

A330920


Number of steps required to reach 1 for repeated applications of the Collatzinspired function f(n) = 6*n+5(n mod 5), or 1 if 1 is never reached.


0



0, 7, 5, 3, 1, 12, 31, 10, 29, 8, 27, 8, 8, 25, 6, 6, 23, 19, 19, 4, 21, 17, 17, 36, 2, 19, 15, 15, 34, 13, 129, 17, 13, 13, 32, 30, 32, 127, 15, 11, 11, 13, 144, 28, 30, 125, 13, 30, 13, 9, 11, 142, 26, 11, 28, 123, 11, 28, 11, 9, 140, 9, 140, 24, 9, 9, 28, 121
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OFFSET

1,2


COMMENTS

Inspired by the Collatz conjecture, I tried to generalize for more divisors than 2. I quickly came up with a formula, and then discovered that Carnielliy had written about it previously in 2015.


LINKS

Table of n, a(n) for n=1..68.
Carnielliy, Walter. Some Natural Generalizations Of The Collatz Problem, Applied Mathematics ENotes, 2015, page 208.


PROG

(Python)
def f(n, d):
"""
A Collatzlike function.
When d == 2 this becomes '3x+1' problem exactly.
"""
i = n % d
if r == 0:
return n / d
else:
# Produce a larger number that is divisible by d.
return (d + 1) * n + d  i
def steps(n, d):
"""
Return the number of steps needed to reach 1, or 1 if a 1 is never reached.
"""
count = 0
seen = set([1])
x = n
# Loop until a cycle is detected.
while x not in seen:
seen.add(x)
x = f(x, d)
count += 1
if x == 1:
return count
else:
# There was a cycle
return 1
# Create a bunch of terms for d=5
S = [steps(x, d=5) for x in xrange(1, 1000)]
print S


CROSSREFS

If you replace d=5 with d=2, this code produces A006577.
Sequence in context: A085927 A180597 A219242 * A155816 A308414 A335985
Adjacent sequences: A330917 A330918 A330919 * A330921 A330922 A330923


KEYWORD

nonn


AUTHOR

Matt Donahoe, Aug 22 2020


STATUS

approved



