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 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. 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified August 4 13:04 EDT 2024. Contains 374921 sequences. (Running on oeis4.)