A187322 a(n) = floor(n/2) + floor(3*n/4).
0, 0, 2, 3, 5, 5, 7, 8, 10, 10, 12, 13, 15, 15, 17, 18, 20, 20, 22, 23, 25, 25, 27, 28, 30, 30, 32, 33, 35, 35, 37, 38, 40, 40, 42, 43, 45, 45, 47, 48, 50, 50, 52, 53, 55, 55, 57, 58, 60, 60, 62, 63, 65, 65, 67, 68, 70, 70, 72, 73, 75, 75, 77, 78, 80, 80, 82, 83, 85, 85, 87, 88, 90, 90, 92, 93, 95, 95, 97
Offset: 0
Links
- Colin Barker, 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
Table[Floor[n/2]+Floor[3n/4], {n,0,120}] LinearRecurrence[{1,0,0,1,-1},{0,0,2,3,5},80] (* Harvey P. Dale, Dec 05 2018 *)
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PARI
a(n) = n\2 + 3*n\4; \\ Altug Alkan, Aug 14 2017
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PARI
concat(vector(2), Vec(x^2*(2 + x + 2*x^2) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^100))) \\ Colin Barker, Aug 14 2017
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Python
def A187322(n): return (n>>1)+(3*n>>2) # Chai Wah Wu, Jan 31 2023
Formula
a(n) = (10*n-5+3*cos(n*Pi)+2*(cos(n*Pi/2)-sin(n*Pi/2)))/8. - Luce ETIENNE, Aug 14 2017
From Colin Barker, Aug 14 2017: (Start)
G.f.: x^2*(2 + x + 2*x^2) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
(End)
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