2
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We know:

Map[Head, {a, a^2, a^3}]
(* {Symbol, Power, Power} *)

and Map[Head, {2, 2^2, 2^3}] all of them will be known as Integers.

But how can we get the result of

Map[Head, {2, 2^2, 2^3}]` 
{Integer, Power, Power} 

I mean how Mathematica recognize a value is a power of two others?

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4
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    $\begingroup$ #[[1, 0]] & /@ {Hold[2], Hold[2^2], Hold[2^3]} gives Integer, Power, Power $\endgroup$
    – LLlAMnYP
    Feb 19, 2016 at 9:11
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    $\begingroup$ Just for the record, after 2^2 evaluates and becomes 4, it is not a Power. $\endgroup$
    – LLlAMnYP
    Feb 19, 2016 at 9:13
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    $\begingroup$ Alternatively SetAttributes[f, HoldAll]; f[x_]:=Head@Unevaluated@x; Map[f, Unevaluated@{2, 2^2, 2^3}] $\endgroup$
    – LLlAMnYP
    Feb 19, 2016 at 9:17
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    $\begingroup$ Have a look at PrimePowerQ[]. $\endgroup$ Feb 19, 2016 at 11:27

1 Answer 1

1
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You can use solutions from How to map Unevaluated over a list to prepare your list to be mapped with Head. E.g.:

Map[Head, Unevaluated /@ Unevaluated @ {1, 2^2, 2^3}]
{Integer, Power, Power}
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