A087229 Exponent of p=2 in 12n+4 = 3(4n+1)+1.

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

Author

Labos Elemer, Aug 28 2003

Keywords

• nonn

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.

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..21845
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Crossrefs

Cf. A007814, A085058, A017569.

Programs

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 *)

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

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

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