A304174 -id:A304174 - cp's OEIS frontend

A304524 Consider the ratio res(p) = 2^A006666(p) / (p*3^A006667(p)) where p is prime. The prime numbers in this sequence are those for which res(p) sets a new record.

2, 3, 7, 37, 43, 229, 271, 379, 673, 839, 1987, 5297, 25111, 44641, 50221, 94057, 334423, 1189057, 1759579, 2505337, 28153249, 46869157, 87780541, 584543567, 768901097

Offset: 1

Michel Lagneau, May 14 2018

Is the sequence finite?
In the general case, the residue of a number n in the 3x+1 problem is defined as the ratio res(n) = 2^A006666(n) / (n*3^A006667(n)) (see A127789).
Conjecture: for all prime p, res(p) < res(993) = 2^61/(3^32*993) = 1.253142... (see A304174).

			From _Jon E. Schoenfield_, May 23 2018: (Start)
Let D = A006666(p) and U = A006667(p); then res(p) = 2^D/(p*3^U). It seems clear that res(993) - res(p) is converging toward a positive value:
.
          p |   D |  U |     res(p)      | res(993)-res(p)
  ----------+-----+----+-----------------+----------------
          2 |   1 |  0 | 1               | 0.2531421443...
          3 |   5 |  2 | 1.1851851851... | 0.0679569592...
          7 |  11 |  5 | 1.2039976484... | 0.0491444959...
         37 |  15 |  6 | 1.2148444741... | 0.0382976702...
         43 |  20 |  9 | 1.2389111604... | 0.0142309838...
        229 |  24 | 10 | 1.2407145246... | 0.0124276197...
        271 |  29 | 13 | 1.2425797507... | 0.0105623936...
        379 |  39 | 19 | 1.2480350469... | 0.0051070974...
        673 |  43 | 21 | 1.2494773856... | 0.0036647587...
        839 |  56 | 29 | 1.2514151532... | 0.0017269911...
       1987 |  62 | 32 | 1.2525114739... | 0.0006306704...
       5297 |  65 | 33 | 1.2529055685... | 0.0002365758...
      25111 |  72 | 36 | 1.2529406796... | 0.0002014647...
      44641 |  76 | 38 | 1.2529625095... | 0.0001796348...
      50221 |  73 | 36 | 1.2529656281... | 0.0001765162...
      94057 |  85 | 43 | 1.2529812032... | 0.0001609411...
     334423 |  90 | 45 | 1.2529882803... | 0.0001538640...
    1189057 |  95 | 47 | 1.2529909733... | 0.0001511710...
    1759579 | 113 | 58 | 1.2529910420... | 0.0001511023...
    2505337 | 104 | 52 | 1.2529915763... | 0.0001505680...
   28153249 | 117 | 58 | 1.2529917096... | 0.0001504347...
   46869157 | 132 | 67 | 1.2529917720... | 0.0001503722...
   87780541 | 144 | 74 | 1.2529919281... | 0.0001502162...
  584543567 | 161 | 83 | 1.2529919325... | 0.0001502118...
  768901097 | 182 | 96 | 1.2529919396... | 0.0001502047...
(End)

Eric Roosendaal, On the 3x+1 Problem

Cf. A006666, A006667, A127789, A304174.

lst={2};Print["a(n)"," ","A006667(a(n))"," ","A006666(a(n))","       ","res(a(n))"];q=1;Collatz[n_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,Prime[n],#>1&];nn=10000;t={};n=0;While[Length[t]5000,Break[]];q=Prime[n]*3^od/2^ev]];lst

a(23)-a(24) from Jon E. Schoenfield, May 19 2018

A381762 Numbers k such that S(k) sets a new record, where S(k) denotes the sum of the reciprocals of odd elements in the Collatz sequence which starts at k.

Original entry on oeis.org

1, 3, 7, 9, 559, 745, 993

Offset: 1

Barak Manos, Mar 06 2025

This sequence is conjectured to be finite.

Cf. A304174.

Cf. A127789.

f[n_] := Total[1/Select[NestWhileList[If[OddQ[#], 3*# + 1, #/2] &, n, # > 1 &], OddQ]]; seq[lim_] := Module[{s = {}, fm = 0, f1}, Do[f1 = f[n]; If[f1 > fm, fm = f1; AppendTo[s, n]], {n, 1, lim}]; s]; seq[1000] (* Amiram Eldar, Mar 06 2025 *)

from fractions import Fraction
def S(n):
    arr = []
    while True:
        n //= (n - (n & (n - 1)))
        arr.append(n)
        if n == 1:
            break
        n = 3 * n + 1
    return sum(Fraction(1, x) for x in arr)
m = 0
for n in range(1, 1000):
    k = S(n)
    if m < k:
        m = k
        print(n)

cp's OEIS Frontend

A304524 Consider the ratio res(p) = 2^A006666(p) / (p*3^A006667(p)) where p is prime. The prime numbers in this sequence are those for which res(p) sets a new record.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions