

A101461


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.


2



1, 1, 1, 2, 3, 5, 9, 14, 28, 48, 90, 165, 297, 572, 1001, 2002, 3640, 7072, 13260, 25194, 48450, 90440, 177650, 326876, 653752, 1225785, 2414425, 4601610, 8947575, 17298645, 33266625, 65132550, 124062000, 245642760, 463991880, 927983760
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OFFSET

0,4


COMMENTS

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.
The largest difference between adjacent binomial coefficients on nth row of Pascal's triangle.  Vladimir Reshetnikov, Sep 16 2019


LINKS



FORMULA

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.


PROG

(Haskell)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



