A139272 a(n) = n*(8*n-5).

0, 3, 22, 57, 108, 175, 258, 357, 472, 603, 750, 913, 1092, 1287, 1498, 1725, 1968, 2227, 2502, 2793, 3100, 3423, 3762, 4117, 4488, 4875, 5278, 5697, 6132, 6583, 7050, 7533, 8032, 8547, 9078, 9625, 10188, 10767, 11362, 11973, 12600

Offset: 0

Author

Omar E. Pol, Apr 26 2008

Keywords

• nonn
• easy

Comments

Sequence found by reading the line from 0, in the direction 0, 3, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139276 in the same spiral.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
• Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(n*k-k+6)/2, this sequence is the case k=16: see Comments lines of A226492.

Programs

Mathematica

s=0;lst={s};Do[s+=n++ +3;AppendTo[lst, s], {n, 0, 7!, 16}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
Table[n*(8*n -5), {n,0,50}] (* G. C. Greubel, Jul 18 2017 *)
LinearRecurrence[{3,-3,1},{0,3,22},50] (* Harvey P. Dale, Jan 13 2024 *)

PARI

a(n)=n*(8*n-5) \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 8*n^2 - 5*n.

Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008

a(n) = 16*n + a(n-1) - 13 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010

From G. C. Greubel, Jul 18 2017: (Start)

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

G.f.: x*(13*x + 3)/(1-x)^3.

E.g.f.: (8*x^2 + 3*x)*exp(x). (End)

Sum_{n>=1} 1/a(n) = ((sqrt(2)-1)*Pi + 8*log(2) - 2*sqrt(2)*log(sqrt(2)+1))/10. - Amiram Eldar, Mar 17 2022