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%I #38 Dec 25 2023 01:51:45
%S 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,
%T 30,31,1,11,17,35,36,37,19,13,5,41,42,43,11,5,23,47,6,49,25,17,13,53,
%U 6,55,7,19,29,59,30,61,31,7,1,65,66,67,17,23,35,71,36
%N a(n) is obtained from n by removing factors of 2 and 3 that do not contribute to a factor of 6.
%C 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.
%H Alois P. Heinz, <a href="/A214681/b214681.txt">Table of n, a(n) for n = 1..10000</a>
%F 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.
%F Sum_{k=1..n} a(k) ~ (7/24) * n^2. - _Amiram Eldar_, Dec 25 2023
%e For n=4, v_2(4)=2, v_3(4)=0, and v_6(4)=0, so a(4) = 4*1/(4*1) = 1.
%e For n=36, v_2(36)=2, v_3(36)=2, and v_6(36)=2, so a(36) = 36*36/(4*9) = 36.
%e For n=17, a(17) = 17 since 17 has no factors of 6, 2 or 3.
%p a:= proc(n) local i, m, r; m:=n;
%p for i from 0 while irem(m, 6, 'r')=0 do m:=r od;
%p while irem(m, 2, 'r')=0 do m:=r od;
%p while irem(m, 3, 'r')=0 do m:=r od;
%p m*6^i
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jul 04 2013
%t 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 *)
%o (Sage)
%o n=100 #change n for more terms
%o C=[]
%o b=6
%o P = factor(b)
%o for i in [1..n]:
%o prod = 1
%o for j in range(len(P)):
%o prod = prod * ((P[j][0])^(Integer(i).valuation(P[j][0])))
%o C.append((b^(Integer(i).valuation(b)) * i) /prod)
%Y Cf. A214682, A214685.
%K easy,nonn
%O 1,5
%A _Tom Edgar_, Jul 25 2012
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