Does anyone happen to know what the limit of the sum of the nth catalan number over 4^n is? inf ___ \ C(n)
---- /__ 4^n n=0
Thanks! -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039
about : inf ___ \ C(n)
---- = 2. /__ 4^n n=0
The limit is 2. C(n) is slightly lower than 4^n, while the principal term is 4^n, eventually the sum converge to 2, I am not a combinatorist or a wilf-zeilbergerologist but I can say that there is probably a variety of ways to show that. As a number theorist I would suggest, the series slowly increases so it means you can apply lots of tests for series on it validating the convergence of the sum. simon plouffe
You've already had plenty of answers, but may like one more. See A000346 in OEIS and the sum of the first n terms is /2n+1\ / n 2 - | | / 4 which --> 2 \ n+1/ / R. On Fri, 30 Jan 2004, Mike Stay wrote:
Does anyone happen to know what the limit of the sum of the nth catalan number over 4^n is?
inf ___ \ C(n)
---- /__ 4^n n=0
Thanks!
-- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039
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rkg> sum of the first n terms is [...] Good point, Macsyma could (c131) nusum(binomial(2*k,k)/4^k/(k+1),k,0,n) binomial(2 n + 1, n + 1) (d131) 2 - ------------------------ n 4 in 1977!-). Apropos Catalan numbers, a few weeks ago I conjectured the integerhood of 2 binomial(n, i) (n + 2 i + 3)! ------------------------------- . (i + 1)! (i + 2)! (n + 3)! This is the special case C(3,i,n-i) of k /===\ (n + k)! (n + m)! | | (j - 1)! (n + j m + m)! C(m, k, n) := ----------------- | | ----------------------- , n! (n + m + k)! | | (m + j - 1)! (n + j m)! j = 1 which I also conjecture integral. C(0, k, n) = 1, C(1, k, n) = binomial(n + k, k), and Catalan(n) = C(2, n, 0). n + 2 k + 2 (n + 2 k + 1)! C(2, k, n) = ----------------------- -------------- , (n + k + 1) (n + k + 2) k! (k + 1)! n! which may provide a stepping stone to the more general proof.
-----Original Message----- Behalf Of R. William Gosper Sent: Wednesday, February 04, 2004 2:57 AM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] limit question
Apropos Catalan numbers, a few weeks ago I conjectured the integerhood of
2 binomial(n, i) (n + 2 i + 3)! ------------------------------- . (i + 1)! (i + 2)! (n + 3)!
This is the special case C(3,i,n-i) of
k /===\ (n + k)! (n + m)! | | (j - 1)! (n + j m + m)! C(m, k, n) := ----------------- | | ----------------------- , n! (n + m + k)! | | (m + j - 1)! (n + j m)! j = 1 which I also conjecture integral.
This is true. Consider the following partition p(m,k,n)=(n+m,m,...,m) of n+m*(k+1), where m is repeating k times. It is easy to see that C(m,k,n) equals the dimension of the irreducible representation of S_(n+m*(k+1)) corresponding to p(m,k,n) calculated using hook length formula. Another formula for C(m,k,n) is m - 1 /====\ (n + m k + m)! | | i! -------------- | | ------------------------ n! | | (k + i)! (n + k + i + 1) | | i = 0 Alec Mihailovs http://webpages.shepherd.edu/amihailo/
participants (5)
-
Alec Mihailovs -
Mike Stay -
R. William Gosper -
Richard Guy -
Simon Plouffe