Divisor sum simplification
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How does this : $$D(n)=displaystylesumlimits_{i=1}^n i leftlfloorfrac{n}{i}rightrfloor$$ become $$D(n)=displaystylesumlimits_{i=1}^{n/(u+1)} i leftlfloorfrac{n}{i}rightrfloor + sum_{d=1}^u d left( sum_{1+lfloor n/(d+1)rfloor}^{lfloor n/d rfloor}kright), u=lfloor sqrt{n}rfloor$$
sequences-and-series number-theory floor-function divisor-sum
asked Nov 17 at 17:31
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How does this : $$D(n)=displaystylesumlimits_{i=1}^n i leftlfloorfrac{n}{i}rightrfloor$$ become $$D(n)=displaystylesumlimits_{i=1}^{n/(u+1)} i leftlfloorfrac{n}{i}rightrfloor + sum_{d=1}^u d left( sum_{1+lfloor n/(d+1)rfloor}^{lfloor n/d rfloor}kright), u=lfloor sqrt{n}rfloor$$
sequences-and-series number-theory floor-function divisor-sum
asked Nov 17 at 17:31
Aizen
136
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How does this : $$D(n)=displaystylesumlimits_{i=1}^n i leftlfloorfrac{n}{i}rightrfloor$$ become $$D(n)=displaystylesumlimits_{i=1}^{n/(u+1)} i leftlfloorfrac{n}{i}rightrfloor + sum_{d=1}^u d left( sum_{1+lfloor n/(d+1)rfloor}^{lfloor n/d rfloor}kright), u=lfloor sqrt{n}rfloor$$
sequences-and-series number-theory floor-function divisor-sum
asked Nov 17 at 17:31
Aizen
136
How does this : $$D(n)=displaystylesumlimits_{i=1}^n i leftlfloorfrac{n}{i}rightrfloor$$ become $$D(n)=displaystylesumlimits_{i=1}^{n/(u+1)} i leftlfloorfrac{n}{i}rightrfloor + sum_{d=1}^u d left( sum_{1+lfloor n/(d+1)rfloor}^{lfloor n/d rfloor}kright), u=lfloor sqrt{n}rfloor$$
sequences-and-series number-theory floor-function divisor-sum
sequences-and-series number-theory floor-function divisor-sum
asked Nov 17 at 17:31
Aizen
136
asked Nov 17 at 17:31
Aizen
136
asked Nov 17 at 17:31
Aizen
136
asked Nov 17 at 17:31
Aizen
136
asked Nov 17 at 17:31
Aizen
136
136
add a comment |
add a comment |
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