A254068 -id:A254068 - cp's OEIS frontend

A253720 a(n) = length of row n in A253676 and A254068, assuming the 3x+1 (or Collatz) conjecture.

1, 2, 5, 3, 4, 2, 6, 5, 7, 5, 18, 5, 6, 3, 8, 4, 7, 4, 19, 6, 5, 2, 7, 4, 20, 6, 8, 19, 3, 5, 16, 18, 21, 7, 15, 4, 20, 5, 9, 8, 17, 18, 10, 8, 8, 5, 10, 18, 21, 6, 3, 7, 9, 3, 5, 19, 11, 8, 14, 8, 6, 4, 10, 17, 22, 7

Offset: 1

L. Edson Jeffery, May 02 2015

nonn

Cf. A253676, A254068.

(* Row lengths of A253676 and A254068: *)
v[n_] := IntegerExponent[n, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; s[x_] := (3 + (3/2)^v[1 + f[x]]*(1 + f[x]))/6; A253676[n_] := NestWhileList[s[4*# - 3] &, n, # > 1 &]; Table[Length[A253676[n]], {n, 1, 66}]

For n>1, k>=1, a(n) = a((8+(3*n-2)*4^k)/12).

A257480 S(n) = (3 + (3/2)^v(1 + F(4n - 3))(1 + F(4n - 3)))/6, n >= 1, where F(x) = (3x + 1)/2^v(3*x + 1) for x odd, and v(y) denotes the 2-adic valuation of y.

Original entry on oeis.org

1, 1, 5, 2, 4, 1, 8, 5, 7, 5, 41, 5, 10, 2, 17, 14, 13, 4, 32, 8, 16, 1, 26, 14, 19, 8, 68, 11, 22, 5, 35, 41, 25, 7, 59, 14, 28, 5, 44, 23, 31, 41, 365, 17, 34, 5, 53, 41, 37, 10, 86, 20, 40, 2, 62, 32, 43, 17, 149

Offset: 1

L. Edson Jeffery, Apr 26 2015

nonn

In the following, let F^(k)(x) denote k-fold iteration of F and defined by the recurrence F^(k)(x) = F(F^(k-1)(x)), k > 0, with initial condition F^(0)(x) = x, and let S^(k)(n) denote k-fold iteration of S and defined by the recurrence S^(k)(n) = S(S^(k-1)(n)), k > 0, with initial condition S^(0)(n) = n, where F and S are as defined above.
Theorem 1: For each x, there exists a j>0 such that F^(j)(x) == 1 (mod 4).
Theorem 2: S(n) = m if and only if S(4*n-2) = m.
Conjecture 1: For each n, there exists a k such that S^(k)(n) = 1.
Theorem 3: Conjecture 1 is equivalent to the 3x+1 conjecture.
Theorem 4: The sequence {log(S(n))/log(n)}_{n>1} is bounded with least upper bound equal to log(3)/log(2).
[I have proved Theorems 1--4 (along with several lemmas) and am trying to finish typesetting the draft containing the proofs but had been too ill to finish that work until now. The draft also contains the derivation of the function S from properties of the known function F (A075677). When that paper is completed (hopefully within two weeks) I will then upload it to the links section and delete this comment.]

K. H. Metzger, Untersuchungen zum (3n+1)-Algorithmus, Teil II: Die Konstruktion des Zahlenbaums, PM (Praxis der Mathematik in der Schule) 42, 2000, 27-32.

I. Korec and Štefan Znám, A Note on the 3x+1 Problem, Amer. Math. Monthly 94, 1987, pp. 771-772.
J. C. Lagarias, The 3x + 1 Problem and Its Generalizations, Amer. Math. Monthly 92, 1985, pp. 3-23.
J. C. Lagarias, The 3x+1 Problem: An Annotated Bibliography (1963-2000), arXiv:math/0309224 [math.NT], 2003-2011.
J. C. Lagarias, The 3x+1 Problem: an annotated bibliography, II (2000-2009), arXiv:math/0608208 [math.NT], 2006-2012.

Cf. A006370, A070165, A075677, A256598.
Cf. A241957, A254067, A254311, A257499, A257791 (all used in the proof of Thm 4).
Cf. A253676 (iteration of S terminating at the first occurrence of 1, assuming the 3x+1 conjecture).
Cf. A253720, A254068, A254070, A254131, A254312, A255138, A255168, A258415.

v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; s[n_] := (3 + (3/2)^v[1 + f[4*n - 3]]*(1 + f[4*n - 3]))/6; Table[s[n], {n, 59}]

a(n) = my(x=3*n-2, v=valuation(x, 2)); x>>=v; v=valuation(x+1, 2); (((x>>v)+1)*3^(v-1)+1)/2; \\ Ruud H.G. van Tol, Jul 30 2023

A254070 a(n) = -1 + (3/2)^(-1 + v(1 + F(4n - 3)))(1 + F(4n - 3)), where v(y) is the 2-adic valuation of y, F(x) = (3x + 1)/2^v(3*x + 1), and x == 1 (mod 2).

Original entry on oeis.org

1, 1, 17, 5, 13, 1, 29, 17, 25, 17, 161, 17, 37, 5, 65, 53, 49, 13, 125, 29, 61, 1, 101, 53, 73, 29, 269, 41, 85, 17, 137, 161, 97, 25, 233, 53, 109, 17, 173, 89, 121, 161, 1457, 65, 133, 17, 209, 161, 145, 37, 341, 77, 157, 5, 245, 125, 169, 65, 593, 89, 181, 53, 281, 485, 193, 49, 449, 101, 205, 13

Offset: 1

L. Edson Jeffery, May 03 2015

nonn

a(n) is the first successor in the 3x+1 trajectory of 4*n-3 that is congruent to 1 mod 4. - Ruud H.G. van Tol, Jul 16 2023

Ruud H.G. van Tol, Table of n, a(n) for n = 1..10000

Cf. A257480, A253676, A254068.

v[y_] := IntegerExponent[y, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; s[n_] := -1 + (3/2)^(-1 + v[1 + f[4*n - 3]])*(1 + f[4*n - 3]); Table[s[n], {n, 70}] (* L. Edson Jeffery, Mar 29 2021 *)

a(n) = my(x=3*n-2, v=valuation(x,2)); x>>=v; v=valuation(x+1, 2)-1; ((x>>v)+1)*3^v-1; \\ Ruud H.G. van Tol, Jul 16 2023

a(n) = 4*A257480(n) - 3. - L. Edson Jeffery, Mar 29 2021

New name by L. Edson Jeffery, Mar 29 2021

cp's OEIS Frontend

A253720 a(n) = length of row n in A253676 and A254068, assuming the 3x+1 (or Collatz) conjecture.

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A257480 S(n) = (3 + (3/2)^v(1 + F(4n - 3))(1 + F(4n - 3)))/6, n >= 1, where F(x) = (3x + 1)/2^v(3*x + 1) for x odd, and v(y) denotes the 2-adic valuation of y.

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A254070 a(n) = -1 + (3/2)^(-1 + v(1 + F(4n - 3)))(1 + F(4n - 3)), where v(y) is the 2-adic valuation of y, F(x) = (3x + 1)/2^v(3*x + 1), and x == 1 (mod 2).

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cp's OEIS Frontend

A253720 a(n) = length of row n in A253676 and A254068, assuming the 3x+1 (or Collatz) conjecture.

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

A257480 S(n) = (3 + (3/2)^v(1 + F(4*n - 3))*(1 + F(4*n - 3)))/6, n >= 1, where F(x) = (3*x + 1)/2^v(3*x + 1) for x odd, and v(y) denotes the 2-adic valuation of y.

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Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

PARI

A254070 a(n) = -1 + (3/2)^(-1 + v(1 + F(4*n - 3)))*(1 + F(4*n - 3)), where v(y) is the 2-adic valuation of y, F(x) = (3*x + 1)/2^v(3*x + 1), and x == 1 (mod 2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A257480 S(n) = (3 + (3/2)^v(1 + F(4n - 3))(1 + F(4n - 3)))/6, n >= 1, where F(x) = (3x + 1)/2^v(3*x + 1) for x odd, and v(y) denotes the 2-adic valuation of y.

A254070 a(n) = -1 + (3/2)^(-1 + v(1 + F(4n - 3)))(1 + F(4n - 3)), where v(y) is the 2-adic valuation of y, F(x) = (3x + 1)/2^v(3*x + 1), and x == 1 (mod 2).