

A280963


Numbers n such that for all divisors of n, ratios of 2 consecutive divisors of n will always reduce to lowest terms to a fraction with numerator=denominator+2.


1



1, 3, 9, 15, 27, 75, 81, 99, 243, 255, 315, 375, 729, 783, 1089, 1875, 2187, 4335, 6561, 6723, 9375, 9999, 11979, 19683, 22707, 46875, 59049, 65535, 73695, 99855, 131769, 177147, 234375, 531441, 558009, 658503, 1009899, 1171875, 1188099, 1252815, 1449459, 1594323
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OFFSET

1,2


COMMENTS

This sequence is similar to A140110. Both sequences concern numbers such that consecutive divisors of these numbers have a ratio which is of the form (k+1)/k for A140110 and (k+2)/k for this sequence.
So for each q >= 1, one can define a corresponding sequence where the said ratio is of the form (k+q)/k. It appears that such sequences are reduced to a single term 1 when q+1 is not prime. On the other hand when p=q1 is prime (see A006093), then these sequences include the terms 1, p, p^2, .... so they are infinite.
The sequence of powers of 3 (A000244) is a subsequence. And all terms except 1 are divisible by p, here 3.


LINKS



EXAMPLE

9 is in the sequence for the following reason. Divisors of 9 are {1,3,9}; ratios formed by pairing adjacent divisors are 3/1,9/3, both reduce to 3/1. The difference between numerator and denominator is 2 in both cases.  Michael De Vlieger, Jan 11 2017


MATHEMATICA

Select[Range[10^6], Times @@ Boole@ Map[Denominator@ #  Numerator@ # == 2 &, Divide @@@ Partition[Divisors@ #, 2, 1]] == 1 &] (* Michael De Vlieger, Jan 11 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) + 2, return(0)); ); return(1); }


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



