A270440 Least k such that binomial(k, 2) >= binomial(2*n, n).

2, 3, 4, 7, 13, 23, 44, 84, 161, 313, 609, 1189, 2327, 4562, 8958, 17614, 34673, 68318, 134724, 265878, 525066, 1037554, 2051390, 4057939, 8030892, 15900354, 31493446, 62400953, 123682583, 245223436, 486342641, 964809156, 1914483817, 3799849586, 7543612064, 14979070587, 29749371096, 59095356237, 117410567231

Offset: 0

Author

Danny Rorabaugh, Mar 17 2016

Keywords

• nonn

Comments

Open question: Does binomial(a(n), 2) = binomial(2*n, n) for any n > 2? An affirmative answer would settle whether there exists an odd term greater than 3 in A003016.

binomial(a(n),2) > binomial(2*n,n) for 2 < n <= 800000. - Chai Wah Wu, Mar 22 2016

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000217, A000984, A003015, A100967.

Programs

Mathematica

Table[SelectFirst[Range[10^7], Binomial[#, 2] >= Binomial[2 n, n] &], {n, 0, 22}] (* Michael De Vlieger, Mar 17 2016, Version 10 *)

PARI

a(n) = {my(c = binomial(2*n, n)); my(k = 0); while (binomial(k,2) < c, k++); k;} \\ Michel Marcus, Mar 17 2016

Python

from _future_ import division
from gmpy2 import iroot
A270440_list, b = [], 8
for n in range(1001):
    q, r = iroot(b+1,2)
    A270440_list.append(int((q+1)//2 + (0 if r else 1)))
    b = b*2*(2*n+1)//(n+1) # Chai Wah Wu, Mar 22 2016

Sage

def k2_2nn(M): # Produces the first M terms.
    K, n, center, k, triangle = [], 0, 1, 1, 0
    while len(K)
				

Formula

Conjecture: a(n) ~ 2^(n + 1/2) / (Pi*n)^(1/4). - Vaclav Kotesovec, Mar 23 2016

a(n) = ceiling(((8*binomial(2*n,n)+1)^(1/2)+1)/2). The above conjecture is true asymptotically. Using Stirling's formula for the approximation of n!, we get binomial(2*n,n) ~ 2^(2*n)/(Pi*n)^(1/2) and inserting this in the formula for a(n) results in the above approximation for a(n). - Chai Wah Wu, Mar 23 2016