%I #16 Jul 07 2020 20:51:14
%S 1,4,1,5,11,4,11,1,9,19,5,16,29,11,25,4,19,36,11,29,1,20,41,9,31,55,
%T 19,44,5,31,59,16,45,76,29,61,11,44,79,25,61,4,41,80,19,59,101,36,79,
%U 11,55,101,29,76,1,49,99,20,71,124,41,95,9,64,121,31,89,149,55,116,19,81,145
%N a(n) is the number k for which there exists a unique pair (j,k) of positive integers such that (j + k + 1)^2 - 4*k = 5*n^2.
%C The j's that match these k's comprise A005752.
%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.
%H Clark Kimberling, <a href="http://www.fq.math.ca/Papers1/45-4/quartkimberling04_2007.pdf">The equation m^2 - 4k = 5n^2 and unique representations of positive integers</a>, Fibonacci Quart. 45(4) (2007), 304-312.
%F Let r = (1/2)*sqrt(5). If n is odd, then a(n) = ([n*r+1/2] + 1/2)^2 - (5/4)*n^2; if n is even, then a(n) = (1 + [n*r])^2 - (5/4)*n^2, where [ ] is the floor function.
%e 1 = ([r+1/2] + 1/2)^2 - (5/4)*1^2,
%e 4 =(1+[2*r])^2 - (5/4)*2^2,
%e 1 = ([3*r+1/2] + 1/2)^2 - (5/4)*3^2, etc.
%e Moreover,
%e for n = 1, the unique (j,k) is (1,1): (1 + 1 + 1)^2 - 4*1 = 5*1;
%e for n = 2, the unique (j,k) is (1,4): (1 + 4 + 1)^2 - 4*4 = 5*4;
%e for n = 3, the unique (j,k) is (5,1): (5 + 1 + 1)^2 - 4*1 = 5*9.
%t r = Sqrt[5]/2; Table[If[OddQ@ n, (Floor[n r + 1/2] + 1/2)^2 - (5/4) n^2, (1 + Floor[n r])^2 - (5/4) n^2], {n, 73}] (* _Michael De Vlieger_, Mar 06 2016 *)
%o (PARI) a(n) = my(fnt = floor(n*(sqrt(5)+1)/2)); fnt^2 + (2-n)*fnt - n^2 - n + 1; \\ _Michel Marcus_, Mar 05 2016
%Y Cf. A005752.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jul 09 2006