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A214681
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a(n) is obtained from n by removing factors of 2 and 3 that do not contribute to a factor of 6.
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3
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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
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1,5
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(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)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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