A372289 a(n) = n*2^e + (4^e - 1)/3, where e is the 2-adic valuation of n.

1, 5, 3, 21, 5, 13, 7, 85, 9, 21, 11, 53, 13, 29, 15, 341, 17, 37, 19, 85, 21, 45, 23, 213, 25, 53, 27, 117, 29, 61, 31, 1365, 33, 69, 35, 149, 37, 77, 39, 341, 41, 85, 43, 181, 45, 93, 47, 853, 49, 101, 51, 213, 53, 109, 55, 469, 57, 117, 59, 245, 61, 125, 63, 5461

Offset: 1

Author

Antti Karttunen and Ali Sada, Apr 26 2024

Keywords

• nonn

Comments

Construction: take the binary expansion of n, and substitute "01" for all trailing 0-bits that follow after its odd part (= A000265(n)). See the examples.

Examples

			For n=4, "100" in binary, when we substitute 01's for the two trailing 0's, we obtain 21, "10101" in binary, therefore a(4) = 21.
For n=11, "1011" in binary, there are no trailing 0's, and thus no changes, therefore a(11) = 11.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000, May 08 2025
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000265, A005408 (odd bisection), A007814, A371094 [= a(3n+1)].

Programs

Maple

a := proc(n) padic[ordp](n, 2): n*2^% + ((2^%)^2 - 1)/3 end:
seq(a(n), n = 1..64);  # Peter Luschny, Apr 27 2024

Mathematica

a[n_]:=n*(2^IntegerExponent[n, 2]) + ((4^IntegerExponent[n, 2]) - 1)/3; Array[a, 75] (* Stefano Spezia, Apr 26 2024 *)

PARI

A372289(n) = { my(e=valuation(n,2)); n*2^e + (4^e-1)/3 }

Python

def A372289(n): return (n<<(e:=(~n & n-1).bit_length()))+((1<<(e<<1))-1)//3 # Chai Wah Wu, Apr 26 2024

Formula

For n >= 0, a(2n+1) = 2n+1.