Conjecture: Any Fibonacci number can be expressed as a sum of consecutive numbers on a twelve-hour clock, if not already present on the clock.

fibonacci-numberssequences-and-series

So, this is an observation I made today while drinking a cup of coffee… I'm not very sure if everyone's already aware of this, or even if this is something noteworthy, but I found it pretty interesting. Thus, I thought of sharing my observation, here, on StackExchange.

We're all familiar with the Fibonacci Series:
$$1,1,2,3,5,8,13,21,34,55,89,144,233,….$$

PLEASE NOTE that my observations satisfy these numbers, I haven't tried it out for numbers past $233$.

As the title suggests, we need a twelve-hour clock too, so let me insert that here as well 🙂enter image description here

So, what exactly is my observation?

Any Fibonacci number can be expressed as a sum of consecutive numbers on a clock, if not already present on the clock.

Here, since the numbers are on a clock, consecutive doesn't have to mean numbers along one direction, of course.

So, $1,1,2,3,5,8$ are already present on the clock.
$$13=6+7$$
$$21=10+11$$
$$34=10+11+12+1$$
$$55=1+2+3+4+5+6+7+8+9+10$$
$$89=1+2+3+4+5+6+7+8+9+10+11+12+11$$
$$144=1+2+3+4+5+6+7+8+9+10+11+12+11+10+9+8+7+6+5+4+3+2+1$$
$$233=1+2+3+4+5+6+7+8+9+10+11+12+1+2+3+4+5+6+7+8+9+10+11+12+12+11+10+9+8+7+6+5+4+3+2$$

Going by the trend, I think this should work for higher Fibonacci Numbers as well.

So, this is what I observed. Again, not sure if this is something very trivial, or has already been noticed by someone.

Kindly let me know what you think of this…

Cheers!!

Best Answer

Notice that a number $x$ can be expressed as a sum of consecutive numbers on a clock if and only if $x + 78$ can be expressed as such. To see why this is, we consider the example of $$17 = 8 + 9$$ To express $95 = 17 + 78$ as a sum of consecutive clock numbers, we simply go around the clock an extra time, giving us $$95 = 8+9 + \underbrace{10+11+12+1+2+3+4+5+6+7+8+9}_{78}$$ For the other direction of the biconditional, notice that if $y > 78$, then in order to express $y$ as a sum of consecutive clock numbers, we'll need to have at least 12 terms in the summation (as the sum of any 11 consecutive clock numbers is less than 78, the sum of all 12). To write $y-78$ as a sum of consecutive clock numbers, we express $y$ as such and then strip off the first twelve terms.

This means that we can determine all of the numbers that satisfy your condition just by considering the numbers 1 through 78. As Greg Martin points out, the better question may be which numbers do not satisfy your condition. The numbers between 1 and 78 inclusive that do not satisfy your condition are $$16, 31, 32, 37, 41, 46, 47, 62$$ I haven't been able to come up with a nice mathematical criterion that gives these numbers; here's a program that prints them out (I gave no regard to efficiency when writing this):

for n in range(1, 78):
    satisfies_property = False
    for k in range(1, 13):
        tot = 0
        s = k
        while tot < n:
            tot += s
            s = (s+1)%12
        if tot == n:
            satisfies_property = True
    if not satisfies_property:
        print(n)

Notably, because we have $$F_{17} = 1597 = 37 + 20 \cdot 78$$ we see that not all Fibonacci numbers can be expressed as the sum of consecutive numbers on a clock. The next few Fibonacci numbers that do not satisfy your condition are $F_{19} = 4181$, $F_{23} = 28657$, and $F_{37} = 24157817$, which are congruent to $47$, $31$, and $47$ modulo 78, respectively. To make a more global statement, there is in fact a positive fraction of the Fibonacci sequence that violates this condition, which follows from the existence of the Pisano period of 78 (which is 168).

There is a natural generalization of your problem to the case when instead of having twelve numbers on the clock, we have some arbitrary number $M$. Then the sum of all of the numbers is $\frac{M(M+1)}2$ instead of 78, so a number $x$ can be expressed as a sum of consecutive numbers on this clock if and only if $x + \frac{M(M+1)}2$ can. Again, we can computationally determine which numbers $x$ between 1 and $\frac{M(M+1)}2$ violate this condition. Here is a table for different values of $M$: \begin{align*} 1 \leq M \leq 5 &: && \emptyset \\ M = 6 &: && \{8,13\} \\ M = 7 &: && \emptyset \\ M = 8 &: && \{17,19\} \\ M = 9 &: && \{16,29\} \\ M = 10 &: && \{23,32\} \\ M = 11 &: && \{16,23,29,37,43,50\} \\ M = 12 &: && \{16,31,32,37,41,46,47,62\} \\ M = 13 &: && \{29,32,43,48,59,62\} \\ M = 14 &: && \{16, 31, 32, 41, 43, 47, 58, 62, 64, 73, 74, 89\} \\ M = 15 &: && \{31, 37, 41, 47, 53, 58, 59, 61, 62, 67, 73, 79, 83, 89\} \end{align*} For general values of $M$, it seems like about one-fifth of all numbers cannot be expressed as a sum of consecutive numbers on an $M$-clock.

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