OFFSET
0,3
COMMENTS
Otherwise said, a(n+1) is either the smallest number not occurring earlier if this number is smaller than a(n)/2, or else a(n+1) is the least unused number >= 2 a(n).
This is a simpler variant of A139080. Here the well-definedness of the infinite sequence is guaranteed.
This is a permutation of the nonnegative integers. Indeed, any number m appears either "early" as successor of m/2, or "late" as successor a(n+1) of some a(n) >= 2m (in which case n+1 = 3k-1, see below).
After the two initial terms, the following 3-periodic pattern repeats: a(3k-1) = the smallest term not occurring earlier, a(3k) = 2 a(3k-1), and a(3k+1) = 2 a(3k), since at this point the smallest unused number is necessarily > a(3k)/2 = a(3k-1), and therefore, using that number, the maximum would be smaller than twice the minimum.
In spite of the simplicity of the pattern, it seems not easy to give an explicit formula of a(n), i.e., for a(3k-1), cf. formulas. I conjecture the following properties:
(i) For all k>0, a(3k+2) = a(3k-1)+2 except for k in S = {1, 8, 9, 13, 14, 22, 23, 31, 32, 40, 41, 49, 50,...} where a(3k+2) = a(3k-1)+1.
(ii) Let D = (7, 1, 4, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, ...) be the first differences of the sequence S. Then D(2k) = 1 for all k>0.
(iii) D(2k+1) = 4 or 8 for all k>0; except for the first one, the 4's always come twice in a row (e.g., for k in {8, 9}, {13, 14}, {22, 23}, ...).
(iv) The length of runs of 8's in the subsequence D(2k+1) are: 6 (k=2..7), 3 (k=10..12), 7 (k=15..21), 7 (k=24..30), 7 (k=33..39), 7 (k=42..48), 7 (k=51..57), 7 (k=60..66), 3 (k=69..71), 3 (k=74..76), 7,... It seems that except for the first two terms, this consists of runs of 7 of various length, but always interrupted by two 3's.
LINKS
FORMULA
For all k>0, a(3k-1) = least number not occurring earlier; a(3k) = 2 a(3k-1); a(3k+1) = 2 a(3k).
EXAMPLE
After a(0)=0, a(1)=1 is the least unused number such that max{1,0} >= 2 min{1,0} = 0.
After a(2)=2, a(3)=4 is the least unused number such that max{2,4} >= 2 min{2,4} = 4.
After a(3)=4, a(4)=8 is the least unused number such that max{8,4} >= 2 min{8,4} = 8, because the only smaller unused number, 3, would not satisfy the requirement.
After a(4)=8, a(5)=3 is the least unused number such that max{8,3} >= 2 min{8,3} = 6.
MATHEMATICA
f[n_] := Block[{s = {1}}, For[i = 2, i <= n, i++, k = 1; While[Nand[! MemberQ[s, k], Max[k, s[[i - 1]]] >= 2 Min[k, s[[i - 1]]]], k++]; AppendTo[s, k]]; s]; f@ 86 (* Michael De Vlieger, Apr 25 2015 *)
PROG
(PARI) {a=vector(2000); u=[]; a[1]=1; for(n=2, #a, u=setunion(u, [a[n-1]]); while(#u>1&&u[2]==u[1]+1, u=u[2..-1]); if( u[1]*2+1 < a[n-1], a[n]=u[1]+1, a[n]=a[n-1]*2; while(setsearch(u, a[n]), a[n]++)))}
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Apr 25 2015
STATUS
approved