A201009 Numbers m such that the set of distinct prime divisors of m is equal to the set of distinct prime divisors of the arithmetic derivative m'.

1, 4, 16, 27, 108, 144, 256, 432, 500, 784, 972, 1323, 1728, 2700, 2916, 3125, 3456, 5292, 8788, 11664, 12500, 13068, 15376, 16875, 19683, 20736, 23328, 25000, 27648, 28125, 31212, 34300, 47916, 54000, 57132, 65536, 72000, 78732, 97556, 102400, 103788, 104544

Offset: 1

Author

Paolo P. Lava, Jan 09 2013

Keywords

• nonn

Comments

A027748(n,k) = A027748(A003415(n),k) for k=1..A001221(n). - Reinhard Zumkeller, Jan 16 2013

A051674 is a subsequence of this sequence.

Examples

			n = 1728 = 2^6*3^3, n' = 6912 = 2^8*3^3 have the same prime factors 2 and 3.
		

Links

• Paolo P. Lava and Donovan Johnson, Table of n, a(n) for n = 1..500 (first 100 terms from Paolo P. Lava)

Crossrefs

Cf. A001221, A003415, A027748, A051674, A055744, A081377, A110751, A110819.

Programs

Haskell

a201009 = a201009_list
a201009_list = 1 : filter
   (\x -> a027748_row x == a027748_row (a003415 x)) [2..]
-- Reinhard Zumkeller, Jan 16 2013

Maple

with(numtheory);
A201009:=proc(q)
local a,b,k,n;
for n from 1 to q do
  a:=ifactors(n)[2]; b:=ifactors(n*add(op(2,p)/op(1,p),p=ifactors(n)[2]))[2];
  if nops(a)=nops(b) then
    if product(a[k][1],k=1..nops(a))=product(b[k][1],k=1..nops(a)) then print(n);
fi; fi; od; end:
A201009(100000); # Paolo P. Lava, Jan 09 2013

Python

from sympy import primefactors, factorint
A201009 = [n for n in range(1,10**5) if primefactors(n) == primefactors(sum([int(n*e/p) for p,e in factorint(n).items()]) if n > 1 else 0)] # Chai Wah Wu, Aug 21 2014