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

0,4

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

Links

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

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.

Prog

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

Crossrefs

Sequence in context: A352079 A105044 A026008 * A085897 A361906 A306829

Adjacent sequences: A101458 A101459 A101460 * A101462 A101463 A101464

Keyword

nonn

Author

Henry Bottomley, Jan 20 2005

Status

approved