OFFSET
1,2
COMMENTS
Essentially the same as the quarter-squares A002620.
Nonsquare terms of this sequence are given by A002378. - Max Alekseyev, Nov 27 2006
This also gives the number of ways to make change for "c" cents using only pennies, nickels and dimes. You must first set n=floor(c/5), to account for the 5-repetitive nature of the task. - Adam Sasson, Feb 09 2011
These are the segment boundaries of Oppermann's conjecture (1882): n^2-n < p < n^2 < p < n^2+n. - Fred Daniel Kline, Apr 07 2011
a(n) is the number of triples (w,x,y) having all terms in {0..n} and w=2*x+y. - Clark Kimberling, Jun 04 2012
a(n+1) is also the number of points with integer coordinates inside a rectangle isosceles triangle with hypotenuse [0,n] (see A115065 for an equilateral triangle). - Michel Marcus, Aug 05 2013
a(n) = degree of generating polynomials of Galois numbers in (n+1)-dimensional vector space, defined as total number of subspaces in (n+1) space over GF(n) (see Mathematica procedure), when n is a power of a prime. - Artur Jasinski, Aug 31 2016, corrected by Robert Israel, Sep 23 2016
Also number of pairs (x,y) with 0 < x <= y <= n, x + y > n. - Ralf Steiner, Jan 05 2020
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..300
Wikipedia, Oppermann's conjecture.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
FORMULA
a(n) = (n + n mod 2)*(n + 2 - n mod 2)/4.
Numbers of the form m^2 or m^2 - m. - Don Reble, Oct 17 2003
a(1) = 1, a(2) = 2, a(n) = n + a(n - 2). - Alonso del Arte, Jun 18 2005
From Bruno Berselli, Feb 09 2011: (Start)
G.f.: x/((1+x)*(1-x)^3).
a(n) = (2*n*(n+2)-(-1)^n+1)/8. (End)
G.f.: G(0)/(2*(1-x^2)*(1-x)), where G(k) = 1 + 1/(1 - x*(2*k+1)/(x*(2*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
a(n) = (C(n+2,2) - floor((n+2)/2))/2. - Mircea Merca, Nov 23 2013
a(n) = ((-1)^n*(-1 + (-1)^n*(1 + 2*n*(2 + n))))/8. - Fred Daniel Kline, Jan 06 2015
a(n) = Product_{k=1...n-1} (1 + 2 / (k + k mod 2)), n >= 1. - Fred Daniel Kline, Oct 30 2016
E.g.f.: (1/4)*(x*(3 + x)*cosh(x) + (1 + 3*x + x^2)*sinh(x)). - Stefano Spezia, Jan 05 2020
a(n) = (n*(n+2)+(n mod 2))/4. - Chai Wah Wu, Jul 27 2022
Sum_{n>=1} 1/a(n) = Pi^2/6 + 1. - Amiram Eldar, Sep 17 2022
a(n) = A024206(n) + 1. - Ya-Ping Lu, Dec 29 2023
MAPLE
f:= gfun:-rectoproc({a(n)=n+a(n-2), a(1)=1, a(2)=2}, a(n), remember):
map(f, [$1..100]); # Robert Israel, Aug 31 2016
MATHEMATICA
a[1] := 1; a[2] := 2; a[n_] := n + a[n - 2]; Table[a[n], {n, 57}] (* Alonso del Arte *)
GaloisNumber[n_, q_] :=
Sum[QBinomial[n, m, q], {m, 0, n}]; aa = {}; Do[
sub = Table[GaloisNumber[m, n], {n, 0, 200}];
pp = InterpolatingPolynomial[sub, x]; pol = pp /. x -> n + 1;
coef = CoefficientList[pol, n];
AppendTo[aa, Length[coef] - 1], {m, 2, 25}]; aa (* Artur Jasinski, Aug 31 2016 *)
Select[Range[900], Divisible[#, Ceiling[Sqrt[#]]]&] (* or *) LinearRecurrence[ {2, 0, -2, 1}, {1, 2, 4, 6}, 60] (* Harvey P. Dale, Nov 06 2016 *)
PROG
(Magma) [ n: n in [1..841] | n mod Ceiling(Sqrt(n)) eq 0 ]; // Bruno Berselli, Feb 09 2011
(PARI) a(n)=(n+n%2)*(n+2-n%2)/4 \\ Charles R Greathouse IV, Apr 03 2012
(PARI) j=0; for(k=1, 850, s=sqrtint(4*k+1); if(s>j, j=s; print1(k, ", "))) \\ Hugo Pfoertner, Sep 17 2018
(Haskell)
a087811 n = (n + n `mod` 2) * (n + 2 - n `mod` 2) `div` 4
-- Reinhard Zumkeller, Oct 27 2012
(Python)
def A087811(n): return n*(n+2)+(n&1)>>2 # Chai Wah Wu, Jul 27 2022
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Reinhard Zumkeller, Oct 16 2003
STATUS
approved