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%I #55 Jan 10 2023 01:32:50
%S 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,
%T 28,29,15,31,16,33,17,35,36,37,19,39,20,41,21,43,44,45,23,47,48,49,25,
%U 51,52,53,27,55,28,57,29
%N Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.
%C 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.
%H Charles R Greathouse IV, <a href="/A214682/b214682.txt">Table of n, a(n) for n = 1..10000</a>
%F 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.
%F For n odd, a(n)=n since n has no factors of 2 (or 4).
%F From _Peter Munn_, Nov 29 2020: (Start)
%F a(A003159(n)) = n.
%F a(A036554(n)) = n/2.
%F a(n) = n/A056832(n) = n/A059895(n, 2) = min(n, A073675(n)).
%F a(A059897(n, k)) = A059897(a(n), a(k)).
%F (End)
%F 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
%F Sum_{k=1..n} a(k) ~ (5/12) * n^2. - _Amiram Eldar_, Nov 10 2022
%F Dirichlet g.f.: zeta(s-1)*(2^s+1)/(2^s+2). - _Amiram Eldar_, Dec 30 2022
%e For n=8, v_4(8)=1, v_2(8)=3, so a(8)=(8*4^1)/(2^3)=4.
%e For n=12, v_4(12)=1, v_2(12)=2, so a(12)=(12*4^1)/(2^2)=12.
%t a[n_] := n/(2^Mod[IntegerExponent[n, 2], 2]); Array[a, 100] (* _Amiram Eldar_, Dec 09 2020 *)
%o (Sage)
%o C = []
%o for i in [1..n]:
%o C.append(i*(4^(Integer(i).valuation(4))/(2^(Integer(i).valuation(2)))
%o (PARI) a(n)=n>>(valuation(n,2)%2) \\ _Charles R Greathouse IV_, Jul 26 2012
%o (Python)
%o def A214682(n): return n>>1 if (~n&n-1).bit_length()&1 else n # _Chai Wah Wu_, Jan 09 2023
%Y Cf. A214681, A214685.
%Y Range of values: A003159.
%Y Missing values: A036554.
%Y A056832, A059895, A073675 are used in a formula defining this sequence.
%Y A059897 is used to express relationship between terms of this sequence.
%Y Cf. A007814 (v_2(n)), A235127 (v_4(n)).
%K easy,nonn,mult
%O 1,3
%A _Tyler Ball_, Jul 25 2012
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