login
A152741
13 times triangular numbers.
6
0, 13, 39, 78, 130, 195, 273, 364, 468, 585, 715, 858, 1014, 1183, 1365, 1560, 1768, 1989, 2223, 2470, 2730, 3003, 3289, 3588, 3900, 4225, 4563, 4914, 5278, 5655, 6045, 6448, 6864, 7293, 7735, 8190, 8658, 9139, 9633, 10140, 10660, 11193, 11739, 12298, 12870
OFFSET
0,2
COMMENTS
Sequence found by reading the line from 0, in the direction 0, 13,... and the same line from 0, in the direction 0, 39,..., in the square spiral whose vertices are the generalized 15-gonal numbers. - Omar E. Pol, Oct 03 2011
Sum of the numbers from 6n to 7n. - Wesley Ivan Hurt, Dec 22 2015
FORMULA
a(n) = 13*n*(n+1)/2 = 13 * A000217(n).
a(n) = a(n-1)+13*n (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
a(n) = A069126(n+1) - 1. - Omar E. Pol, Oct 03 2011
From Wesley Ivan Hurt, Dec 22 2015: (Start)
G.f.: 13*x/(1-x)^3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>2.
a(n) = Sum_{i=6n..7n} i. (End)
E.g.f.: 13*x*(2+x)*exp(x)/2. - G. C. Greubel, Sep 01 2018
From Amiram Eldar, Feb 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2/13.
Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/13.
Product_{n>=1} (1 - 1/a(n)) = -(13/(2*Pi))*cos(sqrt(21/13)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (13/(2*Pi))*cos(sqrt(5/13)*Pi/2). (End)
MAPLE
A152741:=n->13*n*(n+1)/2: seq(A152741(n), n=0..60); # Wesley Ivan Hurt, Dec 22 2015
MATHEMATICA
Table[13*n*(n-1)/2, {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 06 2011 *)
CoefficientList[Series[13 x/(1 - x)^3, {x, 0, 50}], x] (* Wesley Ivan Hurt, Dec 22 2015 *)
PROG
(Magma) [13*n*(n+1)/2 : n in [0..60]]; // Wesley Ivan Hurt, Dec 22 2015
(PARI) a(n)=13*n*(n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Omar E. Pol, Dec 12 2008
STATUS
approved