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%I #25 Sep 08 2022 08:46:18
%S 1,4,7,9,12,14,17,20,22,25,28,30,33,35,38,41,43,46,49,51,54,56,59,62,
%T 64,67,69,72,75,77,80,83,85,88,90,93,96,98,101,103,106,109,111,114,
%U 117,119,122,124,127,130,132,135,138,140,143,145,148,151,153,156,158,161,164,166,169,172,174,177,179
%N Largest integer m for which binomial(m,n-1) > binomial(m-1,n).
%C Equivalently, a(n) is also the largest integer m for which C(m-1,n-2) + C(m-1,n-1) > C(m-1,n) in the (m-1)st row of Pascal's triangle.
%C Equivalently, a(n) is the largest integer m for which m*n > (m-n)*(m-n+1). - _Robert Israel_, Dec 05 2016
%H Robert Israel, <a href="/A279054/b279054.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael Penn, <a href="https://www.youtube.com/watch?v=vTPwcn9-c3w">Putnam Exam | 2016: A2</a>, Youtube video, 2020.
%F a(n) = (3n-1+sqrt(5n^2-2n+1))/2 - 1 if 5n^2-2n+1 is a perfect square, else a(n) = floor((3n-1+sqrt(5n^2-2n+1))/2).
%F a(n) = ceiling((3*n - 3 + sqrt(5*n^2-2*n+1))/2). - _Robert Israel_, Dec 05 2016
%F (3/2+sqrt(5)/2)*n - 2 < f(n) < (3/2+sqrt(5)/2)*t. - _Robert Israel_, Dec 22 2016
%e a(1) = 1, since C(m,0) = 1 > m-1 = C(m-1,1) when m = 1 and C(m,0) <= C(m-1,1) when m >= 2.
%e a(2) = 4, since C(m,1) = m > (m-1)(m-2)/2 = C(m-1,2) when 1 <= m <= 4 and C(m,1) < C(m-1,2) when m >= 5.
%e a(3) = 7, since C(m,2) = m(m-1)/2 >= (m-1)(m-2)(m-3)/6 = C(m-1,3) when 1 <= m <= 7 and C(m,2) < C(m-1,3) when m >= 8.
%p seq(ceil((3*n - 3 + sqrt(5*n^2-2*n+1))/2), n=1..100); # _Robert Israel_, Dec 05 2016
%t Table[Ceiling[(3 n - 3 + Sqrt[5 n^2 - 2 n + 1]) / 2], {n, 60}] (* _Vincenzo Librandi_, Dec 05 2016 *)
%o (Magma) [Ceiling((3*n-3+Sqrt(5*n^2-2*n+1))/2): n in [1..70]]; // _Vincenzo Librandi_, Dec 05 2016
%K nonn
%O 1,2
%A _Timothy L. Tiffin_, Dec 04 2016
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