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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.

%I A140110 #26 Jan 14 2017 07:45:58
%S A140110 1,2,4,6,8,12,16,18,20,24,32,36,42,48,54,64,72,96,100,108,120,128,144,
%T A140110 156,162,168,180,192,216,240,256,272,288,294,324,342,360,384,432,480,
%U A140110 486,500,512,576,600,648,720,768,840,900,960,972,1024,1080,1152,1176
%N 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.
%C A140110 All numbers in sequence greater than 2 must leave remainders of 0, 4, 6, or 8 when divided by 12.
%C A140110 Every term > 2 appears to be a multiple of 4 or 6. - _John W. Layman_, Jul 03 2008
%C A140110 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
%C A140110 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
%H A140110 Michel Marcus, <a href="/A140110/b140110.txt">Table of n, a(n) for n = 1..359</a>
%e A140110 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.
%t A140110 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 *)
%o A140110 (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
%K A140110 nonn
%O A140110 1,2
%A A140110 _J. Lowell_, Jun 21 2008
%E A140110 More terms from _John W. Layman_, Jul 03 2008