%I #68 Sep 04 2022 12:32:35
%S 0,8,11,19,22,30,33,41,44,52,55,63,66,74,77,85,88,96,99,107,110,118,
%T 121,129,132,140,143,151,154,162,165,173,176,184,187,195,198,206,209,
%U 217,220,228,231,239,242,250,253,261,264,272,275,283,286,294,297,305
%N Numbers that are congruent to {0, 8} mod 11.
%C Union of A008593 and A017485. - _Michel Marcus_, Jun 15 2014
%C This sequence mimics in some sense the ceiling function of n/2 (the seq. A110654) relative to variations from a main class of recurrence relations; in order to get the ceiling function of n/2 (see Formula section), the vector v must be [0,1] instead of [3,8]. - _R. J. Cano_, Jun 15 2014
%H R. J. Cano, <a href="/A243520/b243520.txt">Table of n, a(n) for n = 0..9999</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = -5/4*(-1)^n + 11*n/2 + 5/4.
%F From _R. J. Cano_, Jun 15 2014: (Start)
%F a(n) = 5*n + 2*(n mod 2) + ceiling(n/2).
%F If n=0 then a(n) is zero, else a(n) = a(n-1) + v[n mod 2], where v is [3,8]. (End)
%F G.f.: x*(8 + 3*x) / ((1 + x)*(1 - x)^2). [_Bruno Berselli_, Jun 16 2014]
%F a(n) = sum( A010706(i), i=0..n ) - 3. [_Bruno Berselli_, Jun 16 2014]
%F E.g.f.: (11*x*exp(x) + 5*sinh(x))/2. - _David Lovler_, Sep 04 2022
%p A243520:=n->5*n + 2*(n mod 2) + ceil(n/2); seq(A243520(n), n=0..50); # _Wesley Ivan Hurt_, Jun 21 2014
%t Flatten[Table[11 n + {0, 8}, {n, 0, 32}]] (* _Alonso del Arte_, Jun 15 2014 *)
%o (PARI) a(n)=5*n+2*(n%2)+ceil(n/2); /* or */ a(n)=if(!n,0,a(n-1)+[3,8][1+n%2]); \\ _R. J. Cano_, Jun 15 2014
%o (Magma) &cat [[11*n,11*n+8]: n in [0..30]]; // [_Bruno Berselli_, Jun 16 2014]
%Y Cf. A010706, A110654, A078707.
%K nonn,easy
%O 0,2
%A _Viet Quoc Le Tran_, Jun 14 2014
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