# Novel integer sequence problem?

In the following sequence, what should be the next number?

1 1 1 2 5 1 1 2 2 4 5 8 17 50 11 ?

• I notice that rot13(Rirel ahzore vf n fhz bs pbafrphgvir cerivbhf ahzoref). Commented Mar 20, 2020 at 20:57
• @DrXorile Not for long it does... Commented Apr 18, 2020 at 11:18

14

The pattern here looks at how often the decimals remain the same at a certain depth when dividing (feel free to edit this answer to make it clearer), from numbers below 1.

It helps to consider 2 numbers/rows at a time:

 1/1 = 1.00, when we divide 1 by 2, the 0 changes to a 5, so [1]
|
1/2 = 0.50, when we divide 1 by 3, the 5 changes to a 3, so [1]
|
1/3 = 0.33, when we divide 1 by 4, the 3 changes to a 2, so [1]
|
1/4 = 0.25, when we divide 1 by 5, the 2 stays a 2, -> continue
=
1/5 = 0.20, when we divide 1 by 6  the 2 changes to a 1, so [2]
|
1/6 = 0.16, when we divide 1 by 7, the 1 stays a 1, -> continue
=
1/7 = 0.14, when we divide 1 by 8, the 1 stays a 1, -> continue
=
1/8 = 0.12, when we divide 1 by 9, the 1 stays a 1, -> continue
=
1/9 = 0.11, when we divide 1 by 10,the 1 stays a 1, -> continue
=
1/10 = 0.10, when we divide 1 by 11,the 1 changes to a 9, so [5]
|
1/11 = 0.09, when we divide 1 by 12,the 9 changes to a 8, so [1]
|
1/12 = 0.08, when we divide 1 by 13,the 8 changes to a 7, so [1]

... and so on ...


Notably, after 1/10, we move to <0.1, so then you check the next significant number, in the 0.01 - 0.09 range.

You can generate the sequence using this (unoptimized) Python code:

counts = []
last = None
ending = ""
it = 2
count = 1
while True:
v = 1 / it
ending = "0"
sv = str(v).replace("0.", "").lstrip("0")
if sv:
ending = sv[0]
if ending != last:
counts.append(count)
print("new:", it, "[{}]".format(count))
last = ending
count = 1
else:
count += 1
#print(v)
it += 1


Here is a plot:

I found it to be a rather weird sequence, but perhaps it reminds the community of similar sequences? I hope you enjoy it!