A253676 -id:A253676 - cp's OEIS frontend

A254068 Irregular triangle T read by rows in which the entry in row n and column k is given by T(n,k) = 4*A253676(n,k) - 3, k = 1..A253720(n), assuming the 3x+1 (or Collatz) conjecture.

1, 5, 1, 9, 17, 13, 5, 1, 13, 5, 1, 17, 13, 5, 1, 21, 1, 25, 29, 17, 13, 5, 1, 29, 17, 13, 5, 1, 33, 25, 29, 17, 13, 5, 1, 37, 17, 13, 5, 1, 41, 161, 121, 137, 233, 593, 445, 377, 425, 2429, 3077, 577, 433, 325, 61, 53, 5, 1, 45, 17, 13, 5, 1

Offset: 1

L. Edson Jeffery, May 03 2015

Definitions: Let v(y) denote the 2-adic valuation of y. Let N_1 denote the set of odd natural numbers. Let F : N_1 -> N_1 be the map defined by F(x) = (3*x + 1)/2^v(3*x + 1) (cf. A075677). 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.
This triangle can be constructed by restricting the initial values to the numbers 4*n - 3, iterating F until 1 is reached (assuming the 3x+1 conjecture) and removing all iterates not congruent to 1 modulo 4. Equivalently, for each n, this is accomplished by iterating (until 1 is reached, assuming the 3x+1 conjecture) the function S defined in A257480 to get the triangle A253676, and finally taking T(n,k) = 4*A253676(n,k) - 3.
Conjecture: For each natural number n, there exists a k >= 0, such that F^k(4*n - 3) = 1.
Theorem 1: Conjecture 1 is equivalent to the 3x+1 (or Collatz) conjecture.
Proof: See A257480.

			T begins:
   1
   5   1
   9  17  13   5   1
  13   5   1
  17  13   5   1
  21   1
  25  29  17  13   5   1
  29  17  13   5   1
  33  25  29  17  13   5   1
  37  17  13   5   1
  41 161 121 137 233 593 445 377 425 2429 3077 577 433 325 61 53 5 1
  45  17  13   5   1
  49  37  17  13   5   1
  53   5   1
  57  65  49  37  17  13   5   1

Cf. A014682, A070165, A075677, A256598, A257480, A253676, A254070.

v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; s[n_] := NestWhileList[(3 + (3/2)^v[1 + f[4*# - 3]]*(1 + f[4*# - 3]))/6 &, n, # > 1 &]; t = Table[4*s[n] - 3, {n, 1, 15}]; Flatten[t] (* Replace Flatten with Grid to display the triangle *)

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

Original entry on oeis.org

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

A254068 Irregular triangle T read by rows in which the entry in row n and column k is given by T(n,k) = 4*A253676(n,k) - 3, k = 1..A253720(n), assuming the 3x+1 (or Collatz) conjecture.

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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(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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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).

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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.

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).