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We have recurrence
$$T(n,k)=frac{n}{lfloorfrac{n+k+1}{2}rfloor}(T(n-1,k)+T(n-1,k-1))$$
so the closed form
$$T(n,k)=binom{n}{k}binom{n-k}{lfloorfrac{n-k}{2}rfloor}$$
How can I prove it?
recurrence-relations binomial-coefficients floor-function
asked yesterday
user514787
683210
put on hold as off-topic by Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut
If this question can be reworded to fit the rules in the help center, please edit the question.
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We have recurrence
$$T(n,k)=frac{n}{lfloorfrac{n+k+1}{2}rfloor}(T(n-1,k)+T(n-1,k-1))$$
so the closed form
$$T(n,k)=binom{n}{k}binom{n-k}{lfloorfrac{n-k}{2}rfloor}$$
How can I prove it?
recurrence-relations binomial-coefficients floor-function
asked yesterday
user514787
683210
put on hold as off-topic by Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
We have recurrence
$$T(n,k)=frac{n}{lfloorfrac{n+k+1}{2}rfloor}(T(n-1,k)+T(n-1,k-1))$$
so the closed form
$$T(n,k)=binom{n}{k}binom{n-k}{lfloorfrac{n-k}{2}rfloor}$$
How can I prove it?
recurrence-relations binomial-coefficients floor-function
asked yesterday
user514787
683210
We have recurrence
$$T(n,k)=frac{n}{lfloorfrac{n+k+1}{2}rfloor}(T(n-1,k)+T(n-1,k-1))$$
so the closed form
$$T(n,k)=binom{n}{k}binom{n-k}{lfloorfrac{n-k}{2}rfloor}$$
How can I prove it?
recurrence-relations binomial-coefficients floor-function
recurrence-relations binomial-coefficients floor-function
asked yesterday
user514787
683210
asked yesterday
user514787
683210
asked yesterday
user514787
683210
asked yesterday
user514787
683210
asked yesterday
user514787
683210
683210
put on hold as off-topic by Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, amWhy, Eevee Trainer, Carl Schildkraut
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
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