A088177 a(1)=1, a(2)=1; for n>2, a(n) is the smallest positive integer such that the products a(i)*a(i+1), i=1..n-1, are all distinct.

1, 1, 2, 2, 3, 1, 5, 2, 4, 3, 3, 5, 4, 4, 6, 3, 7, 1, 11, 2, 7, 4, 8, 5, 5, 6, 6, 7, 5, 9, 3, 11, 4, 12, 5, 10, 7, 7, 8, 8, 9, 6, 11, 5, 13, 1, 17, 2, 13, 3, 17, 4, 13, 6, 14, 7, 9, 9, 10, 8, 11, 7, 13, 8, 12, 9, 11, 10, 10, 12, 11, 11, 13, 9, 14, 8, 16, 9

Offset: 1

Author

John W. Layman, Sep 22 2003

Keywords

• nonn
• nice

Comments

A088178 is the sequence of distinct products a(i)a(i+1), i=1,2,3,... and appears to be a permutation of the natural numbers.

It appears that for k>2 the k-th occurrence of 1 lies between the first occurrences of primes p(2*k-4) and p(2*k-3). For instance, the 5th occurrence of 1 lies between the first occurrences of 13 and 17, the 6th and 7th primes, respectively. - John W. Layman, Nov 16 2011

Note that a(n) = 1 for infinitely many n, because the sequence a(n) is not bounded and beside every new prime number must be the number 1. - Thomas Ordowski, Sep 04 2014. [This seems a rather sketchy argument, but I have a more complete proof using arguments similar to those we used in A098550. - N. J. A. Sloane, Oct 18 2021]

Example: ..., 5, 13, 1, 17, 2, 13, 3, 17, 4; ...

General: ..., k, p, 1, q, 2, p, 3, q, ..., k-1; ...

- Thomas Ordowski, Sep 08 2014

Examples

			Given that the sequence begins 1,1,2,2,... then a(5)=3, since either of the choices a(5)=1 or a(5)=2 would lead to a repetition of one of the previous products 1,2,4 of adjacent pairs of terms.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000, (first 1000 terms from T. D. Noe)

Crossrefs

Cf. A088178, A307720, A307730, A341492, A348437, A348438, A348439.

Records: A348440, A348441.

Programs

Maple

A[1]:= 1: A[2]:= 1: S:= {1}:
for n from 3 to 100 do
  Sp:= select(type,map(s -> s/A[n-1],S),integer);
  if nops(Sp) = Sp[-1] then A[n]:= Sp[-1]+1
  else A[n]:= min({$1..Sp[-1]} minus Sp)
  fi;
  S:= S union {A[n-1]*A[n]};
od:
seq(A[n],n=1..100); # Robert Israel, Aug 28 2014

Mathematica

t = {1, 1}; Do[AppendTo[t, 1]; While[Length[Union[Most[t]*Rest[t]]] < n - 1, t[[-1]]++], {n, 3, 100}]; t (* T. D. Noe, Nov 16 2011 *)

Python

from itertools import islice
def A088177(): # generator of terms
    yield 1
    yield 1
    p, a = {1}, 1
    while True:
        n = 1
        while n*a in p:
            n += 1
        p.add(n*a)
        a = n
        yield n
A088177_list = list(islice(A088177(),20)) # Chai Wah Wu, Oct 21 2021

Formula

a(n)*gcd(a(n-1),a(n+1)) = gcd(A088178(n-1),A088178(n)). - Thomas Ordowski, Jun 29 2015