%I
%S 1,2,3,5,4,7,6,10,8,13,9,15,11,18,12,20,14,23,16,26,17,28,19,31,21,34,
%T 22,36,24,39,25,41,27,44,29,47,30,49,32,52,33,54,35,57,37,60,38,62,40,
%U 65,42,68,43,70,45,73,46,75,48,78,50,81,51,83,53,86,55,89,56,91,58,94
%N [t], 1+[t], [2t], 2+[2t], [3t], 3+[3t], ..., where t=tau = (1+sqrt(5))/2 and []=floor.
%C The same sequence can be defined as follows: "a(1) = 1 and, for n>1, a(n) = a(n1) + n/2 if n is even, otherwise a(n) = smallest positive integer which has not yet appeared in the sequence." This was originally a separate entry in the database, contributed by _John W. Layman_, Jul 08 2004. _Antti Karttunen_ noticed on Jul 10 2004 that the two entries appeared to be identical. This was finally proved by _Clark Kimberling_, Aug 22 2007.
%C A permutation of the positive integers. Bisections are the lower and upper Wythoff sequences.
%C The consecutive pairs (1,2), (3,5), (4,7), (6,10), ... are the muchstudied Wythoff pairs, arising in connection with Wythoff's game.
%C Conjecture: For even n, the ratio a(n)/a(n1) is asymptotic to (1 + sqrt(5))/2 as n becomes large. (At n=3000, the ratio is 1.61804697, compared to the exact value 1.61803399.)  _John W. Layman_, Jul 08 2004
%C A more general conjecture may be stated as follows: Define {a(n)} by a(1)=1 and, for n>1, a(n) = a(n1)+floor(kn) if n is even, else a(n)=smallest positive integer which has not yet appeared in the sequence, where k is a positive real number. Then a(2n)/a(2n1) is asymptotic to k+sqrt(k^2+1) for large n.  _John W. Layman_, Jul 08 2004
%H MathWorld, <a href="http://mathworld.wolfram.com/WythoffsGame.html">Wythoff's game</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F a(n) = n*(1+(1)^n)/4+floor((2*n+1(1)^n)*(1+sqrt(5))/8).  _Wesley Ivan Hurt_, Apr 10 2015
%p A072061:=n>n*(1+(1)^n)/4+floor((2*n+1(1)^n)*(1+sqrt(5))/8): seq(A072061(n), n=1..100); # _Wesley Ivan Hurt_, Apr 10 2015
%t Table[n*(1 + (1)^n)/4 + Floor[(2 n + 1  (1)^n) (1 + Sqrt[5])/8], {n, 100}] (* _Wesley Ivan Hurt_, Apr 10 2015 *)
%o (MAGMA) [n*(1+(1)^n)/4+Floor((2*n+1(1)^n)*(1+Sqrt(5))/8) : n in [1..100]]; // _Wesley Ivan Hurt_, Apr 10 2015
%o (PARI) lista(nn) = {v = []; for (n=1, nn, v = concat(v, nt = floor(n*(1+sqrt(5))/2)); v = concat(v, n+nt);); v;} \\ _Michel Marcus_, Apr 14 2015
%Y Cf. A000201, A001950, A026272, A072062, A094077.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jun 11 2002, Aug 17 2007
%E Edited by _N. J. A. Sloane_, Jul 26 2008
