OFFSET
0,5
COMMENTS
Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).
Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008
Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010
The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).
FORMULA
G.f.: x^3 / ((1-x^3)*(1-x)^2).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = round(n*(n-1)/6) = round((n^2-n-1)/6) = floor(n*(n-1)/6) = ceiling((n+1)*(n-2)/6). - Mircea Merca, Nov 28 2010
a(n) = a(n-3) + n - 2, n > 2. - Mircea Merca, Nov 28 2010
a(n) = A214734(n, 1, 3). - Renzo Benedetti, Aug 27 2012
a(3n) = A000326(n), a(3n+1) = A005449(n), a(3n+2) = 3*A000217(n) = A045943(n). - Philippe Deléham, Mar 26 2013
a(n) = (3*n*(n-1) - (-1)^n*((1+i*sqrt(3))^(n-2) + (1-i*sqrt(3))^(n-2))/2^(n-3) - 2)/18, where i=sqrt(-1). - Bruno Berselli, Nov 30 2014
Sum_{n>=3} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 17 2022
MAPLE
seq(floor(n*(n-1)/6), n=0..60); # Robert Israel, Nov 27 2014
MATHEMATICA
Table[n, {n, 0, 19}, {3}] // Flatten // Accumulate (* Jean-François Alcover, Jun 05 2013 *)
PROG
(Sage) [floor(binomial(n, 2)/3) for n in range(0, 60)] # Zerinvary Lajos, Dec 01 2009
(Magma) [Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011
(PARI) a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013
(GAP) List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Hieronymus Fischer, Jun 01 2007
STATUS
approved