%I #80 Jan 02 2023 12:30:51

%S 1,3,4,8,10,15,18,24,28,35,40,48,54,63,70,80,88,99,108,120,130,143,

%T 154,168,180,195,208,224,238,255,270,288,304,323,340,360,378,399,418,

%U 440,460,483,504,528,550,575,598,624,648,675,700,728,754,783,810,840

%N a(n) = a(n-1) + floor((n-1)/2) - (-1)^n + 2 for n>=2, a(0)=1, a(1)=3.

%H Enrique Pérez Herrero, <a href="/A265611/b265611.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%H Peter Luschny, <a href="http://list.seqfan.eu/oldermail/seqfan/2015-December/015868.html">Looking for an interpretation</a>, seqfan mailing list.

%F O.g.f.: (x^4-2*x^3+2*x^2-x-1)/(x^4-2*x^3+2*x-1).

%F E.g.f.: 1-(5/8)*exp(-x)+(1/8)*(5+14*x+2*x^2)*exp(x).

%F 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]

%F a(2*n+1) = (n+1)*(n+3) = A005563(n+1).

%F a(n+1) - a(n) = floor(n/2) + 2 + (-1)^n - 0^n.

%F 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]

%F 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

%F Sum_{n>=0} 1/a(n) = 85/36. - _Enrique Pérez Herrero_, Dec 18 2015

%F 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.

%F 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

%F a(n) = (floor((n+3)/2)-1)*(ceiling((n+3)/2)+1) for n>0. - _Wesley Ivan Hurt_, Mar 30 2017

%p A265611 := proc(n) iquo(n+1,2); %*(%+irem(n+1,2)+2)+0^n end:

%p seq(A265611(n), n=0..55);

%t Join[{1}, Table[(2 n (n + 6) - 5 (-1)^n + 5)/8, {n, 1, 60}]] (* _Bruno Berselli_, Dec 18 2015 *)

%o (Sage)

%o # The initial values x, y = 0, 1 give the quarter-squares A002620.

%o def A265611():

%o x, y = 1, 2

%o while True:

%o yield x

%o x, y = x + y, x//y + 1

%o a = A265611(); print([next(a) for i in range(60)])

%o (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

%o (Magma) [1] cat [(2*n*(n+6)-5*(-1)^n+5)/8: n in [1..60]]; // _Bruno Berselli_, Dec 18 2015

%Y Cf. A002620, A005563, A028552, A052928, A132411, A198442, A217748.

%Y Cf. A084964 and A097065, after the first 3: a(n+1) - a(n) for n>0.

%Y Cf. A055998, after 3: a(n+1) + a(n) for n>0.

%Y Cf. A063929: a(2*n+1) gives the second column of the triangle; for n>0, a(2*n) gives the third column.

%K nonn,easy

%O 0,2

%A _Peter Luschny_, Dec 17 2015