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A372289
a(n) = n*(2^e) + ((4^e)-1)/3, where e is the 2-adic valuation of n.
3
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.
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
KEYWORD
nonn
AUTHOR
Antti Karttunen and Ali Sada, Apr 26 2024
STATUS
approved