A187324 a(n) = floor(n/2) + floor(n/3) - floor(n/4).
0, 0, 1, 2, 2, 2, 4, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 9, 11, 11, 11, 12, 13, 13, 14, 14, 15, 16, 16, 16, 18, 18, 18, 19, 20, 20, 21, 21, 22, 23, 23, 23, 25, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 30, 32, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 37, 39, 39, 39, 40, 41, 41, 42, 42, 43, 44, 44, 44, 46, 46, 46, 47, 48
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (0,0,1,1,0,0,-1).
Programs
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Magma
[Floor(n/2)+Floor(n/3)-Floor(n/4): n in [0..85] ]; // Vincenzo Librandi, Jul 18 2011
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Mathematica
Table[Floor[n/2]+Floor[n/3]-Floor[n/4], {n,0,120}]
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Python
def A187324(n): return (n>>2)+bool(n&2)+n//3 # Chai Wah Wu, Jan 31 2023
Formula
a(n) = floor(n/2) + floor(n/3) - floor(n/4).
G.f.: x^2*(1 + 2*x + 2*x^2 + x^3 + x^4) / ( (1+x)*(x^2+1)*(1+x+x^2)*(x-1)^2 ). - R. J. Mathar, Mar 08 2011