[math-fun] First, second, third, fourth... absolute differences
Hello Math-Fun, Is this of interest? http://www.cetteadressecomportecinquantesignes.com/1234diff.htm Best, É.
It is always possible to get a sequence whose nth absolute difference is the sequence itself. Taking n = 3, for example, we have an array: a1 a2 a3 ... b1 b2 b3 ... c1 c2 c3 ... a1 a2 a3 ... We can take a1, b1, and c1 arbitrary non-negative numbers. Now we must have a1 = |c1-c2|, so either c2 = c1 - a1 or c2 = c1 + a1. The case c1 - a1 may be negative, in which case it can't be used, but c1 + a1 will always work. Likewise b2 = b1 - c1 or b2 = b1 + c1, with the latter always possible; and a2 = a1 - b1 or a2 = a1 + b1. Repeat for each diagonal column. This shows that there are infinitely many such sequences. Now, if we always add in column i-1, we will have b(i-1) < a(i-1), and then a(i) can be either a(i-1) - b(i-1) or a(i-1) + b(i-1). Thus there is always at least one choice in every two columns. It folllows that there are, in fact, uncountably many such sequences. (This argument does not apply to the case n = 1, where the only solutions are a(m) = k*2^m and a(m) = k^2^m for m < M, 0 thereafter.) The presence of uncountably many solutions means that it is very unlikely that there is any formula simpler than the definition. For the case n = 3, one solution is the following, a permutation of the positive Fibonacci numbers: 1, 2, 1, 3, 8, 5, 13, 34, 21, 55, 144, 89, 233, 610, 377, 987, 2584, 1597, 4181, 10946, 6765, 17711, 46368, ... Each triple is 3 Fibonacci numbers later than the previous one. The pattern is F(3i+1), F(3i+3), F(3i+2), F(3(i+1)+1), .... Starting at slightly different offsets, we can get: 0, 1, 1, 2, 5, 3, 8, 21, 13, 34, 89, 55, 144, 377, 233, 610, 1597, 987, 2584, 6765, 4181, 10946, 28657, 17711, ..., which is F(3i), F(3i+2), F(3i+1), F(3(i+1)), ..., or 1, 0, 1, 3, 2, 5, 13, 8, 21, 55, 34, 89, 233, 144, 377, 987, 610, 1597, 4181, 2584, 6765, 17711, 10946, 28657, ..., which is F(3i+1), F(3i), F(3i+2), F(3(i+1)+1), .... Of course, any other Fibonacci-type sequence can be achieved in similar manner. Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
That second case should of course be a(m) = k*2^m for m < M, 0 thereafter. Franklin T. Adams-Watters -----Original Message----- From: franktaw@netscape.net ... (This argument does not apply to the case n = 1, where the only solutions are a(m) = k*2^m and a(m) = k^2^m for m < M, 0 thereafter.) ... ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
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Eric Angelini -
franktaw@netscape.net