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

1,5

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.

Links

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

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

Example

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.

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 *)

Prog

(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)

Crossrefs

Cf. A214682, A214685.

Sequence in context: A189240 A081820 A306324 * A019978 A030178 A038458

Adjacent sequences: A214678 A214679 A214680 * A214682 A214683 A214684

Keyword

easy,nonn

Author

Tom Edgar, Jul 25 2012

Status

approved