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I would like to find the powers of a prime in the unique prime factorization of an $n$. I want a function $f[n,p]$ such that $n,p$ are given and I need to know what the power of $p$ is. For instance

n=60;
f[n,2]=2;
f[n,3]=1;
f[n,5]=1;
f[n,7]=0;
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    $\begingroup$ maybe f[n_, p_] := Cases[FactorInteger[n], {p, e_} :> e]? $\endgroup$
    – kglr
    Oct 28, 2016 at 21:06
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    $\begingroup$ If you want efficiency when dealing with large inputs, just use division. getPower[n_, p_ /; Abs[p] != 1] := Module[{i = 0, q = n}, While[Divisible[q, p], i++; q /= p]; i] $\endgroup$ Oct 28, 2016 at 21:24

2 Answers 2

6
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If you really mean, as you say, that $n$ and $p$ are given, so you could simply use IntegerExponent[n,p].

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  • $\begingroup$ I can't believe I missed that. Again (yes, it's happened before). $\endgroup$ Oct 29, 2016 at 18:59
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f[n_, p_] := 
 Block[{s}, s = Select[FactorInteger[n], MemberQ[#, p] &]; 
  If[s == {}, 0, s[[1, 2]]]]

or

f[n_, p_] := First @ (Cases[FactorInteger[n], {p, e_} :> e] /. {} -> {0})

Then

f[60, 2]
f[60, 3]
f[60, 5]
f[60, 7]

2

1

1

0

f[11^17 13^15, 5]
f[11^17 13^15, 13]

0

15

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