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I've been playing around with sequences a bit. In particular with using ## with unary and binary operators.

Let's start simple, the following all make some kind of sense:

 + ## & [a,b] (* a + b *) x + ## & [a,b] (* x + a + b *) x * ## & [a,b] (* x * a * b *) x ^ ## & [a,b] (* x ^ a ^ b *)

Now here is a slightly weird case:

 - ## & [a,b] (* -a*b *) x - ## & [a,b] (* x - a*b *)

I guess, this sort of makes sense if - is actually interpreted as something like +(-1)*. But it also means that +##-## is generally non-zero.

But now here's the real puzzle:

x / ## & [a,b] (* x a^(1/b) *) x / ## & [a,b,c] (* x a^b^(1/c) *)

Wh... what? Can anyone explain what's happening here or at least give some justification like the one for subtraction? Answers which correct my explanation for subtraction are also welcome!

(No, I would never use this stuff in production code. But knowing what exactly is going on under the hood could come in handy some time.)

Bonus Question: Are there any other operators that yield unexpected and potentially "useful" results? (I mean, !## will yield Not[a,b] but that's neither very unexpected nor useful.)

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    $\begingroup$ Related: (31797) $\endgroup$ Commented Jan 8, 2015 at 19:07
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    $\begingroup$ Users interested in such question should consider Trace[x/## &[a, b, c]]. $\endgroup$ Commented Aug 4, 2015 at 11:19

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The documentation for Minus states that

-x is converted to Times[-1,x] on input.

So -Sequence[a,b] == Times[-1,Sequence[a,b]] == Times[-1,a,b] by this definition. Similarly the documentation for Divide states that

x/y is converted to x y^-1 on input.

and therefore x / Sequence[a,b] == x Sequence[a,b]^-1. Sequence[a,b]^c == Power[a, Power[b,c]]. When c == -1 you get Power[b, -1] == 1/b.

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    $\begingroup$ Accepting this, because it covers both Minus and Divide. $\endgroup$ Commented Jan 11, 2015 at 15:33
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x/## & // FullForm 
Function[Times[x,Power[SlotSequence[1],-1]]]

and Power[a,b,c...] == Power[a, Power[b, c...]] so now it should be clear.

This syntax is mentioned in the last bullet point in details of Power documentation.

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    $\begingroup$ beat me by a femtosecond $\endgroup$ Commented Jan 8, 2015 at 18:08
  • $\begingroup$ Ah, thanks a lot. I was thinking it would probably be analogous to the minus case, but couldn't work it out. I think linking to the documentation for Divide would be more useful than linking to the documentation for Power though. $\endgroup$ Commented Jan 8, 2015 at 18:12

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