Recurrence relation for 2nCn












0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $184756.00000013$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

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 $184756.00000013$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $184756.00000013$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $184756.00000013$.






share|cite|improve this answer











$endgroup$



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 $184756.00000013$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    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


















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