A255586 Composite k such that Sum_{i=1..t-1} d(i+1)/d(i) is an integer, where d(1), ..., d(t) are the divisors of k in ascending order.

4, 8, 9, 16, 18, 25, 27, 32, 48, 49, 50, 64, 81, 98, 108, 121, 125, 128, 162, 169, 242, 243, 256, 289, 338, 343, 361, 375, 512, 529, 578, 625, 722, 729, 841, 961, 1024, 1029, 1058, 1250, 1331, 1369, 1458, 1681, 1682, 1849, 1920, 1922, 2048, 2187, 2197, 2209

Offset: 1

Author

Michel Lagneau, Feb 27 2015

Keywords

• nonn

Comments

The sequence is infinite because the powers of 2 (A000079) are in the sequence.

The prime powers with even exponents (A056798) are in the sequence.

The cubes of primes (A030078) are in the sequence.

The numbers of the form 2p^2 (A079704) with p prime are in the sequence.

The corresponding integers are 4, 6, 6, 8, 9, 10, 9, 10, 14, 14, 11, 12, 12, 13, 17, 22, 15, 14, 16, 26, 17, 15, 16, 34, 19, ...

Examples

			18 is in the sequence because the divisors of 18 are {1, 2, 3, 6, 9, 18} and 2/1 + 3/2 + 6/3 + 9/6 + 18/9 = 9 is an integer.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000079, A030078, A079704, A227993, A255585.

Programs

Mathematica

lst={};Do[s=0;Do[s=s+Divisors[n][[i+1]]/Divisors[n][[i]],{i,1,Length[Divisors[n]]-1}];If[IntegerQ[s]&&!PrimeQ[n],AppendTo[lst,n]],{n,2300}];lst
Select[Range[2210],CompositeQ[#]&&IntegerQ[Total[#[[2]]/#[[1]]&/@Partition[ Divisors[ #],2,1]]]&] (* Harvey P. Dale, Jul 09 2019 *)