OFFSET
1,2
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.
EXAMPLE
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}
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 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
L. Edson Jeffery, May 03 2015
STATUS
approved