OFFSET
1,1
COMMENTS
If we insert an initial 0, and alternate the signs: 0,2,-5,11,-18,28,..., we get a sequence where the average of the first n terms is an integer, with no repeats: specifically A001057(n-1). The sum of the first n terms is (-1)^(n-1)*A093353(n-1). - Franklin T. Adams-Watters, May 20 2010
Suppose that n cards have the numbers 1..2n written on them randomly, one number to a side, and are set out on a table randomly. You have the task of maximizing the sum of the visible numbers by flipping cards. If you have no information other than the numbers on the upper faces, and may not flip any particular card more than once, a(n) is the largest sum you can guarantee in the worst case. - Andrew Woods, Jun 06 2013
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
FORMULA
a(n) = Sum_{i=1..n} A042964(i).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - R. H. Hardin, Nov 13 2011
a(n) = ceiling((2*n+1)*n/2). - Andrew Woods, Jun 06 2013
G.f.: x*(2+x+x^2) / ((1-x)^3*(x+1)). - Colin Barker, Jun 04 2014
a(n) = round(n/(1-exp(-1/n))). - Richard R. Forberg, Jan 28 2015
MATHEMATICA
a[n_]:=Ceiling[((2n+1)n/2)]; Array[a, 60] (* Vincenzo Librandi, Jul 02 2019 *)
LinearRecurrence[{2, 0, -2, 1}, {2, 5, 11, 18}, 60] (* G. C. Greubel, Jul 02 2019 *)
PROG
(PARI) Vec(x*(x^2+x+2)/((1-x)^3*(x+1)) + O(x^60)) \\ Colin Barker, Jun 04 2014
(Magma) [Ceiling((2*n+1)*n/2): n in [1..60]]; // Vincenzo Librandi, Jul 02 2019
(Sage) [ceiling(n*(1+2*n)/2) for n in (1..60)] # G. C. Greubel, Jul 02 2019
(GAP) a:=[2, 5, 11, 18];; for n in [5..60] do a[n]:=2*a[n-1]-2*a[n-3] + a[n-4]; od; a; # G. C. Greubel, Jul 02 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jaroslav Krizek, Dec 18 2009
STATUS
approved