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A244414
Remove highest power of 6 from n.
5
1, 2, 3, 4, 5, 1, 7, 8, 9, 10, 11, 2, 13, 14, 15, 16, 17, 3, 19, 20, 21, 22, 23, 4, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 1, 37, 38, 39, 40, 41, 7, 43, 44, 45, 46, 47, 8, 49, 50, 51, 52, 53, 9, 55, 56, 57, 58, 59, 10, 61, 62, 63, 64, 65, 11
OFFSET
1,2
COMMENTS
This is instance g = 6 of the g-family of sequences, call it r(g,n), where for g >= 2 the highest power of g is removed from n. See the crossrefs.
The present sequence is not multiplicative: a(6) = 1 not a(2)*a(3) = 6. In the prime factor decomposition one has to consider a(2^e2*3^e^3) as one entity, also for e2 >= 0, e3 >= 0 with a(1) = 1, and apply the rule given in the formula section. With this rule the sequence will be multiplicative in an unusual sense. - Wolfdieter Lang, Feb 12 2018
LINKS
FORMULA
a(n) = n/6^A122841(n), n >= 1.
For n >= 2, a(n) is sort of multiplicative if a(2^e2*3^e3) = 2^(e2 - m)*3^(e3 - m) with m = m(e2, e3) = min(e2, e3), for e2, e3 >= 0, a(1) = 1, and a(p^e) = p^e for primes p >= 5.
From Peter Munn, Jun 04 2020: (Start)
Proximity to being multiplicative may be expressed as follows:
a(n * A007310(k)) = a(n) * a(A007310(k));
a(n^2) = a(n)^2;
a(n) = a(A007913(n)) * a(A008833(n)).
(End)
Sum_{k=1..n} a(k) ~ (3/7) * n^2. - Amiram Eldar, Nov 20 2022
EXAMPLE
a(1) = 1 = 1/6^A122841(1) = 1/6^0.
a(9) = a(2^0*3^2), min(0,2) = 0, a(9) = 2^(0-0)*3^(2-0) = 1*9 = 9.
a(12) = a(2^2*3^1), m = min(2,1) = 1, a(12) = 2^(2-1)*3^(1-1) = 2^1*1 = 2.
a(30) = a(2*3*5) = a(2^1*3^1)*a(5) = 1*a(5) = 5.
MATHEMATICA
a[n_] := n/6^IntegerExponent[n, 6]; Array[a, 66] (* Robert G. Wilson v, Feb 12 2018 *)
PROG
(PARI) a(n) = n/6^valuation(n, 6); \\ Joerg Arndt, Jun 28 2014
CROSSREFS
A007310, A007913, A008833 are used to express relationship between terms of this sequence.
Sequence in context: A325976 A375125 A319654 * A214394 A255691 A030105
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jun 27 2014
EXTENSIONS
Incorrect multiplicity claim corrected by Wolfdieter Lang, Feb 12 2018
STATUS
approved