A163756 14 times triangular numbers.

0, 14, 42, 84, 140, 210, 294, 392, 504, 630, 770, 924, 1092, 1274, 1470, 1680, 1904, 2142, 2394, 2660, 2940, 3234, 3542, 3864, 4200, 4550, 4914, 5292, 5684, 6090, 6510, 6944, 7392, 7854, 8330, 8820, 9324, 9842, 10374, 10920, 11480, 12054, 12642, 13244, 13860

Offset: 0

Author

Vincenzo Librandi, Aug 03 2009

Keywords

• nonn
• easy

Comments

Sequence found by reading the line from 0, in the direction 0, 14, ... and the same line from 0, in the direction 0, 42, ..., in the square spiral whose vertices are the generalized 16-gonal numbers. - Omar E. Pol, Oct 03 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000217, A002378, A069127.

Cf. A274978 (generalized 16-gonal numbers).

Programs

Mathematica

Table[7*n*(n-1),{n,100}] (* Vladimir Joseph Stephan Orlovsky, Jul 06 2011 *)
14*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,14,42},50] (* Harvey P. Dale, May 11 2021 *)

PARI

a(n)=7*n*(n+1) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 7*n*(n+1) = 14*A000217(n).

G.f.: 14*x/(1-x)^3.

a(n) = 7*A002378(n) = 2*A024966(n) = A069127(n+1) - 1. - Omar E. Pol, Oct 03 2011

E.g.f.: 7*x*(x + 2)*exp(x). - G. C. Greubel, Aug 02 2017

From Amiram Eldar, Feb 21 2023: (Start)

Sum_{n>=1} 1/a(n) = 1/7.

Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/7.

Product_{n>=1} (1 - 1/a(n)) = -(7/Pi)*cos(sqrt(11/7)*Pi/2).

Product_{n>=1} (1 + 1/a(n)) = (7/Pi)*cos(sqrt(3/7)*Pi/2). (End)

Extensions

Extended by R. J. Mathar, Aug 06 2009