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

1,2

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.

Links

Table of n, a(n) for n=1..64.

Index entries for sequences related to binary expansion of n

Formula

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

Example

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.

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A049457 A061037 A070262 * A171621 A084183 A099730

Adjacent sequences: A372286 A372287 A372288 * A372290 A372291 A372292

Keyword

nonn

Author

Antti Karttunen and Ali Sada, Apr 26 2024

Status

approved