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A347409
Longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.
5
0, 1, 4, 2, 4, 4, 4, 3, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 6, 4, 5, 4, 4, 4, 5, 4, 4, 5, 5, 5, 4, 4, 5, 4, 4, 4, 4, 4, 5, 6, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 5, 5, 5, 5, 6, 4, 4, 4, 4, 4, 5, 5, 4, 5, 4, 8, 4, 4, 4, 4, 4, 5, 5, 5, 6, 8, 4, 4
OFFSET
1,3
COMMENTS
If the longest run of halving steps occurs as the final part of the trajectory, a(n) = A135282(n), otherwise a(n) > A135282(n).
Every nonnegative integer appears in the sequence at least once, since for any k >= 0 a(2^k) = k.
Conjecture: every integer >= 4 appears in the sequence infinitely many times.
EXAMPLE
a(15) = 5 because the Collatz trajectory starting at 15 contains, as the longest halving run, a 5-step subtrajectory (namely, 160 -> 80 -> 40 -> 20 -> 10 -> 5).
MATHEMATICA
nterms=100; Table[c=n; sm=0; While[c>1, If[OddQ[c], c=3c+1, If[(s=IntegerExponent[c, 2])>sm, sm=s]; c/=2^s]]; sm, {n, nterms}]
PROG
(PARI) a(n)=my(nb=0); while (n != 1, if (n % 2, n=3*n+1, my(x = valuation(n, 2)); n /= 2^x; nb = max(nb, x)); ); nb; \\ Michel Marcus, Sep 03 2021
(Python)
def A347409(n):
m, r = n, 0
while m > 1:
if m % 2:
m = 3*m + 1
else:
s = bin(m)[2:]
c = len(s)-len(s.rstrip('0'))
m //= 2**c
r = max(r, c)
return r # Chai Wah Wu, Sep 29 2021
CROSSREFS
Cf. A347668 (indices of records), A347669 (indices of first occurrences), A348007.
Sequence in context: A317951 A095382 A135282 * A179411 A103859 A253808
KEYWORD
nonn,easy
AUTHOR
Paolo Xausa, Aug 30 2021
STATUS
approved