A271723 Numbers k such that 3*k - 8 is a square.

3, 4, 8, 11, 19, 24, 36, 43, 59, 68, 88, 99, 123, 136, 164, 179, 211, 228, 264, 283, 323, 344, 388, 411, 459, 484, 536, 563, 619, 648, 708, 739, 803, 836, 904, 939, 1011, 1048, 1124, 1163, 1243, 1284, 1368, 1411, 1499, 1544, 1636, 1683, 1779, 1828, 1928, 1979, 2083, 2136, 2244, 2299

Offset: 1

Author

Juri-Stepan Gerasimov, Apr 13 2016

Keywords

• nonn
• easy

Comments

Square roots of resulting squares gives A001651. - Ray Chandler, Apr 14 2016

Examples

			a(1) = 3 because 3*3 - 8 = 1^2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A001651.

Cf. numbers n such that 3*n + k is a square: this sequence (k=-8), A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6), A271741 (k=7), A067725 (k=9).

Programs

Magma

[n: n in [1..2400] | IsSquare(3*n-8)];

Maple

seq(seq(((3*m+k)^2+8)/3, k=1..2),m=0..50); # Robert Israel, Dec 05 2016

Mathematica

Select[Range@ 2400, IntegerQ@ Sqrt[3 # - 8] &] (* Bruno Berselli, Apr 14 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{3,4,8,11,19},60] (* Harvey P. Dale, Oct 02 2020 *)

Python

from gmpy2 import is_square
[n for n in range(3000) if is_square(3*n-8)] # Bruno Berselli, Dec 05 2016

Python

[(6*(n-1)*n-(2*n-1)*(-1)**n+23)/8 for n in range(1, 60)] # Bruno Berselli, Dec 05 2016

Formula

From Ilya Gutkovskiy, Apr 13 2016: (Start)

G.f.: x*(3 + x - 2*x^2 + x^3 + 3*x^4)/((1 - x)^3*(1 + x)^2).

a(n) = (6*(n - 1)*n - (2*n - 1)*(-1)^n + 23)/8. (End)