A279054 Largest integer m for which binomial(m,n-1) > binomial(m-1,n).

1, 4, 7, 9, 12, 14, 17, 20, 22, 25, 28, 30, 33, 35, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 69, 72, 75, 77, 80, 83, 85, 88, 90, 93, 96, 98, 101, 103, 106, 109, 111, 114, 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

Offset

1,2

Comments

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.

Equivalently, a(n) is the largest integer m for which m*n > (m-n)*(m-n+1). - Robert Israel, Dec 05 2016

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Michael Penn, Putnam Exam | 2016: A2, Youtube video, 2020.

Formula

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).

a(n) = ceiling((3*n - 3 + sqrt(5*n^2-2*n+1))/2). - Robert Israel, Dec 05 2016

(3/2+sqrt(5)/2)*n - 2 < f(n) < (3/2+sqrt(5)/2)*t. - Robert Israel, Dec 22 2016

Example

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.
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.
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.

Maple

seq(ceil((3*n - 3 + sqrt(5*n^2-2*n+1))/2), n=1..100); # Robert Israel, Dec 05 2016

Mathematica

Table[Ceiling[(3 n - 3 + Sqrt[5 n^2 - 2 n + 1]) / 2], {n, 60}] (* Vincenzo Librandi, Dec 05 2016 *)

Prog

(Magma) [Ceiling((3*n-3+Sqrt(5*n^2-2*n+1))/2): n in [1..70]]; // Vincenzo Librandi, Dec 05 2016

Crossrefs

Sequence in context: A186323 A007064 A007073 * A086824 A266936 A080574

Adjacent sequences: A279051 A279052 A279053 * A279055 A279056 A279057

Keyword

nonn

Author

Timothy L. Tiffin, Dec 04 2016

Status

approved