Ordered Cycles of the Last Digit Under Repeated Multiplication (Base 10)

[Ali Sada]

Hi everyone, 

Hope all is well. I am not sure whether this observation is already recorded in the OEIS.

Consider repeated multiplication by a fixed digit in base 10, tracking only the last digit of the product. The last digit of the result depends only on the last digit of the original number and the last digit of the multiplier. When this multiplication is repeated, the last digits eventually form ordered cycles (“loops”). Cycles that differ by direction are treated as distinct.
If the repeated multiplier is 1, the last digit of the result remains the same as the original last digit. This produces 10 distinct loops:
{0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}.
If the repeated multiplier is 2, there is one nontrivial loop:
{2, 4, 8, 6}.
The digits 0 and 5 form trivial fixed-point loops already counted.
If the repeated multiplier is 3, there are two distinct loops:
{1, 3, 9, 7} and {6, 8, 4, 2}.
If the repeated multiplier is 4, there are two distinct loops:
{2, 8} and {4, 6}.
If the repeated multiplier is 5, only the fixed-point loops {0} and {5} occur, yielding no new loops.
If the repeated multiplier is 6, the digits {0}, {2}, {4}, {6}, and {8} are fixed points, yielding no new nontrivial loops.
If the repeated multiplier is 7, there are two loops:
{7, 9, 3, 1}  and {4, 8, 6, 2}, but the second one is already mentioned, so we have only one distinct loop.  
If the repeated multiplier is 8, there is one loop:
{6, 8, 4, 2}, which has already appeared in the same order, so no new loop is introduced.
If the repeated multiplier is 9, there are four ordered loops:
{1, 9}, {7, 3}, {8, 2}, and {4, 6}.
Of these, {8, 2} and {4, 6} have already appeared, leaving two new ordered loops:

{1, 9} and {7, 3}.

Altogether, in base 10, there are 18 distinct ordered loops for the last digit under repeated multiplication by a fixed digit.
The sequence of interest is therefore: the number of distinct ordered last-digit loops in base n.
Is this a new sequence?

Best,

Ali

[jp allouche math]

Hi this might have some sort of relation with A006694 (base 2, not 10)
jp

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[Daniel Mondot]

You're not the first person to notice that...

Not only, when you repeatedly multiply by some number, the last digit (digit 0) forms loops having a cycle c,

But if you look at the next digit (digit 1) it also forms loops having a cycle c*k.

And the next digit  (digit 2) forms loops having a cycle c*k^2, and so on....

Naturally there are some exceptions: for digits>0, the cycles do not start immediately. 

For instance, if the multiplier is 10, and the original number is 1234, the fourth digit starts : 1, 2, 3, 4, before settling on a cycle of zeroes.

Daniel.

[Allan Wechsler]

I would have expected to see, "Number of limit cycles of the process a -> ak, mod n", somewhere on OEIS. If my hand calculations can be trusted, the sequence should begin 1, 2, 4, 5, 9, 8, 16, 14, ... and I can't find any sequence matching this data. I have probably made an arithmetic error, because I would be very surprised if no OEIS contributor before Ali considered this simple question. I fully expect that after somebody fixes my arithmetic we will find that this is "Euler's mu-function" or something like that.

-- Allan

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[Jack Brennen]

I get this for the sequence:

1, 2, 4, 5, 9, 8, 16, 15, 19, 18, 28, 21, 41, 32, 40, 43, 49, 38, 64, 49, 69, 56, 64, 67, 81, 82, 82, 85, 105, 80

and that similarly has no hits in the OEIS database.

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[Dave Consiglio]

This sequence might be interesting in different bases, too.

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[Christopher Landauer]

Hihi, all -

Isn’t this about the multiplicative structure of the ring Z_b (integers mod b) for a radix b, and the powers of an element d in that ring? There will be many many sequences that can be derived from this

For example,
for base 10, the additive group of the ring is Z_2 x Z_5 - then for any element, its powers define a collection of cosets that give the result, presumably easily computed in advance (it seems that the only real consideration for this is gcd(b, d))

It seems that there should be a relatively simple expression for the answer for any b and d

More soon,
chris

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[Kevin Ryde]

Christopher Landauer <topc...@gmail.com> writes:
>
> (integers mod b) for a radix b, and the powers of an element d in that
> ring

A023142 does it with d=10 if that's a starting point.
(That one is related to how many different periodic parts occur
in the decimals of fractions x/n, over all numerators x.
Differs only when n has a factor of 2 or 5, I think.)

[Ali Sada]

Thank you all for your responses. I really appreciate them. I have added the sequence, A390033. I used Jack's data and slightly modified Allan's definition. 

Best,

Ali

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[Sean A. Irvine]

There is now a draft for this:

https://oeis.org/draft/A392620

Please feel free to make additions (programs, better examples, and to check my later terms!).

Sean.

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[Allan Wechsler]

Is this new draft supposed to supersede Ali's earlier announcement of https://oeis.org/draft/A390033 ?

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[Sean A. Irvine]

Whoops, my bad. I'll get rid of my one.

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[Daniel Mondot]

If, instead of looking at the number of distinct cycles, we look at the length of the cycles,

- for the last digit (position 0), we get : A173635

- the next to last digit (position 1), or simply the last 2 digits, we get A220022

- the last 3 digits, we get A220023

- the last 4 digits, we get A220024

- the last 5 digits, we get A220025.

- etc...

Perhaps these sequences should be crossrefed in A390033

And incidentally, I was wrong in my earlier comment (i.e: the cycle being c*k^n where n is the number of digits-1), this appears to only true for some multipliers (e.g.: 2, 4, 6, 8, 10, 12, 14, 16, 17, 20, 22, 28, 30, etc...)

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[Ali Sada]

Thank you everyone. I am sorry I thought I copied and pasted Allan's definition but I got the wrong text (Yes, my ADHD is that bad). Kevin thinks this shouldn't be a "base" sequence and I agree with him. I don't want to make another stupid mistake, so I would really appreciate it if someone help me with the exact definition.

Best,

Ali

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