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

%I A270440 #32 Mar 06 2025 08:18:18
%S A270440 2,3,4,7,13,23,44,84,161,313,609,1189,2327,4562,8958,17614,34673,
%T A270440 68318,134724,265878,525066,1037554,2051390,4057939,8030892,15900354,
%U A270440 31493446,62400953,123682583,245223436,486342641,964809156,1914483817,3799849586,7543612064,14979070587,29749371096,59095356237,117410567231
%N A270440 Least k such that binomial(k, 2) >= binomial(2*n, n).
%C A270440 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.
%C A270440 binomial(a(n),2) > binomial(2*n,n) for 2 < n <= 800000. - _Chai Wah Wu_, Mar 22 2016
%H A270440 Chai Wah Wu, <a href="/A270440/b270440.txt">Table of n, a(n) for n = 0..1000</a>
%F A270440 Conjecture: a(n) ~ 2^(n + 1/2) / (Pi*n)^(1/4). - _Vaclav Kotesovec_, Mar 23 2016
%F A270440 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
%t A270440 Table[SelectFirst[Range[10^7], Binomial[#, 2] >= Binomial[2 n, n] &], {n, 0, 22}] (* _Michael De Vlieger_, Mar 17 2016, Version 10 *)
%o A270440 (Sage)
%o A270440 def k2_2nn(M): # Produces the first M terms.
%o A270440     K, n, center, k, triangle = [], 0, 1, 1, 0
%o A270440     while len(K)<M:
%o A270440         while triangle<center:
%o A270440             triangle, k = triangle + k, k + 1
%o A270440         K.append(k)
%o A270440         center, n = center*(2*n+1)*(2*n+2)/(n+1)^2, n + 1
%o A270440     return K
%o A270440 (PARI) a(n) = {my(c = binomial(2*n, n)); my(k = 0); while (binomial(k,2) < c, k++); k;} \\ _Michel Marcus_, Mar 17 2016
%o A270440 (Python)
%o A270440 from __future__ import division
%o A270440 from gmpy2 import iroot
%o A270440 A270440_list, b = [], 8
%o A270440 for n in range(1001):
%o A270440     q, r = iroot(b+1,2)
%o A270440     A270440_list.append(int((q+1)//2 + (0 if r else 1)))
%o A270440     b = b*2*(2*n+1)//(n+1) # _Chai Wah Wu_, Mar 22 2016
%Y A270440 Cf. A000217, A000984, A003015, A100967.
%K A270440 nonn
%O A270440 0,1
%A A270440 _Danny Rorabaugh_, Mar 17 2016