%I #19 Sep 17 2019 09:02:37
%S 1,1,1,2,3,5,9,14,28,48,90,165,297,572,1001,2002,3640,7072,13260,
%T 25194,48450,90440,177650,326876,653752,1225785,2414425,4601610,
%U 8947575,17298645,33266625,65132550,124062000,245642760,463991880,927983760
%N Row maximum of Catalan triangle with zeros (A053121), i.e., maximum value of (m+1)*binomial(n+1,(nm)/2)/(n+1) for given n with m same parity as n.
%C There are two maximum values when n is of the form k^2 + 2k  1 (i.e., 2 less than a square, A008865 offset) in which case m = k +/ 1. In general m is the integer with the same parity as n closest to sqrt(n+2)  1.
%C The largest difference between adjacent binomial coefficients on nth row of Pascal's triangle.  _Vladimir Reshetnikov_, Sep 16 2019
%H Reinhard Zumkeller, <a href="/A101461/b101461.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = (m+1)*binomial(n+1, (nm)/2)/(n+1) where m = floor(sqrt(n+2)  (1 + (1)^floor(n + sqrt(n+2)  1))/2). a(n) seems to be slightly less than 2^n/n.
%o (Haskell)
%o a101461 = maximum . a053121_row  _Reinhard Zumkeller_, Mar 04 2012
%K nonn
%O 0,4
%A _Henry Bottomley_, Jan 20 2005
