Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$prime numbers in an intervalIn how many different ways can we fully parenthesize the matrix product?Proving combinatorial identity with the product of Stirling numbers of the first and second kindsCalculate $left(beginsmallmatrixn \ rendsmallmatrixright)/k^n$ for very large $n$How do you calculate the width of the Poset Lattice of Divisors?When the product of any two consecutive digits in the number is a prime numberInteresting resolution to considering prime factorisations and finding the number of positive divisors of large numbers??Let $S$ be the set of first 100 natural numbers,$L$ is the least common multiple of all the elements in $S,P$ is the product of all the primes in $S$Calculate $(-1)^k left(beginmatrix -1/2 \ kendmatrixright)$. Is it equal to $frac (2k)!2^2k(k!)^2$?prove $left( beginarrayc 2n \ n endarray right) = 2 left( beginarrayc n \ 2 endarray right) + n^2$Number of divisors of $10!$
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Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$
prime numbers in an intervalIn how many different ways can we fully parenthesize the matrix product?Proving combinatorial identity with the product of Stirling numbers of the first and second kindsCalculate $left(beginsmallmatrixn \ rendsmallmatrixright)/k^n$ for very large $n$How do you calculate the width of the Poset Lattice of Divisors?When the product of any two consecutive digits in the number is a prime numberInteresting resolution to considering prime factorisations and finding the number of positive divisors of large numbers??Let $S$ be the set of first 100 natural numbers,$L$ is the least common multiple of all the elements in $S,P$ is the product of all the primes in $S$Calculate $(-1)^k left(beginmatrix -1/2 \ kendmatrixright)$. Is it equal to $frac (2k)!2^2k(k!)^2$?prove $left( beginarrayc 2n \ n endarray right) = 2 left( beginarrayc n \ 2 endarray right) + n^2$Number of divisors of $10!$
$begingroup$
Show that for every m the product of all prime numbers on the interval $[m+1, 2m]$ is $ le left(beginmatrix 2m \mendmatrixright)$
I don't know how to do this, maybe somehow show that all the prime numbers on that interval are divisors of the combination number or something like that...
combinatorics
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
$endgroup$
add a comment |
$begingroup$
Show that for every m the product of all prime numbers on the interval $[m+1, 2m]$ is $ le left(beginmatrix 2m \mendmatrixright)$
I don't know how to do this, maybe somehow show that all the prime numbers on that interval are divisors of the combination number or something like that...
combinatorics
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
$endgroup$
$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08
add a comment |
$begingroup$
Show that for every m the product of all prime numbers on the interval $[m+1, 2m]$ is $ le left(beginmatrix 2m \mendmatrixright)$
I don't know how to do this, maybe somehow show that all the prime numbers on that interval are divisors of the combination number or something like that...
combinatorics
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
$endgroup$
Show that for every m the product of all prime numbers on the interval $[m+1, 2m]$ is $ le left(beginmatrix 2m \mendmatrixright)$
I don't know how to do this, maybe somehow show that all the prime numbers on that interval are divisors of the combination number or something like that...
combinatorics
combinatorics
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
asked Mar 17 at 23:04
J. LastinJ. Lastin
1439
1439
$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08
add a comment |
$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08
$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08
$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Due to Tchebychev.
The product off all those primes appears in the numerator and not in the denominator of $2mchoose m =frac(2m)!m!m!$ therefore that product divide $2mchoose m$.
More precisely: $forall p, m<ple 2m$, $p|2mchoose m$ therefore : $prod_m<ple 2m, p primep|2mchoose m$. The claim follows
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
$begingroup$
...maybe somehow show that all the prime numbers on that interval are divisors of the combination number...
Yes, they are. How many times does such a prime appear in the numerator $(2m)!=1cdot 2cdot 3cdots (2m-1)cdot 2m$? How many times does it appear in the denominator $(m!)^2=1^2cdot 2^2cdot 3^2cdots (m-1)^2cdot m^2$?
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Due to Tchebychev.
The product off all those primes appears in the numerator and not in the denominator of $2mchoose m =frac(2m)!m!m!$ therefore that product divide $2mchoose m$.
More precisely: $forall p, m<ple 2m$, $p|2mchoose m$ therefore : $prod_m<ple 2m, p primep|2mchoose m$. The claim follows
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
$begingroup$
Due to Tchebychev.
The product off all those primes appears in the numerator and not in the denominator of $2mchoose m =frac(2m)!m!m!$ therefore that product divide $2mchoose m$.
More precisely: $forall p, m<ple 2m$, $p|2mchoose m$ therefore : $prod_m<ple 2m, p primep|2mchoose m$. The claim follows
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
add a comment |
$begingroup$
Due to Tchebychev.
The product off all those primes appears in the numerator and not in the denominator of $2mchoose m =frac(2m)!m!m!$ therefore that product divide $2mchoose m$.
More precisely: $forall p, m<ple 2m$, $p|2mchoose m$ therefore : $prod_m<ple 2m, p primep|2mchoose m$. The claim follows
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
$endgroup$
Due to Tchebychev.
The product off all those primes appears in the numerator and not in the denominator of $2mchoose m =frac(2m)!m!m!$ therefore that product divide $2mchoose m$.
More precisely: $forall p, m<ple 2m$, $p|2mchoose m$ therefore : $prod_m<ple 2m, p primep|2mchoose m$. The claim follows
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
edited Mar 18 at 15:09
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
answered Mar 17 at 23:11
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
1
add a comment |
add a comment |
$begingroup$
...maybe somehow show that all the prime numbers on that interval are divisors of the combination number...
Yes, they are. How many times does such a prime appear in the numerator $(2m)!=1cdot 2cdot 3cdots (2m-1)cdot 2m$? How many times does it appear in the denominator $(m!)^2=1^2cdot 2^2cdot 3^2cdots (m-1)^2cdot m^2$?
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
$begingroup$
...maybe somehow show that all the prime numbers on that interval are divisors of the combination number...
Yes, they are. How many times does such a prime appear in the numerator $(2m)!=1cdot 2cdot 3cdots (2m-1)cdot 2m$? How many times does it appear in the denominator $(m!)^2=1^2cdot 2^2cdot 3^2cdots (m-1)^2cdot m^2$?
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
$begingroup$
...maybe somehow show that all the prime numbers on that interval are divisors of the combination number...
Yes, they are. How many times does such a prime appear in the numerator $(2m)!=1cdot 2cdot 3cdots (2m-1)cdot 2m$? How many times does it appear in the denominator $(m!)^2=1^2cdot 2^2cdot 3^2cdots (m-1)^2cdot m^2$?
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
$endgroup$
...maybe somehow show that all the prime numbers on that interval are divisors of the combination number...
Yes, they are. How many times does such a prime appear in the numerator $(2m)!=1cdot 2cdot 3cdots (2m-1)cdot 2m$? How many times does it appear in the denominator $(m!)^2=1^2cdot 2^2cdot 3^2cdots (m-1)^2cdot m^2$?
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
answered Mar 17 at 23:08
jmerryjmerry
16.5k11633
16.5k11633
add a comment |
add a comment |
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$begingroup$
Consider the prime factorisation of $binom2mm$.
$endgroup$
– Lord Shark the Unknown
Mar 17 at 23:08