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In the OEIS sequence A049210, I noticed an odd notation I haven't seen before:

a(n) = (8*n-1)(!^8), n >= 1, a(0) = 1.

What does the (!^8) mean? My first thought was something to do with factorials, but, to my knowledge, the factorial symbol must always be preceded by a number, like 3! or (5+n)!. I have no idea what a ! would mean just floating around on its own.

I'm pretty sure it doesn't mean "take the factorial of (8*n-1) and raise that value to the 8th power," since that would result in the numbers growing far faster than the sequence should.

In logic and computer science, ! is often used to mean "not," such that !false is true, but I don't think that would have any application here.

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From the context, it seems that we have a recursion $n(!^k)=(n-k)(!^k)\cdot n$ and $m(!^k)=m$ for $1\le m<k$. This makes $n(!^1)=n!$ and $n(!^2)=n!!$.

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We can read octo factorial numbers in the synopsis of the OEIS sequence A049210 which is addressed by OP.

Octo factorial numbers are identified as A051189 and defined as \begin{align*} \left(n!\cdot 8^n\right)_{n\geq 0}=(1, 8, 128, 3072, 98304,\ldots) \end{align*}

Here we have a variant in some weird notation which can be written as \begin{align*} \left(-\prod_{j=0}^n\left(8j-1\right)\right)_{n\geq 0}=(1, 7, 105, 2415, 74865,\ldots) \end{align*}

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Speculation:

What I see when I visit OEIS is

A049210 a(n) = -product_{k=0..n} (8*k-1); octo-factorial numbers.

In the unix bash shell the symbol "!" means "repeat the previous command". Perhaps in this context "(!^8)" means something like "replace $n$ with $n-8$ in the formula for $a(n)$, so defining a recursion. You could check to see if that reproduces the product in the definition.

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  • $\begingroup$ You have to scroll down to the Formula section. $\endgroup$ – Somatic Apr 17 '16 at 17:06

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