A214068 a(n) = floor((3/2)*floor((3/2)*n)).
0, 1, 4, 6, 9, 10, 13, 15, 18, 19, 22, 24, 27, 28, 31, 33, 36, 37, 40, 42, 45, 46, 49, 51, 54, 55, 58, 60, 63, 64, 67, 69, 72, 73, 76, 78, 81, 82, 85, 87, 90, 91, 94, 96, 99, 100, 103, 105, 108, 109, 112, 114, 117, 118, 121, 123, 126
Offset: 0
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
Programs
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Mathematica
f[n_]:=Floor[(3/2)Floor[3n/2]]; t=Table[f[n],{n,0,70}] LinearRecurrence[{1,0,0,1,-1},{0,1,4,6,9},60] (* Harvey P. Dale, Jun 21 2021 *)
Formula
a(n) = (18*n - 5 + 3*(-1)^n + (1 - i)*(-i)^n + (1 + i)*i^n)/8, where i = sqrt(-1).
a(n) = a(n-1) + a(n-4) - a(n-5).
G.f.: (x*(1 + 3*x + 2*x^2 + 3*x^3))/(1 - x - x^4 + x^5).
Comments