A251413 a(n) = 2n-1 if n <= 3, otherwise the smallest odd number not occurring earlier having at least one common factor with a(n-2), but none with a(n-1).

1, 3, 5, 9, 25, 21, 55, 7, 11, 35, 33, 49, 15, 77, 27, 91, 45, 13, 51, 65, 17, 39, 85, 57, 115, 19, 23, 95, 69, 125, 63, 145, 81, 29, 75, 203, 93, 119, 31, 105, 341, 87, 121, 111, 143, 37, 99, 185, 117, 155, 123, 175, 41, 133, 205, 147, 215, 141, 43, 47, 129

Offset: 1

Author

N. J. A. Sloane, Dec 02 2014

Keywords

• nonn

Comments

Conjecture 1: Sequence is a permutation of the odd numbers.

Conjecture 2: The odd primes occur in the sequence in their natural order.

Comments from N. J. A. Sloane, Dec 13 2014: (Start)

The following properties are known (the proofs are analogous to the proofs for the corresponding facts about A098550).

1. The sequence is infinite.

2. At least one-third of the terms are composite.

3. For any odd prime p, there is a term that is divisible by p.

4. Let a(n_p) be the first term that is divisible by p. Then a(n_p) = q*p where q is an odd prime less than p. If p < r are primes then n_p < n_r. (End)

References

• L. Edson Jeffery, Posting to Sequence Fans Mailing List, Dec 01 2014

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..11945
• David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 2015 and J. Int. Seq. 18 (2015) 15.6.7.

Crossrefs

Cf. A098550, A251414.

Programs

Haskell

import Data.List (delete)
a251413 n = a251413_list !! (n-1)
a251413_list = 1 : 3 : 5 : f 3 5 [7, 9 ..] 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 25 2014

Maple

N:= 10^3: # to get a(1) to a(n) where a(n+1) is the first term > N
B:= Vector(N, datatype=integer[4]):
for n from 1 to 3 do A[n]:= 2*n-1: od:
for n from 4 do
  for k from 4 to N do
    if B[k] = 0 and igcd(2*k-1, A[n-1]) = 1 and igcd(2*k-1, A[n-2]) > 1 then
       A[n]:= 2*k-1;
       B[k]:= 1;
       break
    fi
  od:
  if 2*k-1 > N then break fi
od:
seq(A[i], i=1..n-1); # Based on Robert Israel's program for A098550

Mathematica

max = 54; f = True; a = {1, 3, 5}; NN = Range[4, 1000]; s = 2*NN - 1; While[TrueQ[f], For[k = 1, k <= Length[s], k++, If[Length[a] < max, If[GCD[a[[-1]], s[[k]]] == 1 && GCD[a[[-2]], s[[k]]] > 1, a = Append[a, s[[k]]]; s = Delete[s, k]; k = 0; Break], f = False]]]; a (* L. Edson Jeffery, Dec 02 2014 *)

Python

from math import gcd
A251413_list, l1, l2, s, b = [1,3,5], 5, 3, 7, {}
for _ in range(1,10**4):
    i = s
    while True:
        if not i in b and gcd(i,l1) == 1 and gcd(i,l2) > 1:
            A251413_list.append(i)
            l2, l1, b[i] = l1, i, True
            while s in b:
                b.pop(s)
                s += 2
            break
        i += 2 # Chai Wah Wu, Dec 07 2014