A182868 a(n) = -1 + n + 4*n^2.

-1, 4, 17, 38, 67, 104, 149, 202, 263, 332, 409, 494, 587, 688, 797, 914, 1039, 1172, 1313, 1462, 1619, 1784, 1957, 2138, 2327, 2524, 2729, 2942, 3163, 3392, 3629, 3874, 4127, 4388, 4657, 4934, 5219, 5512, 5813, 6122, 6439, 6764, 7097, 7438, 7787, 8144, 8509, 8882, 9263, 9652

Offset: 0

Author

Paul Curtz, Feb 01 2011

Keywords

• sign
• easy

Comments

First quadrisection of A176126(n). Take clockwise (square) spiral from A023443(n)=n-1: a(n) is on the negative x-axis. Fourth quadrisection (-1-n+4*n^2) is on the negative y-axis.

Conjecture: the 4 quadrisections of (the family) A064038, A160050, A176126, A178242 (see A181407) come from square spiral.

a(n) mod 9 has period 9: 8,4,8,2,4,5,5,4,2. a(n) mod 10 has period 10: 9,4,7,8,7,4,9,2,3,2. Each polynomial modulo some constant c has a period of length c (and perhaps shorter ones). - Paul Curtz and Bruno Berselli, Feb 05 2011

Links

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

Programs

Magma

[-1+n+4*n^2: n in [0..700] ] // Vincenzo Librandi, Feb 01 2011

Mathematica

f[n_]:=-1+n+4*n^2;f[Range[0,100]] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2011 *)

PARI

a(n)=-1+n+4*n^2 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = A176126(4*n).

a(n) = 4*n^2 + n - 1.

a(n) = a(n-1) - 3 + 8*n.

a(n) = 2*a(n) - a(n-2) + 8.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

G.f.: -(1 - 7*x - 2*x^2)/(1-x)^3. - Bruno Berselli, Feb 05 2011