

A279054


Largest integer m for which binomial(m,n1) > binomial(m1,n).


1



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
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OFFSET

1,2


COMMENTS

Equivalently, a(n) is also the largest integer m for which C(m1,n2) + C(m1,n1) > C(m1,n) in the (m1)st row of Pascal's triangle.
Equivalently, a(n) is the largest integer m for which m*n > (mn)*(mn+1).  Robert Israel, Dec 05 2016


LINKS

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


FORMULA

a(n) = (3n1+sqrt(5n^22n+1))/2  1 if 5n^22n+1 is a perfect square, else a(n) = floor((3n1+sqrt(5n^22n+1))/2).
a(n) = ceiling((3*n  3 + sqrt(5*n^22*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 > m1 = C(m1,1) when m = 1 and C(m,0) <= C(m1,1) when m >= 2.
a(2) = 4, since C(m,1) = m > (m1)(m2)/2 = C(m1,2) when 1 <= m <= 4 and C(m,1) < C(m1,2) when m >= 5.
a(3) = 7, since C(m,2) = m(m1)/2 >= (m1)(m2)(m3)/6 = C(m1,3) when 1 <= m <= 7 and C(m,2) < C(m1,3) when m >= 8.


MAPLE

seq(ceil((3*n  3 + sqrt(5*n^22*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*n3+Sqrt(5*n^22*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



