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a(n) = floor( (3/2)*floor(5*n/2) ).
3

%I #14 Sep 08 2022 08:46:02

%S 0,3,7,10,15,18,22,25,30,33,37,40,45,48,52,55,60,63,67,70,75,78,82,85,

%T 90,93,97,100,105,108,112,115,120,123,127,130,135,138,142,145,150,153,

%U 157,160,165,168,172,175,180,183,187,190

%N a(n) = floor( (3/2)*floor(5*n/2) ).

%C Also, numbers that are congruent to {0,3,7,10} mod 15. - _Bruno Berselli_, Jul 19 2012

%H Clark Kimberling, <a href="/A214066/b214066.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).

%F From _Bruno Berselli_, Jul 19 2012: (Start)

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

%F a(n) = (30*n+2*i^((n-1)*n)+3*(-1)^n-5)/8, where i=sqrt(-1). (End)

%F a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - _Wesley Ivan Hurt_, Jun 04 2016

%p A214066:=n->floor((3/2)*floor(5*n/2)): seq(A214066(n), n=0..100); # _Wesley Ivan Hurt_, Jun 04 2016

%t f[n_]:=Floor[(3/2)Floor[5n/2]]; t=Table[f[n], {n,0,70}]

%o From _Bruno Berselli_, Jul 19 2012: (Start)

%o (Magma) [n: n in [0..190] | n mod 15 in [0,3,7,10]];

%o (Maxima) makelist((30*n+2*%i^((n-1)*n)+3*(-1)^n-5)/8, n, 0, 51);

%o (PARI) concat(0, Vec((3+4*x+3*x^2+5*x^3)/((1+x)*(1-x)^2*(1+x^2))+O(x^51))) (End)

%Y Cf. A214068.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Jul 18 2012