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Differences between consecutive numbers of quadruples Take 4 integers in a row (like 2 6 4 13). Take the (magnitude) of their differences of consecutive integers (where the last number is subtracted from the first): 4 2 9 11. Continue to do that, and observe – (2 7 2 7; 5 5 5 5) they eventually become all equal. Problem: Can for every number of steps it should take until the differences are all equal a suitable quadruple of four numbers be found? If yes, prove it, and, if possible, exhibit the quadruple for the given number of steps . If no, for which number of steps can we find no such quadruple? 1st Example: A sequence of 4 Steps: 1234 1113 0022 0202 2222 2nd Example: 4 Steps 1 3 7 11 2 4 4 10 2068 2626 4444 3rd Example: 6 Steps 5 3 0 11 2 3 11 6 1854 7313 4224 2020 2222 1 → Maximum of the four numbers becomes smaller or stays equal at each step. Let’s call the number of steps (4, 4, and 6 above) of these sequences their length. The question is: Given any sequence, can we always find another longer sequence? Given a quadruple a = (a1, a2 , a3, a4 ), let us focus on finding an increasing quadruple b = (b1, b2, b3 , b4 ) meaning b1 ≤ b2 ≤ b3 ≤ b4, which has a as its quadruple of differences: a1 = b2 − b1 , a2 = b3 − b2, a3 = b4 − b3, and a4 = b4 − b1 . If we add the first of these three equations, we get: a1 + a2 + a3 = b4 − b1 = a4 . (1) Let’s see if this is fulfilled, for let’s say: a = (1, 2, 3, 4). Hmm, not fulfilled: 4 6= 1 + 2 + 3. However, if we reduced every entry by one:a′ = a − (1, 1, 1, 1), then a′ = (0, 1, 2, 3) and indeed, a′4 = 3 = 0 + 1 + 2. Interesting: the sequence of differences of a′ will be the same as that of a. It therefore has the same length. So, let’s try to find the b for the new a′ : Let’s pick any number for b1, let’s say: 0. Then b2 has to be a′1 , since the difference between b2 and b1 = 0 has to be a′1 . So, also, b2 = 0. After that: pick b3 such that its difference with b2 is a′2 = 1. So: b3 = a′2 + b2 = a′2 + a′1 = 1 + 0 = 1. finally, b4 must be chosen such that its difference with b3 = 1 is a′3 = 2, so b4 = 3. 2 Let’s check if it is consistent with b4 − b1 = a′4 : Yes, 3 − 0 = 3, Check!. This last thing works out because we made sure that (1) is fulfilled. So, the new sequence starts with b = (0, 0, 1, 3) and has length 5. Trying the same trick again, leads us to adding to each of the four numbers 1, to end up at the quadrupel c = (1, 1, 2, 4). Nicely, this quadruple fulfills c4 = c1 + c2 c3 . So, (taking a discrete ’antiderivative’): d1 = 0, d2 = c1 = 1, d3 = c1 + c2 = 2, and d4 = c4 = 4 gives us a quadruple of length l(d) = 6, where d = (0, 1, 2, 4). Finally, however, the end of easy-street seems to have been reached: Observe that, while d4 = 4 > 0 + 1 + 2 = d1 + d2 + d3 , we are not able to fi this by a shift: Adding one to every coordinate of d gives us d′ = (1, 2, 3, 5) which gives that now d′4 = 5 < 6 = 1 + 2 + 3 = d′1 + d′2 + d′3 . We shifted past our goal, but we can’t shift by less than 1. We anticipate, that shifting the integers alone is not enough, but that multiplication may help. If multiplying each element of b with z ∈ Z, a quadruple emerges that has the corresponding multiple differences of coordinates. Repeating the process of taking differences gets us a sequence of equal entries which are multiples of the initial terminal quadruple: (10 10 20 40) Therefore this new sequence has the same length! So, while we couldn’t continue to extend the sequence backwards by shifting the entries of the last quadruple, we could pass to another sequence (by multiplication) where the shift operation works again: So, let us multiply d by 2, to obtain e: e = (0, 2, 4, 8). 3 If we add 1 to the entries of this qudruple, we get e′ with: e′ = (1, 3, 5, 9). Indeed: e′4 = 9 = 1 + 3 + 5 = e′1 + e′2 + e′3 ! So now we can continue to form the quadruple of which this will be the quadruple of differences: f1 = 0, f2 = e′1 = 1, f3 = e′1 + e′2 = 4, f4 = e′4 = 9, which is the last quadruple of a sequence of length 6. Note that through consecutive multiplications and shifts the order of the first coordinate being the smallest and the fourth being the largest isn’t ever changed. Therefore, the condition of the last entry having to be the sum of the first three will always be sufficient for finding a sequence with length increased by one. Here is the link to a program in R and some of its results, computing sequences of arbitrary length; http://web.cs.du.edu/~sobieczk/mathclub/programs/ 4