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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
2
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
0,4
COMMENTS
The minimum number of operations is A375494(n) and that minimum is attained by a(n) different sequences of operations.
PROG
(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]))
CROSSREFS
Cf. A375494 (number of operations), A375496 (indices of 1's).
Sequence in context: A339833 A118745 A235671 * A131034 A346873 A130313
KEYWORD
nonn
AUTHOR
Russell Y. Webb, Aug 18 2024
STATUS
approved