A251555 a(1)=1, a(2)=3, a(3)=2; thereafter a(n) is the smallest number not occurring earlier having at least one common factor with a(n-2), but none with a(n-1).
1, 3, 2, 9, 4, 15, 8, 5, 6, 25, 12, 35, 16, 7, 10, 21, 20, 27, 14, 33, 26, 11, 13, 22, 39, 28, 45, 32, 51, 38, 17, 18, 85, 24, 55, 34, 65, 36, 91, 30, 49, 40, 63, 44, 57, 46, 19, 23, 76, 69, 50, 81, 52, 75, 56, 87, 62, 29, 31, 58, 93, 64, 99, 68, 77, 48, 119, 54
Offset: 1
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
- David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669 [math.NT], 2015 and J. Int. Seq. 18 (2015) 15.6.7.
Programs
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Haskell
import Data.List (delete) a251555 n = a251555_list !! (n-1) a251555_list = 1 : 3 : 2 : f 3 2 [4..] where f u v ws = g ws where g (x:xs) = if gcd x u > 1 && gcd x v == 1 then x : f v x (delete x ws) else g xs -- Reinhard Zumkeller, Dec 24 2014 -
Mathematica
a[1]=1; a[2]=3; a[3]=2; A251555 = Array[a, 3]; a[n_] := a[n] = For[k=2, True, k++, If[FreeQ[A251555, k], If[!CoprimeQ[k, a[n-2]] && CoprimeQ[k, a[n-1]], AppendTo[A251555, k]; Return[k]]]]; A251555 = Array[a, 100] (* Jean-François Alcover, Aug 02 2018 *)
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Python
from math import gcd A251555_list, l1, l2, s, b = [1,3,2], 2, 3, 4, set() for _ in range(10**3): i = s while True: if not i in b and gcd(i,l1) == 1 and gcd(i,l2) > 1: A251555_list.append(i) l2, l1 = l1, i b.add(i) while s in b: b.remove(s) s += 1 break i += 1 print(A251555_list) # Chai Wah Wu, Dec 21 2014
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