A075075 a(1) = 1, a(2) = 2 and then the smallest number not occurring earlier such that every term divides the product of its neighbors: a(n-1)*a(n+1)/a(n) is an integer.

1, 2, 4, 6, 3, 5, 10, 8, 12, 9, 15, 20, 16, 24, 18, 21, 7, 11, 22, 14, 28, 26, 13, 17, 34, 30, 45, 27, 33, 44, 32, 40, 25, 35, 42, 36, 48, 52, 39, 51, 68, 56, 70, 50, 55, 66, 54, 63, 49, 77, 88, 64, 72, 81, 90, 60, 38, 19, 23, 46, 58, 29, 31, 62, 74, 37, 41, 82, 76, 114, 57, 43

Offset: 1

Author

Amarnath Murthy, Sep 09 2002

Keywords

• nice
• nonn
• look

Comments

This is a permutation of natural numbers. [Leroy Quet asks (May 06 2009) if this is a theorem or just a conjecture.]

Every time a(n) divides a(n-1), a(n+1) is the next number that is not already in the sequence. I don't have a proof that a(n) divides a(n-1) infinitely often. - Franklin T. Adams-Watters, Jun 12 2014

It appears that a(n): 1,2,...,3,5,...,7,11,...,prime(2k),prime(2k+1),... - Thomas Ordowski, Jul 10 2015

The primes do appear to occur in increasing order, but prime(2k) is not always followed directly by prime(2k+1). For example, a(72) = 43 = prime(14), but a(125) = 47 = prime(15). - Robert Israel, Jul 10 2015

If a(n) and a(n+1) are primes then a(n) divides a(n-1). - Thomas Ordowski, Jul 10 2015 [Cf. second comment]

a(n) is the least multiple of a(n-1)/gcd(a(n-2),a(n-1)) that has not previously occurred. - Robert Israel, Jul 10 2015

Conjecture: if a(n) divides a(n-1) then a(n+1) is prime. - Thomas Ordowski, Jul 11 2015

It seems that a(n) and a(n+1) are consecutive primes if and only if a(n) divides a(n-1) and a(n) < a(n+1). - Thomas Ordowski, Jul 13 2015

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A075076 (ratios), A160256, A064413 (EKG sequence).

Cf. A160516 (inverse), A185635 (fixed points).

Programs

Haskell

import Data.List (delete)
a075075 n = a075075_list !! (n-1)
a075075_list = 1 : 2 : f 1 2 [3..] where
  f z z' xs = g xs where g (u:us) =
    if (z * u) `mod` z' > 0 then g us else u : f z' u (delete u xs)
-- Reinhard Zumkeller, Dec 19 2012

MATLAB

N = 10^6;
Avail = ones(1,N);
A = zeros(1,N);
A(1) = 1; A(2) = 2;
Avail([1,2]) = 0;
for n=3:N
  q = round(A(n-1)/gcd(A(n-1),A(n-2)));
  b = find(Avail(q*[1:floor(N/q)]),1,'first');
  if numel(b) == 0
     break
  end
  A(n) = q*b;
  Avail(A(n)) = 0;
end
A = A(1:n-1); % Robert Israel, Jul 10 2015

Maple

b:= proc(n) option remember; false end: a:= proc(n) option remember; local k, m; if n<3 then b(n):= true; n else m:= denom(a(n-2) /a(n-1)); for k from m by m while b(k) do od; b(k):= true; k fi end: seq(a(n), n=1..100); # Alois P. Heinz, May 16 2009

Mathematica

f[s_List] := Block[{m = Numerator[ s[[ -1]]/s[[ -2]] ]}, k = m; While[ MemberQ[s, k], k += m]; Append[s, k]]; Nest[f, {1, 2}, 70] (* Robert G. Wilson v, May 20 2009 *)

Python

from math import gcd
A075075_list, l1, l2, m, b = [1,2], 2, 1, 2, {1,2}
for _ in range(10**3):
    i = m
    while True:
        if not i in b:
            A075075_list.append(i)
            l1, l2, m = i, l1, i//gcd(l1,i)
            b.add(i)
            break
        i += m # Chai Wah Wu, Dec 09 2014

Extensions

More terms from Sascha Kurz, Feb 03 2003