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%I #11 Feb 18 2024 07:57:36
%S 0,5,12,17,25,30,37,42,50,55,62,67,75,80,87,92,100,105,112,117,125,
%T 130,137,142,150,155,162,167,175,180,187,192,200,205,212,217,225,230,
%U 237,242,250,255,262,267,275,280,287,292,300
%N a(n) = [(5/2)*[(5/2)*n]], where [ ] = floor.
%C Also, numbers congruent to {0,5,12,17} mod 25.
%H Clark Kimberling, <a href="/A214067/b214067.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 a(n) = (50*n - 7 + 5*(-1)^n + (1 + i)*(-i)^n + (1 - i)*i^n)/8, where i = sqrt(-1).
%F a(n) = a(n-1) + a(n-4) - a(n-5).
%F G.f.: f(x)/g(x), where f(x) = 5*x + 7*x^2 + 5*x^3 + 8*x^4 and g(x) = (1 + x + x^2 + x^3)*(1 - x)^2
%t f[n_]:=Floor[(5/2)Floor[5n/2]];
%t t=Table[f[n],{n,0,70}]
%Y Cf. A214066, A214068.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Jul 20 2012
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