One man's manual calculation exercises

What follows are Dan Polansky's manual calculation exercises intended for pencil and paper, with no use of pocket calculator or computer. They involve primarily decimal addition and subtraction and are very simple in that regard. They are something like brain jogging, exercising attention (or focus and concentration) and short term memory. Admittedly, it is easy (boringly trivial) to generate a calculation exercise, e.g. by writing down a random addition assignment, such as 8457 + 256. However, there would be nothing interesting to it. The exercises are meant to reveal something interesting and involve some kind of systematic process.

Fibonacci descent

In Fibonacci descent (term invented for this purpose without trying to figure out whether it has an established meaning elsewhere), we start with a number and go from it downward as if we were trying to trace Fibonacci sequence from higher item to lower item. Thus, given two numbers m and n, the next number is m - n, and the process stops when m - n is bigger than n.

Let us start with 900 and make a guess for the next number to be 300:

• 900, 600, 300, 300, -

We indicate via "-" that the sequence has stopped. Surely we can do better. We ask: do we need the 2nd item, 600, to be higher or lower to maker the sequence longer? We note that the final 300 needs to be lower, meaning the 300 before it needs to be higher, meaning that the 600 before it needs to be lower. Next guess:

• 900, 590, 310, 280, 30, -

We see again that the 2nd number needs to be lower since: 30 (up), 280 (down), 310 (up), 590 (down). We guess that the move down needs to be larger.

• 900, 560, 340, 220, 120, 100, 20, -

We can do beven better and we see 560 needs to go down again.

• 900, 550, 350, 200, 150, 50, -

Alas, we went down too much. Let us go up a bit.

• 900, 555, 345, 210, 135, 75, 60, 15, -

Seeing that 15 needs to go up, we see that 555 needs to go up also. We start using the small increment of 1.

• 900, 556, 344, 212, 132, 80, 52, 28, 24, 4, -

When calculating the above, we would be doing repeated subtraction as before. However, we could also have applied a delta method: we can note that applying -1 to 555 results in delta sequence -1, 1, -2, 3, -5, 8, and 13 (delta of items in the current sequence against the previous sequence), which the Fibonacci sequence to which an alternating sign was added. That makes the calculation easier and allows for double check since we have two different calculation by which to arrive at the 3rd, 4th, etc. item in the sequence.

Noting that 4 needs to go up if the length is to be improved, we think 556 should go up also. Let us see.

• 900, 557, 343, 214, 129, 85, 44, 41, 3, -

This did not help. Thus, we found the longest sequence and note its length in brackets:

• 900, 556, 344, 212, 132, 80, 52, 28, 24, 4, - (10)

Once we have established one proper (longest) sequence for the starting number 900, we can continue doing the same for 899, 898, 897, etc. We can use the 2nd item in the previous sequence as a basis for our estimate of the 2nd item in our current sequence.

Let us do one more item, for which we copy the 2nd item from the previous sentence. It is actually possible for starting numbers less than 1000 to figure out in the head (wihtout pencil) whether the 2nd item needs to be copied unmodified or decremented:

• 899, 556, 343, 213, 130, 83, 47, 36, 11, -

This could hypothetically improve by decrementing 556 to 555 (since 11 needs to go up). We can figure it out without pencil using deltas. Since, this would mean we would apply the delta sequence -1, 1, -2, 3, -5, 8, -13, 21, leading to 11 + 21 = 32, no improvement, and thus, the sequence above is the longest one, with the length of 9, to be noted as (9):

• 899, 556, 343, 213, 130, 83, 47, 36, 11, - (9)

When we do the calcuation/deliberation in the head, we do not need to have two rows, of course: we first write the row without the final brackets, and then add the final bracket when we figure out that it is the optimal sentence.

Let us do one more and leave the rest as the exercise. We guess that in the downward sequence of sequences starting at less than 1000, the 2nd item is repeated at most twice, and thus we decrement 556 to 555.

• 898, 555, 343, 212, 131, 81, 50, 31, 19, 12, 7, 5, 2, - (13)

Here, we hit upon a starting number that produces a beautifully long sequence.

Sometimes, there is not a single optimal sequence for the given starting pair. For these cases, any of the longest solutions are acceptable.

We can spend hours doing this simple exercise and note whether we concentrate well enough to make no calculation errors.

Fibonacci sequence modulo n

Calculating the usual Fibonacci sequence using pencil and paper gets increasingly laborious given its approximately exponential rate of growth, leading to the number of digits in the numbers to be added steadily increading. However, we may calculate Fibonacci sequence modulo n, which leads to a sequence with a finite cycle given the items are all less than n. We can consider not only the sequence starting at 1, 1 but also those starting at other pairs. We can think of the sequences as rings, given they are cycles and one can start in the middle, and by iterating the additive Fibonacci process, arrive at the middle again.

Let us show some examples. Let us start at 10, since that is easy to calculate in the head.

Fibonacci sequence modulo 10 (100 pairs):

• 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0, 1, 1 (60)

We stopped the sequence when the loop was reached. Above, we write the number of pairs covered in brackets. The number of pairs covered is the number of the above list items minus two.

If we see this as a deterministic dynamical system, the state of the system is given by a pair and the system has finitely many states. Given the modulo of 10, the number of states is the number of pairs, which is 10 squared, which is 100.

We can try to find similar "rings" to cover all possible pairs. We have so far covered 60 pairs, so there are more to cover. In general, some would be cover by multiplying the sequence by a natural number. Since the pair 2, 2, resulting from multiplication of 2, is not yet covered, let us go for it.

• 2, 2, 4, 6, 0, 6, 6, 2, 8, 0, 8, 8, 6, 4, 0, 4, 4, 8, 2, 0, 2, 2 (20)

That was shorter. Let us search for more sequences of the form X, X, which are multiples of the first sequence. Another one is 5, 5 and there are no more.

• 5, 5, 0, 5, 5 (3)

A trivial one results from multiplication by 0:

• 0, 0 (1)

We now exahusted the sequences of the form X, X but have not yet exhausted the pair count 100. Let us consider sequences of the form 1, X. Some are already covered (e.g. 1, 2; 1, 5; 1, 6 etc.), but not all of them. Pair 1, 3 seems not covered, but if we find it is covered, we strike it out as an error.

• 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3 (12)

And we can examine a multiple of it:

• 2, 6, 8, 4, 2, 6 (4)

We have now exhausted all 100 pairs, which we see by adding the counts of exhausted pairs. We can move to another modulus. We move to the next modulus, 11, but examining smaller values may also be of interest.

Fibonacci sequence modulo 11 (121 pairs):

• 1, 1, 2, 3, 5, 8, 2, 10, 1, 0, 1, 1 (10)
• 2, 2, 4, 6, 10, 5, 4, 9, 2, 0, 2, 2 (10)
• 3, 3, 6, 9, 4, 2, 6, 8, 3, 0, 3, 3 (10)
• 4, 4, 8, 1, 9, 10, 8, 7, 4, 0, 4, 4 (10)
• 5, 5, 10, 4, 3, 7, 10, 6, 5, 0, 5, 5 (10)
• 6, 6, 1, 7, 8, 4, 1, 5, 6, 0, 6, 6 (10)
• 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7 (10)
• 8, 8, 5, 2, 7, 9, 5, 3, 8, 0, 8, 8 (10)
• 9, 9, 7, 5, 1, 6, 7, 2, 9, 0, 9, 9 (10)
• 10, 10, 9, 8, 6, 3, 9, 1, 10, 0, 10, 10 (10)
• 1, 4, 5, 9, 3, 1, 4 (5)
• 2, 8, 10, 7, 6, 2, 8 (5)
• 1, 8, 9, 6, 4, 10, 3, 2, 5, 7, 1, 8 (10)

Now we are done: we found 11 rings of the length 10, 2 rings of the length 5 and 1 ring of the length 1 (the trivial ring 0,0); summed up, we covered the sought 121 pairs (11 * 11).

Some initial observations concerning the rings found for mod 10 and mod 11:

• For mod 10, there is one very long ring relative to the total pair count. When we create a multiple of it, there is a "collapse" in that it yields shorter rings.
• For mod 11, the pairs are rather "equitably" distributed among the rings.

Examining other modulus values is the exercise itself; examining the modulus values from 2 to 20 (above, we did 10 and 11) with the use of pencil and paper is easy/comfortable enough. One may ponder what ring lengths are to be expected or even allowed and notice various patterns or principles, which are here left unstated.

Fibonacci ratio limit

Let Fibonacci ratio limit denote the limit of f_n+1 / f_n for x going to infinity, where f_n is the n-th number in the Fibonacci sequence. We can practice long division with pencil and paper by calculating approximations of that limit.

We start by producing some initial Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. Then we calculate the ratio as an approximation:

55 / 34 = 1.617647058...
34
210
204
  60
  34
  260
  238
   220
   204
    160
    136
     240
     238
       200
       170
        300
    	272
    	 280

We make an arbitrary stop before we identify the period since for the purpose of the sought limit those digits are probably pseudorandom noise anyway.

By calculating more Fibonacci numbers and then the ratio, we get a better approximation of the limit. Thus, we can take ..., 34, 55, 89 and calculate 89 / 55, which is now left as the exercise.

This exercise is mainly a practice in long division, for ratios that have some meaning other than being arbitrarily generated. The Fibonacci sequence modulo n above seems much more interesting.