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Topic: sum of lcm's (Read 8748 times) 

newb
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sum of lcm's
« on: Jan 29^{th}, 2010, 1:24am » 
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I have to find: summation (lcm(i,n)) for i=1 to n,n<=10^6 I was thinking of summation i/gcd(i,n) But i can't any efficient way to find this. need help. P.S>I have wrote this question in pure maths section also. Apology,


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towr
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Re: sum of lcm's
« Reply #1 on: Jan 29^{th}, 2010, 2:13am » 
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on Jan 29^{th}, 2010, 1:24am, dormant wrote:P.S>I have wrote this question in pure maths section also. Apology, 
 I've removed the thread in the maths section. Quote:I was thinking of summation i/gcd(i,n) 
 You would want summation n*i/gcd(i,n) After all gcd(i,n)*lcm(i,n)=i*n You can use the Euclidean algorithm to calculate the gcd fairly easily. I'll think about it some more whether there is a better way. [e]See also A051193[/e]

« Last Edit: Jan 29^{th}, 2010, 4:16am by towr » 
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towr
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Re: sum of lcm's
« Reply #2 on: Jan 29^{th}, 2010, 5:01am » 
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If you can find a prime decomposition of n, you can use the fact that sum_of_lcm(n) = n * (a(n)+1)/2 where we have that a(k*l) = a(k)*a(l), and for primes p: a(p) = p*(p1)+1 And since n <= 10^{6}, it's not too difficult to find the prime factorization (just try all numbers under 1000, or do it even a bit smarter).

« Last Edit: Jan 29^{th}, 2010, 5:02am by towr » 
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newb
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Re: sum of lcm's
« Reply #3 on: Jan 29^{th}, 2010, 9:13am » 
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Quote:sum_of_lcm(n) = n * (a(n)+1)/2 where we have that a(k*l) = a(k)*a(l), and for primes p: a(p) = p*(p1)+1 
 how we arrive at this formula? Any special name for this a(k) function?


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newb
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Re: sum of lcm's
« Reply #4 on: Jan 29^{th}, 2010, 9:42am » 
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Towr, Thanx for the link to Encyclopedia, I never thought of searching it there. But, I think a(n) is sum {d*phi(d)} for dn. So ,taking only prime factorization can cause trouble.


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towr
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Re: sum of lcm's
« Reply #5 on: Jan 29^{th}, 2010, 3:16pm » 
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On the page describing the sequence a(n) = sum_{{dn}} {d*phi(d)}, A057660, there is some further useful information. Such as the a(k*l) = a(k)*a(l) I mentioned. But as it turns out that's not entirely correct, it only hold if gcd(k,l)=1. Which means we can use the prime factorization, but we need to be careful if a prime factor occurs more than once; namely, that page also says a(p^{e}) = (p^{2e+1}+1)/(p+1) So, we can use the following algorithm: javascript Code:<script> prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997] function a(n) { res=1; for(i=0; i < prime.length && 1 < n && prime[i] < n; i++) { x = prime[i]; while(n % prime[i]==0) { x *= prime[i]*prime[i]; n /= prime[i]; } res *= (x+1)/(prime[i]+1); } if(n > 1) res*=(n*(n1)+1); return res; } function sum_of_lcm(n) { return n*(a(n)+1)/2; } for(n=1; n <= 20; n++) document.write(n + ": " + sum_of_lcm(n) + "<br>"); document.close(); </script> 


« Last Edit: Jan 29^{th}, 2010, 3:21pm by towr » 
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holtz
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Re: sum of lcm's
« Reply #6 on: May 26^{th}, 2012, 12:44am » 
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Hello, I want to find lcm(m,n)+lcm(m+1,n)+...+lcm(n,n) where m<=n. How can it be done? Can it be done similarly to the method described above? Thanks


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towr
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Re: sum of lcm's
« Reply #7 on: May 26^{th}, 2012, 1:22am » 
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Yes, just take the sum from 1n and subtract the sum of 1m1, this leaves the sum from mn


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holtz
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Re: sum of lcm's
« Reply #8 on: May 26^{th}, 2012, 2:34am » 
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I think it wouldn't work, because: sum of 1..m1 is (m1)*(a(m1)+1)/2 = lcm(1,m1)+lcm(2,m1)+...+lcm(m1,m1) but what we need to find is lcm(1, n)+lcm(2,n)+...+lcm(m1,n) in order to subtract from sum 1..n and get the result. So, how can we calculate lcm(1, n)+lcm(2,n)+...+lcm(m1,n), that is the question.

« Last Edit: May 26^{th}, 2012, 2:35am by holtz » 
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towr
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Re: sum of lcm's
« Reply #9 on: May 26^{th}, 2012, 4:05am » 
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Ah, you're right.. There might still be some way to adapt the algorithm, but I'll have to think on it a bit more..


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akasina9
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Re: sum of lcm's
« Reply #10 on: Nov 16^{th}, 2012, 12:12am » 
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For the sum lcm(m,n)+lcm(m+1,n)+...+lcm(n,n), you can proceed in the same way as calculating the sum lcm(1,n)+lcm(2,n)+...+lcm(n,n), but hold the value of the sum when you hit i=m in a temporary location and subtract it from the final achieved sum.


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towr
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Re: sum of lcm's
« Reply #11 on: Nov 16^{th}, 2012, 8:37am » 
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That's not terribly helpful unless you can calculate both in a fast way.


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akasina9
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Re: sum of lcm's
« Reply #12 on: Nov 17^{th}, 2012, 12:50am » 
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on Nov 16^{th}, 2012, 8:37am, towr wrote:That's not terribly helpful unless you can calculate both in a fast way. 
 We need to compute just one value, namely lcm(1,n)+lcm(2,n)+...+lcm(n,n). We just store the intermediate value. So what are the 2 sums you are referring to? Did I miss something?


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towr
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Re: sum of lcm's
« Reply #13 on: Nov 17^{th}, 2012, 2:54am » 
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Well, just that in that case it's faster to just skip the first m terms. There's no reason to compute them, just to subtract them again later. Unless there's a shortcut to compute the sums.


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birbal
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Re: sum of lcm's
« Reply #14 on: Dec 13^{th}, 2012, 10:15am » 
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on Jan 29^{th}, 2010, 5:01am, towr wrote:If you can find a prime decomposition of n, you can use the fact that sum_of_lcm(n) = n * (a(n)+1)/2 where we have that a(k*l) = a(k)*a(l), and for primes p: a(p) = p*(p1)+1 And since n <= 10^{6}, it's not too difficult to find the prime factorization (just try all numbers under 1000, or do it even a bit smarter). 
 Does sum_of_lcm(n) means summition lcm(i,n) ? How is a(p) defined for nonprimes ?


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towr
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Re: sum of lcm's
« Reply #15 on: Dec 13^{th}, 2012, 11:23am » 
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on Dec 13^{th}, 2012, 10:15am, birbal wrote:Does sum_of_lcm(n) means summition lcm(i,n) ? 
 sum_of_lcm(n) means whatever was asked in the opening post that resembles it most closely Quote:How is a(p) defined for nonprimes ? 
 Exactly as it says in the quoted text: a(k*l) = a(k)*a(l) So just primedecompose n. (But bear in mind reply #5)

« Last Edit: Dec 13^{th}, 2012, 11:24am by towr » 
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