login
a(n) = least k such that s(k) = n, where s = A026272.
6

%I #21 Mar 12 2018 04:57:40

%S 1,2,4,6,7,9,10,12,14,15,17,19,20,22,23,25,27,28,30,31,33,35,36,38,40,

%T 41,43,44,46,48,49,51,53,54,56,57,59,61,62,64,65,67,69,70,72,74,75,77,

%U 78,80,82,83,85,86,88,90,91,93,95,96,98,99

%N a(n) = least k such that s(k) = n, where s = A026272.

%C This is the lower s-Wythoff sequence, where s(n)=n+1.

%C See A184117 for the definition of lower and upper s-Wythoff sequences. The first few terms of a and its complement, b=A026274, are obtained generated as follows:

%C s=(2,3,4,5,6,...);

%C a=(1,2,4,6,7,...)=A026273;

%C b=(3,5,8,11,13,...)=A026274.

%C Briefly: b=s+a, and a=mex="least missing".

%C From _Michel Dekking_, Mar 12 2018: (Start)

%C One has r*(n-2*r+3) = n*r-2r^2+3*r = (n+1)*r-2.

%C So a(n) = (n+1)*r-2, and we see that this sequence is simply the Beatty sequence of the golden ratio, shifted spatially and temporally. In other words: if w = A000201 = 1,3,4,6,8,9,11,12,14,... is the lower Wythoff sequence, then a(n) = w(n+2) - 2.

%C (N.B. As so often, there is the 'offset 0 vs 1 argument', w = A000201 has offset 1; it would have been better to give (a(n)) offset 1, too).

%C This observation also gives an answer to Lenormand's question, and a simple proof of Mathar's conjecture in A059426.

%C (End)

%F a(n) = floor[r*(n-2*r+3)], where r=golden ratio.

%F b(n) = floor[(r^2)*(n+2*r-3)] = floor(n*A104457-A134972+1).

%t r=(1+Sqrt[5])/2;

%t a[n_]:=Floor[r*(n-2r+3)];

%t b[n_]:=Floor[r*r*(n+2r-3)];

%t Table[a[n],{n,200}] (* A026273 *)

%t Table[b[n],{n,200}] (* A026274 *)

%Y Cf. A184117, A026274.

%K nonn

%O 1,2

%A _Clark Kimberling_

%E Extended by _Clark Kimberling_, Jan 14 2011