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a(n) = floor(9*n/5).
6

%I #29 Oct 18 2022 01:43:45

%S 0,1,3,5,7,9,10,12,14,16,18,19,21,23,25,27,28,30,32,34,36,37,39,41,43,

%T 45,46,48,50,52,54,55,57,59,61,63,64,66,68,70,72,73,75,77,79,81,82,84,

%U 86,88,90,91,93,95,97,99,100,102,104,106,108,109,111,113,115,117,118,120,122,124,126,127,129,131,133,135,136,138,140,142,144,145,147,149

%N a(n) = floor(9*n/5).

%C Apart from first term 0, these are the numbers such that the iterated sum-of-digits (A010888) is odd. - _Michel Marcus_, Jun 07 2015

%C Equivalently, numbers congruent to {0, 1, 3, 5, 7} mod 9. - _Bruno Berselli_, Jun 15 2016

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).

%F a(n) = n + floor(4*n/5).

%F a(n) = 2*n - 1 - floor((n - 1)/5). - _Wesley Ivan Hurt_, Jan 02 2017

%F From _Chai Wah Wu_, Oct 17 2022: (Start)

%F a(n) = a(n-1) + a(n-5) - a(n-6) for n > 5.

%F G.f.: x*(2*x^4 + 2*x^3 + 2*x^2 + 2*x + 1)/(x^6 - x^5 - x + 1). (End)

%p A187318:=n->floor(9*n/5): seq(A187318(n), n=0..100); # _Wesley Ivan Hurt_, Jan 02 2017

%t Table[Floor[9 n/5], {n,0,120}]

%o (Magma) [Floor(9*n/5) : n in [0..100]]; // _Wesley Ivan Hurt_, Jan 02 2017

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Mar 08 2011