login
A243520
Numbers that are congruent to {0, 8} mod 11.
2
0, 8, 11, 19, 22, 30, 33, 41, 44, 52, 55, 63, 66, 74, 77, 85, 88, 96, 99, 107, 110, 118, 121, 129, 132, 140, 143, 151, 154, 162, 165, 173, 176, 184, 187, 195, 198, 206, 209, 217, 220, 228, 231, 239, 242, 250, 253, 261, 264, 272, 275, 283, 286, 294, 297, 305
OFFSET
0,2
COMMENTS
Union of A008593 and A017485. - Michel Marcus, Jun 15 2014
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
FORMULA
a(n) = -5/4*(-1)^n + 11*n/2 + 5/4.
From R. J. Cano, Jun 15 2014: (Start)
a(n) = 5*n + 2*(n mod 2) + ceiling(n/2).
If n=0 then a(n) is zero, else a(n) = a(n-1) + v[n mod 2], where v is [3,8]. (End)
G.f.: x*(8 + 3*x) / ((1 + x)*(1 - x)^2). [Bruno Berselli, Jun 16 2014]
a(n) = sum( A010706(i), i=0..n ) - 3. [Bruno Berselli, Jun 16 2014]
E.g.f.: (11*x*exp(x) + 5*sinh(x))/2. - David Lovler, Sep 04 2022
MAPLE
A243520:=n->5*n + 2*(n mod 2) + ceil(n/2); seq(A243520(n), n=0..50); # Wesley Ivan Hurt, Jun 21 2014
MATHEMATICA
Flatten[Table[11 n + {0, 8}, {n, 0, 32}]] (* Alonso del Arte, Jun 15 2014 *)
PROG
(PARI) a(n)=5*n+2*(n%2)+ceil(n/2); \\ R. J. Cano, Jun 15 2014
(PARI) a(n)=if(!n, 0, a(n-1)+[3, 8][1+n%2]); \\ R. J. Cano, Jun 15 2014
(Magma) &cat [[11*n, 11*n+8]: n in [0..30]]; // [Bruno Berselli, Jun 16 2014]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Viet Quoc Le Tran, Jun 14 2014
STATUS
approved