

A026273


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


6



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, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 57, 59, 61, 62, 64, 65, 67, 69, 70, 72, 74, 75, 77, 78, 80, 82, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 99
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OFFSET

1,2


COMMENTS

This is the lower sWythoff sequence, where s(n)=n+1.
See A184117 for the definition of lower and upper sWythoff sequences. The first few terms of a and its complement, b=A026274, are obtained generated as follows:
s=(2,3,4,5,6,...);
Briefly: b=s+a, and a=mex="least missing".
One has r*(n2*r+3) = n*r2r^2+3*r = (n+1)*r2.
So a(n) = (n+1)*r2, 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.
(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).
This observation also gives an answer to Lenormand's question, and a simple proof of Mathar's conjecture in A059426.
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LINKS



FORMULA

a(n) = floor[r*(n2*r+3)], where r=golden ratio.


MATHEMATICA

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*(n2r+3)];
b[n_]:=Floor[r*r*(n+2r3)];
Table[a[n], {n, 200}] (* A026273 *)
Table[b[n], {n, 200}] (* A026274 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



