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A270440
Least k such that binomial(k, 2) >= binomial(2*n, n).
1
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
0,1
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
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^(2n)/(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
MATHEMATICA
Table[SelectFirst[Range[10^7], Binomial[#, 2] >= Binomial[2 n, n] &], {n, 0, 22}] (* Michael De Vlieger, Mar 17 2016, Version 10 *)
PROG
(Sage)
def k2_2nn(M): # Produces the first M terms.
K, n, center, k, triangle = [], 0, 1, 1, 0
while len(K)<M:
while triangle<center:
triangle, k = triangle + k, k + 1
K.append(k)
center, n = center*(2*n+1)*(2*n+2)/(n+1)^2, n + 1
return K
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Danny Rorabaugh, Mar 17 2016
STATUS
approved