Arbitrarily long runs of consecutive integers with the same Collatz height

The height of a positive integer n is the number of steps of the Collatz map C (C(n) = n/2 for even n, C(n) = 3n+1 for odd n) needed to reach 1. Garner (1985) conjectured that there are arbitrarily long runs of consecutive integers with the same height; Gao (1993) recalled the conjecture with the same height and the same total stopping time, that is, the same number of halvings, and attributed it also to Penning. We prove it: for all L and N there is n ≥ N such that n, n+1, …, n+L have the same height and the same total stopping time. They then also have the same height for the shortcut map T(x) = x/2 or (3x+1)/2, and every term of the sequence A078441 of the OEIS exists. The proof combines three facts. By the companion paper on a conjecture of Gao, the proportion of residues r modulo 2^t such that r and r+1 merge under T within t steps with the same number of odd terms tends to 1, so a union bound gives L consecutive such pairs. By a theorem of Terras, all the members of such a block of residue classes are carried after t steps to one common value 3^a u + A, for every u, and 3 does not divide A. Since 2 is a primitive root modulo every power of 3, u can be chosen so that the common value is a power of 2, in the spirit of the lemma of Penning and Gao on the repetition of runs. The argument is formalized in Lean 4 with Mathlib, using only the standard axioms. It gives existence without useful bounds, and it says nothing about whether other integers reach 1.