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%I #31 Feb 22 2023 05:36:47
%S 1,1,2,1,0,2,1,2,1,0,2,1,0,1,1,2,1,0,2,1,2,1,1,2,1,0,1,1,2,1,0,2,1,0,
%T 1,1,2,1,0,2,1,2,1,1,2,1,0,1,1,2,1,0,2,1,2,1,1,2,1,0,2,1,2,1,0,2,1,0,
%U 1,1,2,1,0,2,1,2,1,1,2,1,0,1,1,2,1,0,2,1,0,1,1,2,1,0,2,1,2,1,0,2,1,0,1,1,2,1,0,2,1,2,1,1,2,1,0,2,1,2,1,0,2,1,0,1,1,2,1,0,2,1,2
%N a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,2,1) and []=floor.
%C Write a(n) = [(b*n+c)*r] - b*[n*r] - [c*r]. If r>0 and b and c are integers satisfying b>=2 and 0<=c<=b-1, then 0<=a(n)<=b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 position sequences comprise a partition of the positive integers.
%C Examples:
%C (golden ratio,2,0): A078588, A005653, A005652
%C (golden ratio,2,1): A190427 - A190430
%C (golden ratio,3,0): A140397 - A190400
%C (golden ratio,3,1): A140431 - A190435
%C (golden ratio,3,2): A140436 - A190439
%H G. C. Greubel, <a href="/A190427/b190427.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = [(2*n+1)*r] - 2*[n*r] - 1, where r=(1+sqrt(5))/2.
%e a(1)=[3r]-2[r]-1=4-3-1=1.
%e a(2)=[5r]-2[2r]-1=8-6-1=1.
%e a(3)=[7r]-2[3r]-1=11-8-1=2.
%t r = GoldenRatio; b = 2; c = 1;
%t f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
%t t = Table[f[n], {n, 1, 320}] (* A190427 *)
%t Flatten[Position[t, 0]] (* A190428 *)
%t Flatten[Position[t, 1]] (* A190429 *)
%t Flatten[Position[t, 2]] (* A190430 *)
%t Table[Floor[(2*n+1)*GoldenRatio] - 2*Floor[n*GoldenRatio] -1, {n,1,100}] (* _G. C. Greubel_, Apr 06 2018 *)
%o (Python)
%o from mpmath import mp, phi
%o from sympy import floor
%o mp.dps=100
%o def a(n): return floor((2*n + 1)*phi) - 2*floor(n*phi) - 1
%o print([a(n) for n in range(1, 132)]) # _Indranil Ghosh_, Jul 02 2017
%o (PARI) for(n=1,100, print1(floor((2*n+1)*(1+sqrt(5))/2) - 2*floor(n*(1+sqrt(5))/2) - 1, ", ")) \\ _G. C. Greubel_, Apr 06 2018
%o (Magma) [Floor((2*n+1)*(1+Sqrt(5))/2) - 2*Floor(n*(1+Sqrt(5))/2) - 1: n in [1..100]]; // _G. C. Greubel_, Apr 06 2018
%Y Cf. A001622, A078588, A190428, A190429, A190430.
%K nonn
%O 1,3
%A _Clark Kimberling_, May 10 2011
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