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A087229
Exponent of p=2 in 12n+4 = 3(4n+1)+1.
4
2, 4, 2, 3, 2, 6, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 8, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 6, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 7, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 6, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 10, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 6, 2, 3, 2, 4
OFFSET
0,1
COMMENTS
In the Collatz trajectory of m=4n+1, the first 3x+1 step is followed by a(n) consecutive divide-by-2 steps.
Large shifted overlaps with A085058.
FORMULA
a(n) = A007814(A017569(n)). - Antti Karttunen, Oct 09 2018
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3. - Amiram Eldar, Aug 30 2024
EXAMPLE
n=85: m = 4*n + 1 = 4*85 + 1 = 341, whose Collatz trajectory begins with 341 -> 1024 followed by a(85)=10 consecutive halving steps.
MATHEMATICA
Table[Part[Part[FactorInteger[12*w+4], 1], 2], {w, 0, 100}]
Table[IntegerExponent[12*n + 4, 2], {n, 0, 100}] (* Amiram Eldar, Feb 03 2022 *)
PROG
(PARI) A087229(n) = valuation((12*n)+4, 2); \\ Antti Karttunen, Oct 09 2018
(Perl) # In 4*n+1 (n>0), find position of first bit that will repeat.
sub a {
my $nv= ((shift() << 2) | 1);
my $bp= 1;
while (($nv & 1) xor ($nv & 2)) {
$nv>>= 1;
$bp++;
}
return $bp;
}
# Ruud H.G. van Tol, Sep 27 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Aug 28 2003
EXTENSIONS
Part of the name moved to comments section by Antti Karttunen, Oct 09 2018
a(0) = 2 prepended by Ihar Senkevich, Feb 05 2022
STATUS
approved