A214682 Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.

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

Offset: 1

Author

Tyler Ball, Jul 25 2012

Keywords

• easy
• nonn
• mult

Comments

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.

Examples

			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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A214681, A214685.

Range of values: A003159.

Missing values: A036554.

A056832, A059895, A073675 are used in a formula defining this sequence.

A059897 is used to express relationship between terms of this sequence.

Cf. A007814 (v_2(n)), A235127 (v_4(n)).

Programs

Mathematica

a[n_] := n/(2^Mod[IntegerExponent[n, 2], 2]); Array[a, 100] (* Amiram Eldar, Dec 09 2020 *)

PARI

a(n)=n>>(valuation(n,2)%2) \\ Charles R Greathouse IV, Jul 26 2012

Python

def A214682(n): return n>>1 if (~n&n-1).bit_length()&1 else n # Chai Wah Wu, Jan 09 2023

SageMath

C = []
for i in [1..n]:
    C.append(i*4^(Integer(i).valuation(4))/2^(Integer(i).valuation(2)))

Formula

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).

From Peter Munn, Nov 29 2020: (Start)

a(A003159(n)) = n.

a(A036554(n)) = n/2.

a(n) = n/A056832(n) = n/A059895(n, 2) = min(n, A073675(n)).

a(A059897(n, k)) = A059897(a(n), a(k)). (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