OFFSET
0,2
LINKS
Enrique Pérez Herrero, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Peter Luschny, Looking for an interpretation, seqfan mailing list.
FORMULA
O.g.f.: (x^4-2*x^3+2*x^2-x-1)/(x^4-2*x^3+2*x-1).
E.g.f.: 1-(5/8)*exp(-x)+(1/8)*(5+14*x+2*x^2)*exp(x).
a(2*n) = n*(n+3) + 0^n = A028552(n) + 0^n. [Here 0^0 = 1, otherwise 0^s = 0. - N. J. A. Sloane, Aug 26 2022]
a(2*n+1) = (n+1)*(n+3) = A005563(n+1).
a(n+1) - a(n) = floor(n/2) + 2 + (-1)^n - 0^n.
a(n) = a(-n-6) = (2*n*(n+6) - 5*(-1)^n + 5)/8 for n>0, a(0)=1. [Bruno Berselli, Dec 18 2015]
For n>0, a(n) = n + 1 + Sum_{i=1..n+1} floor(i/2) + (-1)^i = n + floor((n+1)^2/4) + (1 - (-1)^n)/2. - Enrique Pérez Herrero, Dec 18 2015
Sum_{n>=0} 1/a(n) = 85/36. - Enrique Pérez Herrero, Dec 18 2015
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>5. - R. H. Hardin, Dec 21 2015, proved by Susanne Wienand for the algorithm sent to the seqfan mailing list and used in the Sage script below.
a(n) = A002620(n+1) + A052928(n+1) for n>=1. (Note A198442(n) = A002620(n+2) - A052928(n+2) for n>=1.) - Peter Luschny, Dec 22 2015
a(n) = (floor((n+3)/2)-1)*(ceiling((n+3)/2)+1) for n>0. - Wesley Ivan Hurt, Mar 30 2017
MATHEMATICA
Join[{1}, Table[(2 n (n + 6) - 5 (-1)^n + 5)/8, {n, 1, 60}]] (* Bruno Berselli, Dec 18 2015 *)
PROG
(Sage)
# The initial values x, y = 0, 1 give the quarter-squares A002620.
def A265611():
x, y = 1, 2
while True:
yield x
x, y = x + y, x//y + 1
a = A265611(); print([next(a) for i in range(60)])
(PARI) Vec((x^4-2*x^3+2*x^2-x-1)/(x^4-2*x^3+2*x-1) + O(x^1000)) \\ Altug Alkan, Dec 18 2015
(Magma) [1] cat [(2*n*(n+6)-5*(-1)^n+5)/8: n in [1..60]]; // Bruno Berselli, Dec 18 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Luschny, Dec 17 2015
STATUS
approved