A101461 Row maximum of Catalan triangle with zeros (A053121), i.e., maximum value of (m+1)*binomial(n+1,(n-m)/2)/(n+1) for given n with m same parity as n.

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

Offset: 0

Author

Henry Bottomley, Jan 20 2005

Keywords

• nonn

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 n-th row of Pascal's triangle. - Vladimir Reshetnikov, Sep 16 2019

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Programs

Haskell

a101461 = maximum . a053121_row  -- Reinhard Zumkeller, Mar 04 2012

Formula

a(n) = (m+1)*binomial(n+1, (n-m)/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.