# Solve for upperbound of a summation to find the nearest LCM to a given #

GROUPS:
 I've been doing some work with finding the LCM of consecutive integers (1-n) and the number of factors in such an LCM.It's become easy to construct an LCM and find the factors when n is given. However, to deconstruct a given LCM and find n is proving difficult.I eventually would like to find the nearest LCM either above or below a value. It would be nice to have the one above, but I could make due with the one below.I've been provided with a few ways of doing this that I'd like to show and see if anyone can show me how to solve it algebraically or code for my objective.The first uses Prime Pi. Product[Prime^Floor[Log[Prime[i], 11]], {i, PrimePi[11]}][i][i][/i][/i]Out:   27720Product[Floor[1 + Log[Prime[i], 11]], {i, PrimePi[11]}]Out:    96Also, I'd like to note that if PrimePi[] exceeds the value of n (11), then it does not affect the outcome.And for some reason, the "" would not copy and paste behind Prime. I had to enter that manually. A bug perhaps?And my computer let me know :"The Mathematica command giving the prime counting function for a \ number x is PrimePi, which works up to a maximum value of x approx 8*10^(13)."The second uses the Mangoldt function to construct the LCM. Product[E^MangoldtLambda[n],{n, 2, 11}]Out:     27720and then using DivisorSigma [0,LCM] to find divisors / factors :DivisorSigma[0, Product[E^MangoldtLambda[n],{n, 2, 11}]]and then theres what I refer to as the Apply method:Apply[LCM, Range[11]]and then I use DivisorSigma with that like this:DivisorSigma[0, Apply[LCM, Range[11]]]and then there's what I refer to as the PrimeQ method for which I only know how to find factors with given n:Times @@ Map[Floor[1 + Log[#, 11]] &, Select[Range[2, 11], PrimeQ]]and then there's the Cyclotomic methodProduct[Cyclotomic[n, 1],{n, 2, 11}]I've read a couple of equations that I haven't been able to implement in mathematicaOne webpage said that the LCM of (1-n) is equal to E^(n*(1 + \[Omicron]*1))which is too high of math for me to understand. Another one out of my reach is: Psi(x) = ln (1,2,3,...,x)If not psi, then it might be a similar symbol like PolyGammaI figured that I could then take e^Psi(x) to get the LCM of (1-n).I estimated that the target objective, the LCM of (1-n) that is somewhere near 10^(10^10) :Floor[1+Log10[LCM@@Range[2297]]]//AbsoluteTiming{0.0156250,1000}Floor[1+Log10[LCM@@Range[23013]]]//AbsoluteTiming{0.1406250,9998}Floor[1+Log10[LCM@@Range[230075]]]//AbsoluteTiming{2.6406250,99997}Floor[1+Log10[LCM@@Range[2302149]]]//AbsoluteTiming{63.7500000,999997}Floor[1+Log10[Floor[1+Log10[LCM@@Range[23021490000]]]]]//AbsoluteTimingAnd this last bit of code I have tried to run overnight and my 32 bit confuser didn't give.My last thought was that since log_a(xy) = log_a(x) + log_a(y), then I need to solve for n where: Sum of Log_10(e^(MangoldtLambda(n))) => (10^10). I tried:NSolve[[[\!$$\*UnderoverscriptBox[\(\$$, $$x = 2$$, $$\$$]$$Floor[1 + Log10[\([E^MangoldtLambda[x], {x, 2, \}]$$]]\)\)] >= 10^10], x]but the code isn't working for me.