A120869 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 = 13*n^2.

1, 4, 9, 3, 9, 17, 3, 12, 23, 1, 13, 27, 43, 12, 29, 48, 9, 29, 51, 4, 27, 52, 79, 23, 51, 81, 17, 48, 81, 9, 43, 79, 117, 36, 75, 116, 27, 69, 113, 16, 61, 108, 3, 51, 101, 153, 39, 92, 147, 25, 81, 139, 9, 68, 129, 192, 53, 117, 183, 36, 103, 172, 17, 87, 159, 233, 69, 144

Offset

1,2

Comments

The k's that match these j's comprise A120870.

Links

Table of n, a(n) for n=1..68.

Clark Kimberling, The equation (j+k+1)^2-4*k = Q*n^2 and related dispersions, Journal of Integer Sequences, 10 (2007), Article #07.2.7.

Formula

Let r = (1/2)*sqrt(13). If n is odd, then a(n) = 13*(n/2)^2 - ([n*r+1/2] - 1/2)^2; if n is even, then a(n) = 13*(n/2)^2 - [n*r]^2, where [ ] is the floor function. [corrected by Michel Marcus, Jul 08 2020]

(a(n) + A120870(n) + 1)^2 - 4*A120870(n) = 13*n^2. - Petros Hadjicostas, Jul 08 2020

Example

1 = 13*(1/2)^2 - ([1*r + 1/2] - 1/2)^2,
4 = 13*(2/2)^2 - ([2*r])^2,
9 = 13*(3/2)^2 - ([3*r + 1/2]- 1/2)^2, etc.
Moreover,
for n = 1, the unique (j,k) is (1,3): (1 + 3 + 1)^2 - 4*3 = 13*1;
for n = 2, the unique (j,k) is (4,3): (4 + 3 + 1)^2 - 4*3 = 13*4;
for n = 3, the unique (j,k) is (9,1): (9 + 1 + 1)^2 - 4*1 = 13*9.

Prog

(PARI) a(n) = my(r = sqrt(13)/2); if (n%2, 13*(n/2)^2 - (floor(n*r+1/2) - 1/2)^2, 13*(n/2)^2 - floor(n*r)^2); \\ Michel Marcus, Jul 08 2020

Crossrefs

Cf. A120862, A120863, A120870.

Sequence in context: A373016 A359259 A224953 * A070436 A259450 A219731

Adjacent sequences: A120866 A120867 A120868 * A120870 A120871 A120872

Keyword

nonn

Author

Clark Kimberling, Jul 09 2006

Status

approved