Is there a simple formula for $binom{2n}{n} pmod{n^3}$?
7
$begingroup$
Is there a simple formula for the following? $$f(n) = binom{2n}{n} pmod{n^3}$$
I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.
prime-numbers modular-arithmetic binomial-coefficients
edited Jan 11 at 20:58
6005
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
$endgroup$
add a comment |
7
$begingroup$
Is there a simple formula for the following? $$f(n) = binom{2n}{n} pmod{n^3}$$
I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.
prime-numbers modular-arithmetic binomial-coefficients
edited Jan 11 at 20:58
6005
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
$endgroup$
1
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
1
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
1
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
2
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56
add a comment |
7
7
7
4
$begingroup$
Is there a simple formula for the following? $$f(n) = binom{2n}{n} pmod{n^3}$$
I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.
prime-numbers modular-arithmetic binomial-coefficients
edited Jan 11 at 20:58
6005
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
$endgroup$
Is there a simple formula for the following? $$f(n) = binom{2n}{n} pmod{n^3}$$
I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.
prime-numbers modular-arithmetic binomial-coefficients
prime-numbers modular-arithmetic binomial-coefficients
edited Jan 11 at 20:58
6005
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
edited Jan 11 at 20:58
6005
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
edited Jan 11 at 20:58
6005
36.2k751125
edited Jan 11 at 20:58
6005
36.2k751125
edited Jan 11 at 20:58
6005
36.2k751125
36.2k751125
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
asked Jan 11 at 19:25
Ben CrossleyBen Crossley
872318
872318
1
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
1
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
1
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
2
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56
add a comment |
1
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
1
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
1
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
2
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56
1
1
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
1
1
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
1
1
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
2
2
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56
add a comment |
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1
$begingroup$
$f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$.
$endgroup$
– Robert Israel
Jan 11 at 19:54
$begingroup$
Apologies, should be primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 20:25
1
$begingroup$
Might help to look for papers that generalize Lucas's theorem.
$endgroup$
– DanielV
Jan 11 at 21:40
1
$begingroup$
I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3.
$endgroup$
– Ben Crossley
Jan 11 at 23:06
2
$begingroup$
For $p$ prime, $p>3$, ${2p choose p} equiv 2 bmod p^3$ is just a reformulation of Wolstenholme Theorem.
$endgroup$
– René Gy
Jan 12 at 13:56