider the sequence $a_1 = 101, a_2 = 10101, a_3 = 1010101$ and so on. Prove that $a_k$ is composite iff $k\geq 2$ .

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Aha I see it. Any even one a sub 2k can be written in the form (a sub k)^2-100^k which neatly factors to (a sub k+10^k)(a sub k-10^k) for all evens. Sorry for my latex. And the odd case is trivial.

Sal Gard - 5 years, 1 month ago

Nice observation. Kudos.

D K - 5 years, 1 month ago

I have another unanswered discussion question. Can you take a look at it?

D K - 5 years, 1 month ago

That's interesting. I can show that if $k$ is composite, then $a_k$ is composite and has a factor of 101...101.

Chung Kevin - 5 years, 1 month ago

you mean if k is odd ? then ak will have 101 as a factor. but I also need proof for when k is even.

D K - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$