OFFSET
0,3
COMMENTS
Equivalently, numbers congruent to {0, 1, 3, 5, 7, 9} mod 11.
In general, n + floor((k-1)*n/k) provides the numbers congruent to {0, 1, 3, 5, ..., 2*k-3} mod (2*k-1) for k>1.
LINKS
Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).
FORMULA
G.f.: x*(1 + 2 x + 2 x^2 + 2 x^3 + 2 x^4 + 2 x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5)).
a(n) = a(n-1) + a(n-6) - a(n-7).
a(6*k + r) = 11*k + 2*r - (1 - (-1)^a(r))/2, with r = 0..5.
MATHEMATICA
Table[n + Floor[5 n/6], {n, 0, 70}]
PROG
(PARI) vector(70, n, n--; n+floor(5*n/6))
(Sage) [n+floor(5*n/6) for n in range(70)];
(Python) [n+int(5*n/6) for n in range(70)]
(Maxima) makelist(n+floor(5*n/6), n, 0, 70);
(Magma) [n+Floor(5*n/6): n in [0..70]];
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Jun 15 2016
STATUS
approved