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%I #27 Sep 08 2022 08:46:16
%S 0,1,3,5,7,9,11,12,14,16,18,20,22,23,25,27,29,31,33,34,36,38,40,42,44,
%T 45,47,49,51,53,55,56,58,60,62,64,66,67,69,71,73,75,77,78,80,82,84,86,
%U 88,89,91,93,95,97,99,100,102,104,106,108,110,111,113,115,117,119
%N a(n) = n + floor(5*n/6).
%C Equivalently, numbers congruent to {0, 1, 3, 5, 7, 9} mod 11.
%C 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.
%H Bruno Berselli, <a href="/A272915/b272915.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).
%F 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)).
%F a(n) = a(n-1) + a(n-6) - a(n-7).
%F a(6*k + r) = 11*k + 2*r - (1 - (-1)^a(r))/2, with r = 0..5.
%t Table[n + Floor[5 n/6], {n, 0, 70}]
%o (PARI) vector(70, n, n--; n+floor(5*n/6))
%o (Sage) [n+floor(5*n/6) for n in range(70)];
%o (Python) [n+int(5*n/6) for n in range(70)]
%o (Maxima) makelist(n+floor(5*n/6), n, 0, 70);
%o (Magma) [n+Floor(5*n/6): n in [0..70]];
%Y Cf. similar sequences with formula n+floor((k-1)*n/k): A032766 (k=2), A047220 (k=3), A047392 (k=4), A187318 (k=5).
%K nonn,easy
%O 0,3
%A _Bruno Berselli_, Jun 15 2016