A120864 - OEIS

%I #27 Oct 12 2024 14:46:01

%S 2,3,2,12,11,8,3,23,18,11,2,32,23,12,50,39,26,11,59,44,27,8,66,47,26,

%T 3,71,48,23,99,74,47,18,104,75,44,11,107,74,39,2,108,71,32,146,107,66,

%U 23,147,104,59,12,146,99,50,192,143,92,39,191,138,83,26,188,131,72,11,183

%N a(n) is the number j for which there exists a unique pair (j,k) of positive integers such that (j + k + 1)^2 - 4*k = 12*n^2.

%C The k's that match these j's comprise A120865.

%H Vincenzo Librandi, <a href="/A120864/b120864.txt">Table of n, a(n) for n = 1..10000</a>

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling2/kimberling45.html">The equation (j+k+1)^2 - 4*k = Q*n^2 and related dispersions</a>, Journal of Integer Sequences, 10 (2007), Article #07.2.7.

%F a(n) = 3*n^2 - floor(n*sqrt(3))^2.

%F (a(n) + A120865(n) + 1)^2 - 4*A120865(n) = 12*n^2. - _Petros Hadjicostas_, Jul 10 2020

%e a(1) = 3*1^2 - floor(1*sqrt(3))^2 = 2;

%e a(2) = 3*2^2 - floor(2*sqrt(3))^2 = 3;

%e a(3) = 3*3^2 - floor(3*sqrt(3))^2 = 2; etc.

%e Moreover,

%e for n = 1, the unique (j,k) is (2,1): (2+1+1)^2 - 4*1 = 12*1;

%e for n = 2, the unique (j,k) is (3,4): (3+4+1)^2 - 4*4 = 12*4;

%e for n = 3, the unique (j,k) is (2,9): (2+9+1)^2 - 4*9 = 12*9.

%t Table[3n^2-Floor[n*Sqrt[3]]^2,{n,68}] (* _James C. McMahon_, Oct 12 2024 *)

%o (Magma) [3*n^2-Floor(n*Sqrt(3))^2: n in [1..70]]; // _Vincenzo Librandi_, Sep 13 2011

%o (PARI) a(n) = 3*n^2 - sqrtint(3*n^2)^2; \\ _Michel Marcus_, Jul 10 2020

%Y Cf. A120865.

%K nonn

%O 1,1

%A _Clark Kimberling_, Jul 09 2006