

A214681


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


3



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
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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 kadic valuation of n, that is v_k(n) gives the largest power of k, a, such that k^a divides n.


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


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



