A120870 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 = 13*n^2.
3, 3, 1, 12, 9, 4, 23, 17, 9, 36, 27, 16, 3, 39, 25, 9, 53, 36, 17, 69, 49, 27, 3, 64, 39, 12, 81, 53, 23, 100, 69, 36, 1, 87, 51, 13, 107, 68, 27, 129, 87, 43, 153, 108, 61, 12, 131, 81, 29, 156, 103, 48, 183, 127, 69, 9, 153, 92, 29, 181, 117, 51, 211, 144, 75, 4, 173, 101
Offset: 1
Keywords
Examples
3 = ([1*r + 1/2] + 1/2)^2 - (13/4)*1^2, 3 = (1 + [2*r])^2 - (13/4)*2^2, 1 = ([3*r+1/2] + 1/2)^2 - (13/4)*3^2, 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.
Links
- 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.
Programs
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PARI
a(n) = my(r = sqrt(13)/2); if (n%2, (floor(n*r+1/2) + 1/2)^2 - (13/4)*n^2, (1 + floor(n*r))^2 - (13/4)*n^2); \\ Michel Marcus, Jul 08 2020
Formula
Let r = (1/2)*sqrt(13). If n is odd, then a(n) = ([n*r+1/2] + 1/2)^2 - (13/4)*n^2; if n is even, then a(n) = (1 + [n*r])^2 - (13/4)*n^2, where [ ] is the floor function. [corrected by Michel Marcus, Jul 08 2020]
(A120869(n) + a(n) + 1)^2 - 4*a(n) = 13*n^2. - Petros Hadjicostas, Jul 08 2020
Comments