A217278 Sequences A124174 and A006454 interlaced.

0, 0, 1, 3, 10, 15, 45, 120, 351, 528, 1540, 4095, 11935, 17955, 52326, 139128, 405450, 609960, 1777555, 4726275, 13773376, 20720703, 60384555, 160554240, 467889345, 703893960, 2051297326, 5454117903, 15894464365, 23911673955, 69683724540, 185279454480

Offset: 0

Author

Raphie Frank, Sep 29 2012

Keywords

• nonn
• easy

Comments

a(2n) and 2*a(2n) + 1 are triangular.

a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square).

Examples

			a(18) = 35*(52326 - 1540) + 45 = 1777555,
a(19) = 35*(139128 - 4095) + 120 = 4726275.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1).

Crossrefs

Cf. (sqrt(8a(2n) + 1) - 1)/2 = A216134(n) = A216162(2n + 1).

Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).

Cf. (sqrt(8a(2n+1) + 1) - 1)/2 = A006451(n).

Programs

Mathematica

LinearRecurrence[{0,1,0,34,0,-34,0,-1,0,1},{0,0,1,3,10,15,45,120,351,528},40] (* Harvey P. Dale, Aug 04 2019 *)

PARI

concat([0,0], Vec(-x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)) + O(x^100))) \\ Colin Barker, Jun 23 2015

Formula

a(n) = 35*(a(n-4) - a(n-8)) + a(n-12).

lim n --> infinity a(2n)/a(2n - 1) = (3 + sqrt(8))/2.

From Raphie Frank, Dec 21 2015: (Start)

a(2*n + 1) = 1/64*(((4 + sqrt(2)) * (1 - (-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2)) + (4 - sqrt(2)) * (1+(-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2))))^2 - 1.

a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).

(End)