A270889 Integers n such that the circular graph C_n has a square size deficiency.

3, 6, 27, 150, 867, 5046, 29403, 171366, 998787, 5821350, 33929307, 197754486, 1152597603, 6717831126, 39154389147, 228208503750, 1330096633347, 7752371296326, 45184131144603, 263352415571286, 1534930362283107, 8946229758127350, 52142448186480987, 303908459360758566, 1771308307978070403

Offset: 0

Author

John Rafael M. Antalan, Mar 25 2016

Keywords

• nonn
• easy

Comments

Define the size deficiency of a graph G as the number of edges needed to complete G. If G is a cycle graph C_n, this sequence gives the values of n for which C_n has a size deficiency which is a perfect square.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• J. R. M. Antalan and I. F. Callano, On the Size Deficiency of Cycle Graphs and Some Integer Sequences, Asian Journal of Mathematics and Computer Research, 11 (3) (2016),192-200.
• Mathematics Stack Exchange, Solving the nonlinear Diophantine equation x2-3x=2y^2
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Programs

Mathematica

a[0] = 3; a[1] = 6; a[n_] := a[n] = 6 a[n - 1] - a[n - 2] - 6; Table[a@ n, {n, 0, 24}] (* Michael De Vlieger, Mar 25 2016 *)
LinearRecurrence[{7,-7,1},{3,6,27},30] (* Harvey P. Dale, Jan 23 2019 *)

PARI

is(n)=issquare(n*(n-3)/2) \\ Charles R Greathouse IV, Mar 25 2016

PARI

a(n)=([0,1,0;0,0,1;1,-7,7]^n*[3;6;27])[1,1] \\ Charles R Greathouse IV, Mar 25 2016

PARI

Vec(3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, Apr 03 2016

Formula

a(n+2) = 6*a(n+1) - a(n) - 6; a(0) = 3 , a(1) = 6.

G.f.: 3*(1-5*x+2*x^2)/((1-x)*(1-6*x+x^2)). - Joerg Arndt, Mar 25 2016

a(n) = 3 * A055997(n+1). - Joerg Arndt, Mar 25 2016

a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2. - Colin Barker, Apr 03 2016

a(n) = 3*(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4. - Colin Barker, Apr 03 2016