A194715 15 times triangular numbers.

0, 15, 45, 90, 150, 225, 315, 420, 540, 675, 825, 990, 1170, 1365, 1575, 1800, 2040, 2295, 2565, 2850, 3150, 3465, 3795, 4140, 4500, 4875, 5265, 5670, 6090, 6525, 6975, 7440, 7920, 8415, 8925, 9450, 9990, 10545, 11115, 11700, 12300, 12915, 13545, 14190, 14850, 15525

Offset: 0

Author

Omar E. Pol, Oct 03 2011

Keywords

• nonn
• easy

Comments

Sequence found by reading the line from 0, in the direction 0, 15, ... and the same line from 0, in the direction 0, 45, ..., in the square spiral whose vertices are the generalized 17-gonal numbers.

Sum of the numbers from 7*n to 8*n. - Wesley Ivan Hurt, Dec 23 2015

Also the number of 4-cycles in the (n+6)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 28 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
• Eric Weisstein's World of Mathematics, Graph Cycle.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000217, A028895, A035008, A045943, A069128, A163756.

Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017

Programs

Magma

[15*n*(n+1)/2: n in [0..50]]; // Vincenzo Librandi, Oct 04 2011

Maple

A194715:=n->15*n*(n+1)/2: seq(A194715(n), n=0..60); # Wesley Ivan Hurt, Dec 23 2015

Mathematica

15*Accumulate[Range[0, 60]] (* Harvey P. Dale, Feb 12 2012 *)
Table[15 n (n + 1)/2, {n, 0, 60}] (* Wesley Ivan Hurt, Dec 23 2015 *)
15 Binomial[Range[20], 2] (* Eric W. Weisstein, Jul 28 2017 *)
15 PolygonalNumber[Range[0, 20]] (* Eric W. Weisstein, Jul 28 2017 *)

PARI

a(n)=15*n*(n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 15*n*(n+1)/2 = 15*A000217(n) = 5*A045943(n) = 3*A028895(n) = A069128(n+1) - 1.

From Wesley Ivan Hurt, Dec 23 2015: (Start)

G.f.: 15*x/(1-x)^3.

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

a(n) = Sum_{i=7*n..8*n} i. (End)

From Amiram Eldar, Feb 21 2023: (Start)

Sum_{n>=1} 1/a(n) = 2/15.

Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/15.

Product_{n>=1} (1 - 1/a(n)) = -(15/(2*Pi))*cos(sqrt(23/15)*Pi/2).

Product_{n>=1} (1 + 1/a(n)) = (15/(2*Pi))*cos(sqrt(7/15)*Pi/2). (End)

E.g.f.: 15*exp(x)*x*(2 + x)/2. - Elmo R. Oliveira, Dec 25 2024