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A214682
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Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.
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5
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1, 1, 3, 4, 5, 3, 7, 4, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 12, 25, 13, 27, 28, 29, 15, 31, 16, 33, 17, 35, 36, 37, 19, 39, 20, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 28, 57, 29
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OFFSET
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1,3
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COMMENTS
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In this sequence, the number 4 exhibits some characteristics of a prime number since all extraneous 2's have been removed from the prime factorizations of all other numbers.
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LINKS
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FORMULA
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a(n) = (n*4^(v_4(n)))/(2^(v_2(n))) where v_k(n) is the k-adic valuation of n. That is, v_k(n) is the largest power of k, a, such that k^a divides n.
For n odd, a(n)=n since n has no factors of 2 (or 4).
(End)
Multiplicative with a(2^e) = 2^(2*floor(e/2)), and a(p^e) = p^e for odd primes p. - Amiram Eldar, Dec 09 2020
Sum_{k=1..n} a(k) ~ (5/12) * n^2. - Amiram Eldar, Nov 10 2022
Dirichlet g.f.: zeta(s-1)*(2^s+1)/(2^s+2). - Amiram Eldar, Dec 30 2022
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EXAMPLE
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For n=8, v_4(8)=1, v_2(8)=3, so a(8)=(8*4^1)/(2^3)=4.
For n=12, v_4(12)=1, v_2(12)=2, so a(12)=(12*4^1)/(2^2)=12.
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MATHEMATICA
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a[n_] := n/(2^Mod[IntegerExponent[n, 2], 2]); Array[a, 100] (* Amiram Eldar, Dec 09 2020 *)
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PROG
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(Sage)
C = []
for i in [1..n]:
C.append(i*(4^(Integer(i).valuation(4))/(2^(Integer(i).valuation(2)))
(Python)
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CROSSREFS
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A059897 is used to express relationship between terms of this sequence.
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KEYWORD
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easy,nonn,mult
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AUTHOR
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STATUS
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approved
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