OFFSET
1,2
COMMENTS
All numbers in sequence greater than 2 must leave remainders of 0, 4, 6, or 8 when divided by 12.
Every term > 2 appears to be a multiple of 4 or 6. - John W. Layman, Jul 03 2008
Proof of John W. Layman's conjecture that every term after the second is a multiple of 4 or 6. The first divisor of any number is always 1. Because the only divisor of 1 is 1, the second divisor of any member of this sequence greater than 1 is 2. Because the divisors of 2 are 1 and 2, the third divisor of any term of this sequence is always either 3 (proving it is a multiple of 6 because 6 is the LCM of 2 and 3) or 4. Therefore every term greater than 2 in this sequence is a multiple of at least one of 4 and 6. - J. Lowell, Jul 07 2008
It appears that consecutive elements of this sequence (for n>1) are consecutive solutions of the equation n = 1 + Sum_{k=1..tau(n)-1} gcd(d_k(n), d_(k+1)(n)), where d_k(n) denotes the k-th smallest divisor. The conjecture was checked for 10^8 consecutive integers. - Lechoslaw Ratajczak, Jan 09 2017
LINKS
Michel Marcus, Table of n, a(n) for n = 1..359
EXAMPLE
Divisors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60. The "6,10" disqualifies 60 from being in this sequence because 6/10 = 3/5, or more generally, a fraction in lowest terms a/b with b-a greater than 1. Specifically, if 6 is a divisor of a member of this sequence, the next divisor must be 7, 8, 9, or 12.
MATHEMATICA
Select[Range@ 1200, Function[n, Times @@ Boole@ Map[Abs[Numerator@ # - Denominator@ #] == 1 &[#2/#1] & @@ # &, Partition[Divisors@ n, 2, 1]] > 0]] (* Michael De Vlieger, Jan 13 2017 *)
PROG
(PARI) isok(n) = {my(vd = divisors(n)); for (k=1, #vd - 1, r = vd[k+1]/vd[k]; if (numerator(r) != denominator(r) + 1, return(0)); ); return(1); } \\ Michel Marcus, Jan 09 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
J. Lowell, Jun 21 2008
EXTENSIONS
More terms from John W. Layman, Jul 03 2008
STATUS
approved