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editing
approved
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Revision History for A053175
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
FORMULA
a(n) = 2^n*Sum_{k=0..n} C(n,k)*C(2*k,k)*C(2(n-k),n-k), where C(n,k)=n!/(k!*(n-k)!). This formula has been proved via the Zeilberger algorithm (both sides of the equality satisfy the same recurrence relation). a(n)/2^n also has another expression: Sum_{k=0..floor(n/2)} C(n,2k2*k)*C(2*k,k)^2*4^(n-2*k). - Zhi-Wei Sun, Mar 21 2013
a(n) = (-1)^n*Sum_{k=0..n}C(2k,2*k,k)*C(2(n-k),n-k)*C(k,n-k)*(-4)^k. I have proved this new formula via the Zeilberger algorithm. - Zhi-Wei Sun, Nov 19 2014
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approved
editing
STATUS
reviewed
approved
STATUS
proposed
reviewed
STATUS
editing
proposed
MAPLE
a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 8 else (8*(3*n^2 -3*n+1)*a(n-1)-128*(n-1)^2*a(n-2))/n^2 fi end; # Peter Luschny, Jun 26 2009]
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proposed
editing
STATUS
editing
proposed
FORMULA
a(n) = 2^n*sum_Sum_{k=0}^..n } C(n,k)*C(2k,2*k,k)*C(2(n-k),n-k), where C(n,k)=n!/(k!*(n-k)!). This formula has been proved via the Zeilberger algorithm (both sides of the equality satisfy the same recurrence relation). a(n)/2^n also has another expression: sum_Sum_{k=0}^{..floor(n/2)} C(n,2k)*C(2k,2*k,k)^2*4^(n-2k2*k). - Zhi-Wei Sun, Mar 21 2013
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approved
editing
STATUS
editing
approved
LINKS
Xiao-Juan Ji, Zhi-Hong Sun, <a href="http://arxiv.org/abs/1505.00668">Congruences for Catalan-Larcombe-French numbers</a>, arXiv:1505.00668 [math.NT], 2015 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19
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approved
editing