A244414 - OEIS

%I #35 Nov 20 2022 01:55:14

%S 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,

%T 28,29,5,31,32,33,34,35,1,37,38,39,40,41,7,43,44,45,46,47,8,49,50,51,

%U 52,53,9,55,56,57,58,59,10,61,62,63,64,65,11

%N Remove highest power of 6 from n.

%C 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.

%C 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

%H Amiram Eldar, <a href="/A244414/b244414.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = n/6^A122841(n), n >= 1.

%F 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.

%F From _Peter Munn_, Jun 04 2020: (Start)

%F Proximity to being multiplicative may be expressed as follows:

%F a(n * A007310(k)) = a(n) * a(A007310(k));

%F a(n^2) = a(n)^2;

%F a(n) = a(A007913(n)) * a(A008833(n)).

%F (End)

%F Sum_{k=1..n} a(k) ~ (3/7) * n^2. - _Amiram Eldar_, Nov 20 2022

%e a(1) = 1 = 1/6^A122841(1) = 1/6^0.

%e a(9) = a(2^0*3^2), min(0,2) = 0, a(9) = 2^(0-0)*3^(2-0) = 1*9 = 9.

%e a(12) = a(2^2*3^1), m = min(2,1) = 1, a(12) = 2^(2-1)*3^(1-1) = 2^1*1 = 2.

%e a(30) = a(2*3*5) = a(2^1*3^1)*a(5) = 1*a(5) = 5.

%t a[n_] := n/6^IntegerExponent[n, 6]; Array[a, 66] (* _Robert G. Wilson v_, Feb 12 2018 *)

%o (PARI) a(n) = n/6^valuation(n,6); \\ _Joerg Arndt_, Jun 28 2014

%Y Cf. A122841, A000265, A038502, A065883, A132739, A242603.

%Y A007310, A007913, A008833 are used to express relationship between terms of this sequence.

%K nonn,easy

%O 1,2

%A _Wolfdieter Lang_, Jun 27 2014

%E Incorrect multiplicity claim corrected by _Wolfdieter Lang_, Feb 12 2018