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 A139080 a(n) = the smallest positive integer not occurring earlier in the sequence such that floor(max(a(n),a(n-1))/min(a(n),a(n-1))) = 2; a(1) = 1. 3
 1, 2, 4, 8, 3, 6, 12, 5, 10, 20, 7, 14, 28, 11, 22, 9, 18, 36, 13, 26, 52, 19, 38, 15, 30, 60, 21, 42, 16, 32, 64, 23, 46, 17, 34, 68, 24, 48, 96, 33, 66, 25, 50, 100, 35, 70, 27, 54, 108, 37, 74, 29, 58, 116, 39, 78, 31, 62, 124, 43, 86, 40, 80, 160, 55, 110, 41, 82, 164, 56 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Is there always an unused positive integer, a(n), such that floor(max(a(n),a(n-1))/min(a(n),a(n-1))) = 2, or does the sequence terminate? If the sequence is infinite, is it a permutation of the positive integers? The sequence grows in 5 distinct "bands" which are populated respectively by every 5th term; these terms are in proportion (a(5k), a(5k+1), a(5k+2), a(5k+3), a(5k+4)) ~ (4, 8, 3, 6, 12)*k*(1-O(1/log(k))). More precisely, for k>6, we have a(5k+1) = 2 a(5k),  a(5k+2) ~ (3/8)*a(5k+1) is the least number not occurring earlier, a(5k+3) = 2 a(5k+2), a(5k+4) = 2 a(5k+3) and a(5k+5) is the least number greater than a(5k+4)/3 not occurring earlier. - M. F. Hasler, Apr 24 2015 See LINKS for a proof of the preceding statement, which provides an affirmative answer to the questions raised in the first comment. See A257280 for the inverse permutation. See also the variant A257470 and its inverse A257271. - M. F. Hasler, Apr 26 2015 LINKS M. F. Hasler, Table of n, a(n) for n=1,...,10000. M. F. Hasler, A139080, OEIS wiki, April 2015. FORMULA For k>6, when n = 5k, 5k+2 or 5k+3, then a(n+1) = 2 a(n); and with N(n) := N \ {a(k); k < n}, a(5k+2) = min N(5k+2) and a(5k+5) = min { m in N(5k+5), m > a(5k+4)/3 }. - M. F. Hasler, Apr 26 2015 MATHEMATICA f[n_] := Block[{s = {1}, i, k}, For[i = 2, i <= n, i++, k = 1; While[Nand[! MemberQ[s, k], Floor[Max[k, s[[i - 1]]]/Min[k, s[[i - 1]]]] == 2], k++]; AppendTo[s, k]]; s]; f@ 70 (* Michael De Vlieger, Apr 26 2015 *) PROG (PARI) { t=0; last=1; for( n=1, 10000, write("b139080.txt", n, " ", last); t+=1<

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