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Asked: 2026-05-26T05:08:01+00:00

I’m puzzled by Mathematica’s responses to the following: ClearAll[n] #^2 & /@ Range[n] #^2

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I’m puzzled by Mathematica’s responses to the following:

ClearAll[n]
#^2 & /@ Range[n]
#^2 & /@ Range[n] // StandardForm

Range1

It seems that even Mathematica (8.0) doesn’t believe what it has just said:

#^2 & /@ Range[5]
Range[5^2]

Range2

Any thoughts about what is happening?

Edit:

The original context for this question was the following. I had written

PrimeOmega[Range[n]] - PrimeNu[Range[n]]

and since n was going to be very large (2^50), I thought I might save time by rewriting it as:

 PrimeOmega[#] - PrimeNu[#] &/@Range[n]

Thinking back, that probably wasn’t such a good idea. (I could have used Module to ‘compute’ the Range only once.)

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1 Answer

  1. Editorial Team

    Since n is undefined, Range[n] evaluated to itself. Therefore, Map acts on it as on any other symbolic head, mapping your function on its elements – here it is just n

    In[11]:= #^2 & /@ someHead[n]
Out[11]= someHead[n^2]

    EDIT

    Addressing the question in your edit – for numeric n, Range evaluates to a list all right, and you get the expected result (which is, Range[5]^2. It is all about the order of evaluation. To get Range[5^2], you could have used #^2&/@Unevaluated[Range[5]], in which case everything happens just like for symbolic n above) . In fact, Range issues an error message on non-numeric input. Also, it is tangential to the question, but functions like #^2& are Listable, and you don’t have to map them.

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