A214681 a(n) is obtained from n by removing factors of 2 and 3 that do not contribute to a factor of 6.

1, 1, 1, 1, 5, 6, 7, 1, 1, 5, 11, 6, 13, 7, 5, 1, 17, 6, 19, 5, 7, 11, 23, 6, 25, 13, 1, 7, 29, 30, 31, 1, 11, 17, 35, 36, 37, 19, 13, 5, 41, 42, 43, 11, 5, 23, 47, 6, 49, 25, 17, 13, 53, 6, 55, 7, 19, 29, 59, 30, 61, 31, 7, 1, 65, 66, 67, 17, 23, 35, 71, 36

Offset: 1

Author

Tom Edgar, Jul 25 2012

Keywords

• easy
• nonn

Comments

In this sequence, the number 6 exhibits some characteristics of a prime number since we have removed extraneous 2's and 3's from the prime factorizations of numbers.

Examples

			For n=4, v_2(4)=2, v_3(4)=0, and v_6(4)=0, so a(4) = 4*1/(4*1) = 1.
For n=36, v_2(36)=2, v_3(36)=2, and v_6(36)=2, so a(36) = 36*36/(4*9) = 36.
For n=17, a(17) = 17 since 17 has no factors of 6, 2 or 3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007814, A007949, A122841.

Cf. A214682, A214685.

Programs

Maple

a:= proc(n) local i, m, r; m:=n;
      for i from 0 while irem(m, 6, 'r')=0 do m:=r od;
      while irem(m, 2, 'r')=0 do m:=r od;
      while irem(m, 3, 'r')=0 do m:=r od;
      m*6^i
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 04 2013

Mathematica

With[{v = IntegerExponent}, a[n_] := n*6^v[n, 6]/2^v[n, 2]/3^v[n, 3]; Array[a, 100]] (* Amiram Eldar, Dec 09 2020 *)

Sage

n=100 #change n for more terms
C=[]
b=6
P = factor(b)
for i in [1..n]:
    prod = 1
    for j in range(len(P)):
        prod = prod * ((P[j][0])^(Integer(i).valuation(P[j][0])))
    C.append((b^(Integer(i).valuation(b)) * i) /prod)

Formula

a(n) = n*6^(v_6(n))/(2^(v_2(n))*3^(v_3(n))), where v_k(n) is the k-adic valuation of n, that is v_k(n) gives the largest power of k, a, such that k^a divides n.

Sum_{k=1..n} a(k) ~ (7/24) * n^2. - Amiram Eldar, Dec 25 2023