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 Author Comment/Response JD 07/08/13 06:54am Hi. I know what I want to do in plain english, I just can't get mathematica to do it. This is not homework. I found a very interesting way to generate prime numbers and also test primality of any format of number, like that of Mersenne numbers. Part One is I need to know how many factors are in the LCM just above 10^(10^10). I want to find the first solution for i, where the summation of "MangoldtLambda[i]" is greater than (10^10). Or where the summation of the log of "MangoldtLambda[i]" is greater than 10. It's a really big number and I don't expect anyone to lock up his/her computer by crunching all the numbers for me, just give me a more basic example. For those unfamiliar with the MangoldtLambda function, it gives the log of the prime base of a input number. So since 3^2=9, MangoldtLambda[9] gives the log of 3. If not prime or composed of a prime power, MangoldtLambda[i] yields the log of 1 which equals 0. Therefore, composites which are not purely made of one prime variable will add nothing to the sum. The sum of MangoldtLambda[i] from 1 to i, using positive integers in increments of 1 for i, yields the Least Common Multiple of (1,2,3,...,i). The LCM of 1 though 10 is (1x2x3x2x5x1x7x2x3x1). This equals 2520. The sum of MangoldtLambda[i] with a minimum of 1 and max of 10 equals the log of 2520, or ~3.401. Applying a Ceiling function to ~3.401 would yield 4, telling me that the LCM of 1 though 10 has 4 digits. Well, three completely filled placeholders and a partial for the fourth place is how I look at it. So, I want to know when the LCM of (1,2,3,...,i) reaches 10,000,000,000 digits or when does the LCM of (1,2,3,...,i) exceed 10^(10^10). The solution for i will be well below the square root of 10^(10^10), which is 10^(10^5). I don't need to know the final LCM, just how many digits it has and the last number in the product before it gets converted to it's prime base. LCM's tend to jump quite a bit, so if the LCM previous to 10^(10^10) is close to 10^(10^10), the next jump could put it's value out of the ballpark and into outer space. This is what I've been trying: - " In: Solve[10<=Sum[Log MangoldtLambda[i], {i,1,n,1} ],n,Integers][[1]]    Solve::nsmet: This system cannot be solved with the methods available to Solve. >> Out: 10<=Sum[Log MangoldtLambda[i],{i,1,n,1}] " - The [[1]] at the end is designed to give me the first solution. {i,1,n,1} is meant to describe the index of the summation, with a minimum of 1 and a maximum of n, with increments or steps of 1. Log MangoldtLambda is meant to be the log of the log of the prime base of i. And then I'm solving for n, where n is a positive integer that could be expressed as a prime to a power of 1 to unknown. One thing I've noticed is that the log of the log of prime numbers below 11 gives a negative value. It should always be a positive number in order for this method to work. It could be that I do not understand the correct use of logs in this scenario. Still, using - Solve[10000000000<=Sum [MangoldtLambda[i],{i,1,n,1}],n,Integers][[1]] Solve::nsmet: This system cannot be solved with the methods available to Solve. ‡ 10000000000<=Sum[MangoldtLambda[i],{i,1,n,1}] - is not working for me either. URL: ,

 Subject (listing for 'Solving for when a summation reaches a value.') Author Date Posted Solving for when a summation reaches a value. JD 07/08/13 06:54am Re: Solving for when a summation reaches a value. Bill Simpson 07/08/13 4:36pm Re: Re: Solving for when a summation reaches a ... JD 07/08/13 6:42pm Re: Solving for when a summation reaches a value. Bill Simpson 07/09/13 01:15am
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