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

Author

L. Edson Jeffery, May 03 2015

Keywords

• nonn
• tabf

Comments

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.

Examples

			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
		

Crossrefs

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

Programs

Mathematica

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 *)