Recurrence relation for 2nCn
$begingroup$
Does there exist an easy to compute (by hand) recurrence relation for the central column of pascals triangle?
I'm trying to avoid factorials.
A recurrence for {1, 2, 6, 20, 70, 252... }
Thanks in advance, Ben
recurrence-relations binomial-coefficients
$endgroup$
add a comment |
$begingroup$
Does there exist an easy to compute (by hand) recurrence relation for the central column of pascals triangle?
I'm trying to avoid factorials.
A recurrence for {1, 2, 6, 20, 70, 252... }
Thanks in advance, Ben
recurrence-relations binomial-coefficients
$endgroup$
1
$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53
add a comment |
$begingroup$
Does there exist an easy to compute (by hand) recurrence relation for the central column of pascals triangle?
I'm trying to avoid factorials.
A recurrence for {1, 2, 6, 20, 70, 252... }
Thanks in advance, Ben
recurrence-relations binomial-coefficients
$endgroup$
Does there exist an easy to compute (by hand) recurrence relation for the central column of pascals triangle?
I'm trying to avoid factorials.
A recurrence for {1, 2, 6, 20, 70, 252... }
Thanks in advance, Ben
recurrence-relations binomial-coefficients
recurrence-relations binomial-coefficients
asked Dec 10 '18 at 9:39
Ben CrossleyBen Crossley
872318
872318
1
$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53
add a comment |
1
$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53
1
1
$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53
$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular,
$$n a_n+2(1-2n)a_{n-1}=0 qquad text{with} qquad a_0=1$$
Edit
As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice
$$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact)
$$log(a_n)simeq 2log(2)n-log(sqrt{pi n})-frac{5 n left(5208 n^2+4121right)}{6 left(34720 n^4+28920 n^2+771right)}$$ which is equivalent to an $Oleft(frac{1}{n^9}right)$ series expansion.
For example, for $a_{10}$ the result is $184756.00000013$.
$endgroup$
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
add a comment |
Your Answer
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1 Answer
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$begingroup$
This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular,
$$n a_n+2(1-2n)a_{n-1}=0 qquad text{with} qquad a_0=1$$
Edit
As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice
$$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact)
$$log(a_n)simeq 2log(2)n-log(sqrt{pi n})-frac{5 n left(5208 n^2+4121right)}{6 left(34720 n^4+28920 n^2+771right)}$$ which is equivalent to an $Oleft(frac{1}{n^9}right)$ series expansion.
For example, for $a_{10}$ the result is $184756.00000013$.
$endgroup$
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
add a comment |
$begingroup$
This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular,
$$n a_n+2(1-2n)a_{n-1}=0 qquad text{with} qquad a_0=1$$
Edit
As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice
$$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact)
$$log(a_n)simeq 2log(2)n-log(sqrt{pi n})-frac{5 n left(5208 n^2+4121right)}{6 left(34720 n^4+28920 n^2+771right)}$$ which is equivalent to an $Oleft(frac{1}{n^9}right)$ series expansion.
For example, for $a_{10}$ the result is $184756.00000013$.
$endgroup$
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
add a comment |
$begingroup$
This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular,
$$n a_n+2(1-2n)a_{n-1}=0 qquad text{with} qquad a_0=1$$
Edit
As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice
$$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact)
$$log(a_n)simeq 2log(2)n-log(sqrt{pi n})-frac{5 n left(5208 n^2+4121right)}{6 left(34720 n^4+28920 n^2+771right)}$$ which is equivalent to an $Oleft(frac{1}{n^9}right)$ series expansion.
For example, for $a_{10}$ the result is $184756.00000013$.
$endgroup$
This is sequence $A000984$ in $OEIS$. You will find a lot of informations about it and, in particular,
$$n a_n+2(1-2n)a_{n-1}=0 qquad text{with} qquad a_0=1$$
Edit
As I wrote in comments, in the $OEIS$ page, Gerry Martens gave in 2011 the nice
$$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ If you want an approximation which is quite accurate (using Stirling approximation and Padé approximants for making the series more compact)
$$log(a_n)simeq 2log(2)n-log(sqrt{pi n})-frac{5 n left(5208 n^2+4121right)}{6 left(34720 n^4+28920 n^2+771right)}$$ which is equivalent to an $Oleft(frac{1}{n^9}right)$ series expansion.
For example, for $a_{10}$ the result is $184756.00000013$.
edited Dec 17 '18 at 15:05
answered Dec 10 '18 at 9:54
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
add a comment |
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
Thanks Claude! I'll have to try OEIS more often. Didn't think of that!
$endgroup$
– Ben Crossley
Dec 10 '18 at 10:02
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
@BenCrossley. It is an incredible source of information. I use it very often (as many eople) in particular for asymptotics, codes and so on. Cheers.
$endgroup$
– Claude Leibovici
Dec 10 '18 at 10:13
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
Do you know of a way to turn this recurrence into one without divisions? Or into one dependant on multiple previous terms?
$endgroup$
– Ben Crossley
Dec 17 '18 at 13:59
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
@BenCrossley. If you look at the page in OEIS, you will find $$a_n=frac{ (-4)^n sqrt{pi }}{Gamma left(frac{1}{2}-nright) Gamma (n+1)}$$ given by Gerry Martens in 2011
$endgroup$
– Claude Leibovici
Dec 17 '18 at 14:17
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
$begingroup$
That has divisions in it too! No fractions at all. I couldn't see one on the OEIS page.
$endgroup$
– Ben Crossley
Dec 17 '18 at 17:06
add a comment |
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$begingroup$
They satisfy the recurrence $$a_{n+1}= frac{4n+2}{n+1}a_n$$ see this
$endgroup$
– rsadhvika
Dec 10 '18 at 9:53