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A371094
a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.
29
1, 21, 7, 21, 13, 341, 19, 45, 25, 117, 31, 69, 37, 341, 43, 93, 49, 213, 55, 117, 61, 5461, 67, 141, 73, 309, 79, 165, 85, 725, 91, 189, 97, 405, 103, 213, 109, 1877, 115, 237, 121, 501, 127, 261, 133, 1109, 139, 285, 145, 597, 151, 309, 157, 5461, 163, 333, 169, 693, 175, 357, 181, 1493, 187, 381, 193, 789, 199
OFFSET
0,2
COMMENTS
Construction: take the binary expansion of 3n+1 (A016777(n)), and substitute "01" for all trailing 0-bits that follow after its odd part (= A067745(1+n)), of which there are A371093(n) in total. See the examples.
FORMULA
a(n) = A372289(A016777(n)).
a(2n) = A016777(2n) = A016921(n).
EXAMPLE
For n=1, 3*n+1 = 4, "100" in binary, when we substitute 01's for the two trailing 0's, we obtain 21, "10101" in binary, therefore a(1) = 21.
For n=6, 3*6+1 = 19, "10011" in binary, and there are no trailing 0's, and no changes, therefore a(6) = 19.
For n=7, 3*7+1 = 22, "10110" in binary, with one trailing 0, which when replaced with 01 gives us 45, "101101" in binary, therefore a(7) = 45.
For n=229, there are e=4 trailing bit expansions 0 -> 01,
3n+1 = binary 101011 0 0 0 0
a(n) = binary 101011 01010101
MATHEMATICA
Array[#2*(2^#3) + ((4^#3) - 1)/3 & @@ {#1, #2, IntegerExponent[#2, 2]} & @@ {#, 3 #1 + 1} &, 67, 0] (* Michael De Vlieger, Apr 19 2024 *)
PROG
(PARI) A371094(n) = { my(m=1+3*n, e=valuation(m, 2)); ((m*(2^e)) + (((4^e)-1)/3)); };
(Python)
def A371094(n): return ((m:=3*n+1)<<(e:=(~m & m-1).bit_length()))+((1<<(e<<1))-1)//3 # Chai Wah Wu, Apr 28 2024
CROSSREFS
Cf. A016921, A372351 (even and odd bisection), A372290 (numbers occurring in the latter).
Cf. also A302338.
Sequence in context: A351321 A040426 A352137 * A040425 A337422 A040424
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen (proposed by Ali Sada), Apr 19 2024
STATUS
approved