Mixing
general evaluation for $left lfloor frac{3^n}{2^k} right rfloor$ and properties
0
$begingroup$
I considered the identity $sum_{r=0}^{n}binom{n}{r}2^r = 3^n$.
Hence $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor = left lfloor frac{3^n}{2^k} right rfloor$. but now i am stuck.
Can we evaluate this value any better?
Thanks
combinatorics
asked Jan 18 at 22:49
oren1oren1
737
$endgroup$
add a comment |
0
$begingroup$
I considered the identity $sum_{r=0}^{n}binom{n}{r}2^r = 3^n$.
Hence $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor = left lfloor frac{3^n}{2^k} right rfloor$. but now i am stuck.
Can we evaluate this value any better?
Thanks
combinatorics
asked Jan 18 at 22:49
oren1oren1
737
$endgroup$
1
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
1
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11
add a comment |
0
0
0
$begingroup$
I considered the identity $sum_{r=0}^{n}binom{n}{r}2^r = 3^n$.
Hence $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor = left lfloor frac{3^n}{2^k} right rfloor$. but now i am stuck.
Can we evaluate this value any better?
Thanks
combinatorics
asked Jan 18 at 22:49
oren1oren1
737
$endgroup$
I considered the identity $sum_{r=0}^{n}binom{n}{r}2^r = 3^n$.
Hence $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor = left lfloor frac{3^n}{2^k} right rfloor$. but now i am stuck.
Can we evaluate this value any better?
Thanks
combinatorics
combinatorics
asked Jan 18 at 22:49
oren1oren1
737
asked Jan 18 at 22:49
oren1oren1
737
edited Jan 19 at 12:08
oren1
asked Jan 18 at 22:49
oren1oren1
737
asked Jan 18 at 22:49
oren1oren1
737
asked Jan 18 at 22:49
oren1oren1
737
737
1
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
1
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11
add a comment |
1
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
1
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11
1
1
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
1
1
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11
add a comment |
0
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1
$begingroup$
What any better are you thinking of ? This is a quite simple expression.
$endgroup$
– Yves Daoust
Jan 18 at 23:21
1
$begingroup$
I think it should be $lfloor sum_{r=0}^kbinom{n}r 2^{r-k} rfloor$. I am pretty sure this cannot be simplified.
$endgroup$
– Mike Earnest
Jan 18 at 23:45
$begingroup$
I thought it might be possible to find an expression without the floor value function. I guess probably not.
$endgroup$
– oren1
Jan 19 at 10:12
$begingroup$
I edited the LHS (typo) from $left lfloor sum_{r=0}^{k}binom{k}{r}2^{r-k} right rfloor$ to $left lfloor sum_{r=0}^{n}binom{n}{r}2^{r-k} right rfloor$. does it change something?
$endgroup$
– oren1
Jan 19 at 12:11