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I have the expression $\frac{1}{15} \left(-(-1)^n-5\times 2^{n+2}+3\times 2^{2 n+1}+15\right)$ which I want to calculate mod m for a very large n.

My current method is to ask for this exression in FullForm:

Times[Rational[1,15],Plus[15,Times[-1,Power[-1,n]],Times[-5,Power[2,Plus[2,n]]],Times[3,Power[2,Plus[1,Times[2,n]]]]]]

And to conver all Rationals and Powers to PowerMods manually:

Mod[Times[PowerMod[15,-1,m],Plus[15,Times[-1,PowerMod[-1,n,m]],Times[-5,PowerMod[2,Plus[2,n],m]],Times[3,PowerMod[2,Plus[1,Times[2,n]],m]]]],m]

However, this is kind of time-consuming and I'm sure there's a better way. Is there any sort of 'smart' find-and-replace function I can apply to do something like this?

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  • $\begingroup$ m has about 8 digits and n has about $10^{18}$ digits. $\endgroup$ – Jakob Weisblat May 10 '14 at 15:34
  • $\begingroup$ But n can be reduced mod $\lambda(n)$ $\endgroup$ – Jakob Weisblat May 10 '14 at 15:37
  • $\begingroup$ @Kuba I just found it and wrote an answer to my own question using it, right before I saw your comment. :P $\endgroup$ – Jakob Weisblat May 10 '14 at 15:38
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Yes, use a few auxiliary functions and ReplaceAll:

power = PowerMod[#, #2, m] &;

fractionMod = Mod[Numerator[#]*PowerMod[Denominator[#], -1, m], m] &;

newThing = fractionMod[ReplaceAll[oldThing, Power -> power]];
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