A168582 a(n) = (4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12.

1, 1, 3, 7, 17, 33, 59, 95, 145, 209, 291, 391, 513, 657, 827, 1023, 1249, 1505, 1795, 2119, 2481, 2881, 3323, 3807, 4337, 4913, 5539, 6215, 6945, 7729, 8571, 9471, 10433, 11457, 12547, 13703, 14929, 16225, 17595, 19039, 20561

Offset

0,3

Comments

Starting with a(2), the sum of the first and last term in row n-1 of the Janet table A172002.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Formula

a(n+2) = A168388(n) + A168380(n), n >= 0.

a(2n) = A168547(n);

a(2n+1) = A168574(n).

G.f.: (1 - 2*x + x^4 + 2*x^2 + 2*x^3)/((1+x)*(x-1)^4). - R. J. Mathar, Jun 27 2011

E.g.f.: (1/12)*((4*x^3 + 6*x^2 + 6*x + 9)*exp(x) + 3*exp(-x)). - G. C. Greubel, Jul 26 2016

Mathematica

Table[(4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12, {n, 0, 50}] (* G. C. Greubel, Jul 26 2016 *)

Prog

(Magma) [2*n/3 +3/4 -n^2/2 +n^3/3 +(-1)^n/4: n in [0..40]]; // Vincenzo Librandi, Aug 06 2011
(PARI) a(n)=(4*n^3-6*n^2+8*n+9+3*(-1)^n)/12 \\ Charles R Greathouse IV, Jul 26 2016

Crossrefs

Cf. A137928 (first differences).

Sequence in context: A292447 A176690 A295089 * A192958 A351405 A219293

Adjacent sequences: A168579 A168580 A168581 * A168583 A168584 A168585

Keyword

nonn,easy

Author

Paul Curtz, Nov 30 2009

Status

approved