sum(floor(k/a)*floor(k/b),k=1..a*b-1)
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From: M.A.Fajjal on 9 Aug 2010 23:52 Evaluate sum(floor(k/a)*floor(k/b),k=1..a*b-1) where gcd(a,b)=1
From: Raymond Manzoni on 10 Aug 2010 13:55 M.A.Fajjal a écrit : > Evaluate > > sum(floor(k/a)*floor(k/b),k=1..a*b-1) > > where gcd(a,b)=1 I'll conjecture that this is : (a + b + 4 a b + 1) (a - 1) (b - 1) / 12 Hoping it helped a little, Raymond
From: Robert Israel on 10 Aug 2010 20:49 Raymond Manzoni <raymman(a)free.fr> writes: > M.A.Fajjal a écrit : > > Evaluate > > > > sum(floor(k/a)*floor(k/b),k=1..a*b-1) > > > > where gcd(a,b)=1 > > > I'll conjecture that this is : > > (a + b + 4 a b + 1) (a - 1) (b - 1) / 12 > > > Hoping it helped a little, > Raymond I'm assuming a and b are positive integers. Let n = a*b. For i in {0,...,a-1} and j in {0,...,b-1}, let K(i,j) be the unique integer k in {0,...,ab-1} such that k == i mod a and k == j mod b. Note that floor(K(i,j)/a) = (K(i,j)-i)/a and floor(K(i,j)/a) = (K(i,j)-j)/b. So your sum is sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j) - i)(K(i,j) - j)/(ab) = sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j)^2 - i K(i,j) - j K(i,j) + ij)/(ab) sum_{i=0}^{a-1} sum_{j=0}^{b-1} K(i,j)^2 = sum_{k=0}^{ab-1} k^2 = a b (a b - 1) (2 a b - 1)/6 sum_{i=0}^{a-1} sum_{j=0}^{b-1} i K(i,j) = sum_{i=0}^{a-1} sum_{m=0}^{b-1} i (m a + i) = a b (a - 1) (3 a b + a - 2) / 12 similarly sum_{i=0}^{a-1} sum_{j=0}^{b-1} j K(i,j) = a b (b - 1) (3 a b + b - 2) / 12 sum_{i=0}^{a-1} sum_{j=0}^{b-1} i j = a b (a-1) (b - 1)/4 After some simplification (with Maple's help), the result is as Raymond conjectured. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Raymond Manzoni on 11 Aug 2010 06:50 Robert Israel a �crit : > Raymond Manzoni <raymman(a)free.fr> writes: > >> M.A.Fajjal a écrit : >>> Evaluate >>> >>> sum(floor(k/a)*floor(k/b),k=1..a*b-1) >>> >>> where gcd(a,b)=1 >> >> I'll conjecture that this is : >> >> (a + b + 4 a b + 1) (a - 1) (b - 1) / 12 >> >> >> Hoping it helped a little, >> Raymond > > I'm assuming a and b are positive integers. Let n = a*b. > For i in {0,...,a-1} and j in {0,...,b-1}, let K(i,j) be the > unique integer k in {0,...,ab-1} such that k == i mod a and > k == j mod b. Note that floor(K(i,j)/a) = (K(i,j)-i)/a > and floor(K(i,j)/a) = (K(i,j)-j)/b. So your sum is > sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j) - i)(K(i,j) - j)/(ab) > = sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j)^2 - i K(i,j) - j K(i,j) + ij)/(ab) > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} K(i,j)^2 = sum_{k=0}^{ab-1} k^2 > = a b (a b - 1) (2 a b - 1)/6 > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} i K(i,j) > = sum_{i=0}^{a-1} sum_{m=0}^{b-1} i (m a + i) > = a b (a - 1) (3 a b + a - 2) / 12 > > similarly > sum_{i=0}^{a-1} sum_{j=0}^{b-1} j K(i,j) > = a b (b - 1) (3 a b + b - 2) / 12 > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} i j = a b (a-1) (b - 1)/4 > > After some simplification (with Maple's help), the result is > as Raymond conjectured. Your K(i,j) trick is rather nice thanks! Raymond
From: M.A.Fajjal on 11 Aug 2010 05:56
> Raymond Manzoni <raymman(a)free.fr> writes: > > > M.A.Fajjal a écrit : > > > Evaluate > > > > > > sum(floor(k/a)*floor(k/b),k=1..a*b-1) > > > > > > where gcd(a,b)=1 > > > > > > I'll conjecture that this is : > > > > (a + b + 4 a b + 1) (a - 1) (b - 1) / 12 > > > > > > Hoping it helped a little, > > Raymond > > I'm assuming a and b are positive integers. Let n = > a*b. > For i in {0,...,a-1} and j in {0,...,b-1}, let K(i,j) > be the > unique integer k in {0,...,ab-1} such that k == i mod > a and > k == j mod b. Note that floor(K(i,j)/a) = > (K(i,j)-i)/a > and floor(K(i,j)/a) = (K(i,j)-j)/b. So your sum is > sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j) - i)(K(i,j) - > j)/(ab) > = sum_{i=0}^{a-1} sum_{j=0}^{b-1} (K(i,j)^2 - i > K(i,j) - j K(i,j) + ij)/(ab) > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} K(i,j)^2 = > sum_{k=0}^{ab-1} k^2 > = a b (a b - 1) (2 a b - 1)/6 > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} i K(i,j) > = sum_{i=0}^{a-1} sum_{m=0}^{b-1} i (m a + i) > = a b (a - 1) (3 a b + a - 2) / 12 > > similarly > sum_{i=0}^{a-1} sum_{j=0}^{b-1} j K(i,j) > = a b (b - 1) (3 a b + b - 2) / 12 > > sum_{i=0}^{a-1} sum_{j=0}^{b-1} i j = a b (a-1) (b - > 1)/4 > > After some simplification (with Maple's help), the > result is > as Raymond conjectured. > -- > Robert Israel > israel(a)math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada Thanks what about gcd(a,b)=/= 1 where a,b are positive intergers |