A375495 a(n) = number of different ways of selecting the minimum number of operations chosen from f(x) = 3x+1 and g(x) = floor(x/2) needed to reach n when starting from 1.

1, 1, 1, 2, 1, 4, 1, 1, 6, 1, 3, 1, 14, 1, 2, 5, 5, 1, 28, 1, 1, 4, 1, 9, 5, 9, 1, 48, 1, 1, 2, 3, 10, 1, 1, 15, 5, 23, 12, 2, 1, 131, 1, 3, 1, 4, 6, 3, 20, 5, 2, 1, 1, 27, 5, 43, 34, 25, 1, 4, 1, 1, 332, 1, 5, 5, 2, 1, 10, 8, 12, 3, 37, 5, 5, 4, 10, 2, 1, 1, 39, 5, 63, 68, 67

Offset: 0

Author

Russell Y. Webb, Aug 18 2024

Keywords

• nonn

Comments

The minimum number of operations is A375494(n) and that minimum is attained by a(n) different sequences of operations.

Crossrefs

Cf. A375494 (number of operations), A375496 (indices of 1's).

Programs

Python

from itertools import product
seq = [None for _ in range(200)]
num = [   0 for _ in range(len(seq))]
for L in range(0, 23):
   for P in product((True, False), repeat=L):
      x = 1
      for upward in P:
         x = 3*x+1 if upward else x//2
      if x < len(seq):
         if num[x] == 0 or L < seq[x]:
            seq[x], num[x] = L, 1
         elif L == seq[x]:
            num[x] += 1
print(', '.join([str(x) for x in num]))