A140110 Numbers n such that for all divisors of n, ratios of 2 consecutive divisors of n will always reduce to lowest terms as a ratio of consecutive integers.

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 42, 48, 54, 64, 72, 96, 100, 108, 120, 128, 144, 156, 162, 168, 180, 192, 216, 240, 256, 272, 288, 294, 324, 342, 360, 384, 432, 480, 486, 500, 512, 576, 600, 648, 720, 768, 840, 900, 960, 972, 1024, 1080, 1152, 1176

Offset: 1

Author

J. Lowell, Jun 21 2008

Keywords

• nonn

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

Examples

			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.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..359

Programs

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 *)

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

Extensions

More terms from John W. Layman, Jul 03 2008