A325307 a(n) = k / (sum of prime factors of k, counted with multiplicity) where k is the n-th number for which the ratio is an integer.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 5, 1, 6, 1, 1, 1, 6, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 11, 1, 1, 15, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 11, 18, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 20, 1, 1, 25

Offset: 1

Author

Sven Kunze, Apr 20 2019

Keywords

• nonn

Examples

			a(8) = 2 comes from k = A036844(8) = 16 = 2^4. The sum of prime factors is 2 + 2 + 2 + 2 = 8, so k/(sum of prime factors of k) = 2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001414 (sum of prime factors), A036844 (values of k).

Programs

Maple

Res:= NULL: count:= 0:
for n from 2 while count < 200 do
  F:= ifactors(n)[2];
  r:= n /add(t[1]*t[2],t=F);
  if r::integer then count:= count+1; Res:= Res, r fi
od:
Res; # Robert Israel, Jul 29 2019

Mathematica

Select[Array[#/Total@ Flatten[ConstantArray[#1, #2] & @@ # & /@ FactorInteger[#]] &, 450, 2], IntegerQ] (* Michael De Vlieger, Apr 21 2019 *)

Python

maxNum = 455
ratios = []
for i in range(2, maxNum):
  sumFactors = A001414(i)
  if i % sumFactors == 0:
    ratio = i // sumFactors
    ratios.append(ratio)
print(ratios) # ratios is an array that contains the sequence