A347669 -id:A347669 - cp's OEIS frontend

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.

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

Paolo Xausa, Aug 30 2021

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.

			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).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A070165, A135282.

Cf. A347668 (indices of records), A347669 (indices of first occurrences), A348007.

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}]

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

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

A347668 Indices of records in A347409.

Original entry on oeis.org

1, 2, 3, 15, 21, 75, 151, 1365, 5461, 7407, 14563, 87381, 184111, 932067, 5592405, 13256071, 26512143, 357913941, 1431655765, 3817748707, 22906492245, 91625968981, 244335917283, 1466015503701, 5212499568715, 10424999137431, 93824992236885

Offset: 1

Paolo Xausa, Sep 10 2021

Conjecture 1: A347409(a(n)) is even for n >= 11. Conjecture 2: all even numbers > 2 appear as A347409(a(n)) for some n. - Chai Wah Wu, Sep 29 2021

If conjectures 1 and 2 are true, then A347409(a(n)) = 2n - 6 for n >= 11, and hence a(n) <= (4^(n-3)-1)/3 for n >= 11 since A347409((4^(n-3)-1)/3) = 2n - 6. - Charles R Greathouse IV, Oct 25 2022

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A070165, A135282, A347409, A347669, A002450.

A347409[n_]:=(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)
upto=100000;a={};rec=-1;Do[If[(r=A347409[i])>rec,rec=r;AppendTo[a,i]],{i,upto}];a

f(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; \\ A347409
lista(nn) = my(r=-1, m); for (n=1, nn, if ((m=f(n)) > r, print1(n, ", "); r = m);); \\ Michel Marcus, Sep 10 2021

a(15) from Michel Marcus, Sep 10 2021
a(16)-a(17) from Alois P. Heinz, Sep 10 2021
a(18)-a(20) from Michael S. Branicky, Sep 28 2021
a(21)-a(22) from Michael S. Branicky, Sep 30 2021
a(23) from Michael S. Branicky, Oct 04 2021
a(24)-a(27) from Kevin P. Thompson, Apr 14 2022

cp's OEIS Frontend

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.

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