n : 0 1 2 3 4 5 ... n*n*n: 0 1 8 27 64 125 ...which tabulates the function starting at zero. The table of sequence values can be formed in more than one way using arithmetic operations. We will be generalizing trigonometric functions so we construct trigonometric sequence tables to illustrate our construction methods and also to test potential constructions of generalized sequences. One good example of a hyperbolic sequence is the following table

n : 0 1 2 3 4 5 6 7 8 9 10 ... s(n): 0 1 3 8 21 55 144 377 987 2584 6765 ...constructed starting with initial terms s(0) = 0 and s(1) = 1 . Here, each term is a constant c times the previous term minus the term before that. The recursion equation is s(n) = c*s(n-1) - s(n-2) for all n. For example, n = 2 , gives s(2) = c*s(1) - s(0) = c, hence c must = 3. Another example, n = 5 , gives s(5) = 3*21 - 8 = 55. Note the identity s(n) = F(2*n) where F() denotes the Fibonacci sequence. We will call this sequence "Fib2" for future reference. Any number could have been used as the constant c to construct a sequence. If the constant is greater than 2 , the sequence is hyperbolic, but if less than 2 in absolute value then the sequence is circular. The linear case is if the constant is 2 and hence s(n) = n by induction.

The trigonometric sequences can be constructed in other ways. For example, we have the equation s(4)*s(1) = s(3)*s(2) - s(2)*s(1) and for Fib2 this is 8*3 - 3*1 = 21 . We solve this equation to get s(4) = 21 . We have also the equation s(5)*s(1) = s(3)*s(3) - s(2)*s(2) and for Fib2 this is 8*8 - 3*3 = 55 which we solve to get s(5) = 55 . These equations and related equations depend on triples of pairs of positive integers which form a pair of patterns

[4,1] [3,2] [2,1] and [3,1] [2,2] [1,1] [6,1] [4,3] [3,2] and [5,1] [3,3] [2,2] ... ... ... ... ... ... [2n+2,1] [n+2,n+1] [n+1,n] and [2n+1,1] [n+1,n+1] [n,n]In other words, the following equations

s(2*n+2)*s(1) = s(n+2)*s(n+1) - s(n+1)*s(n) , and s(2*n+1)*s(1) = s(n+1)*s(n+1) - s(n)*s(n) ,hold for all positive n . These equations can be solved for s(n) where n is greater than 2 given any values for s(2) and s(1) but s(1) must be non-zero. When s(1) = 1 and s(2) = x , these sequences are related to Chebyshev polynomials of the second kind as follows

s(n) = U(n-1, x/2).Now we seek a simple generalization of trigonometric sequences.

Suppose we now require the number to be a sum of four positive squares in more than one way. In this case, the smallest such number is 28 and we have the following three sums of squares

[5,1,1,1] [4,2,2,2] [3,3,3,1] 28 = 25+1+1+1 = 16+4+4+4 = 9+9+9+1 .Above each square appears its square root. The three square root quadruples have many arithmetic properties. For example, if we take the product of the four numbers of each quadruple we get

[5,1,1,1] [4,2,2,2] [3,3,3,1] 5 = 5*1*1*1 32 = 4*2*2*2 27 = 3*3*3*1and note that 5 = 32 - 27 . Now we use each number of each quadruple to index into the Fib2 sequence mentioned earlier. Recall that s(2) = 3 , s(3) = 8 , s(4) = 21 , s(5) = 55 , yielding

[5,1,1,1] [4,2,2,2] [3,3,3,1] 55 = 55*1*1*1 567 = 21*3*3*3 512 = 8*8*8*1as the calculated product of corresponding terms and 55 = 567 - 512 . Thus, these quadruples have nice unexpected arithmetic properties when indexed into trigonometric sequences.

Let us now explore these properties further. First, we find that the next smallest number which is the sum of four positive squares also in three ways is 42. We have the following sums

[6,2,1,1] [5,3,2,2] [4,4,3,1] 42 = 36+4+1+1 = 25+9+4+4 = 16+16+9+1and again the square root quadruples appear above. Note again that we get the equality of difference of products as given by

[6,2,1,1] [5,3,2,2] [4,4,3,1] 12 = 6*2*1*1 60 = 5*3*2*2 48 = 4*4*3*1with 12 = 60 - 48 . Also, taking the product of corresponding indexed terms of the Fib2 sequence gives the expected result

[6,2,1,1] [5,3,2,2] [4,4,3,1] 432 = 144*3*1*1 3960 = 55*8*3*3 3528 = 21*21*8*1which is 432 = 3960 - 3528 . This is an encouraging observation.

Let us briefly summarize our observations. We have found triples of quadruples of positive integers with the following properties :

- (Lagrange) The sum of squares of one quadruple is equal to the sum of squares of the each of the other two quadruples.
- The product of the first quadruple is the product of the second quadruple minus the product of the third quadruple.
- (Weierstrass) Same as property 2 except we use the product of the corresponding indexed terms of trigonometric sequences.

A reasonable goal is now to find all Pfaffian triples and all the sequences of numbers satisfying the Weierstrass property. The next sum of four positive squares in more than two ways is 52, where

[7,1,1,1] [5,5,1,1] [5,3,3,3] [4,4,4,2] 52 = 49+1+1+1 = 25+25+1+1 = 25+9+9+9 = 16+16+16+4We seek to select a Pfaffian triple of quadruples. In this case it is

[7,1,1,1] [5,3,3,3] [4,4,4,2] 7 = 7*1*1*1 135 = 5*3*3*3 128 = 4*4*4*2with 7 = 135 - 128 . Can we generalize this? There are a few clues here. For example, the first and third Pfaffian triples compared

[5,1,1,1] [4,2,2,2] [3,3,3,1] [7,1,1,1] [5,3,3,3] [4,4,4,2]might suggest there is a simple arithmetic progression here. Does

[9,1,1,1] [6,4,4,4] [5,5,5,3]give us another? Yes, it does, and the general pattern is given by

[2*n+1,1,1,1] [n+2,n,n,n] [n+1,n+1,n+1,n-1]This gives us one infinite set of triples. Are there any others? Yes. For example, starting from the second triple we get a pattern

[6,2,1,1] [5,3,2,2] [4,4,3,1] [8,2,1,1] [6,4,3,3] [5,5,4,2] [10,2,1,1] [7,5,4,4] [6,6,5,3] ... ... ... [2*n+2,2,1,1] [n+3,n+1,n,n] [n+2,n+2,n+1,n-1]which is similar to the first pattern. From just these two patterns, we can construct numeric sequences in a very simple way.

a(5)*a(1)*a(1)*a(1) = a(4)*a(2)*a(2)*a(2) - a(3)*a(3)*a(3)*a(1)and if we know the values of a(1),a(2),a(3),a(4), then the equation is linear in a(5) and we solve for it (assuming non-zero a(1)). Now that we know a(5), the next equation is linear in a(6)

a(6)*a(2)*a(1)*a(1) = a(5)*a(3)*a(2)*a(2) - a(4)*a(4)*a(3)*a(1)and we solve for it (assuming non-zero a(1) and a(2)). So, given all the equations from the two patterns we can solve for all of the terms of the sequence given a(1),a(2),a(3),a(4). This gives a construction of an infinite sequence we will call an elliptic sequence.

What about other patterns? For example, startng from the first two Pfaffian triples leads to this pattern

[5,1,1,1] [4,2,2,2] [3,3,3,1] [6,2,1,1] [5,3,2,2] [4,4,3,1] [7,3,1,1] [6,4,2,2] [5,5,3,1] ... ... ... [n,n-4,1,1] [n-1,n-3,2,2] [n-2,n-2,3,1]but does the sequence fit into the other patterns? Yes, but the proof is not obvious and requires work. What do non-trivial examples of elliptic sequences look like? A simple example is

n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ... a(n): 0 1 1 1 -1 -2 -3 -1 7 11 20 -19 -87 -191 -197 ...which satisfies a(n)*a(n-4) = a(n-1)*a(n-3) - a(n-2)*a(n-2) and many other equations. A closely similar example is

n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ... a(n): 0 1 1 -1 1 2 -1 -3 -5 7 -4 -23 29 59 129 ...which satisfies a(n)*a(n-4) = a(n-1)*a(n-3) + a(n-2)*a(n-2) and other equations. In fact, the odd indexed terms alternate in sign, and with the sign removed the resulting positive sequence

n: 0 1 2 3 4 5 6 7 8 9 10 11 12 ... a(n): 1 1 1 1 2 3 7 23 59 314 1529 8209 83313 ...satisfies the same recursion equation. It is called the Somos-4 sequence -- the simplest member of a large family of sequences.

All this and more is part of "the Elliptic Realm". The elliptic sequences are simply related to Jacobi theta functions and Weierstrass sigma functions evaluated at muliples of a number. Some special cases were called "elliptic divisibility sequences" by Morgan Ward in 1948. I have demonstrated that they can easily be discovered and constructed using only simple tools, and without knowledge of elliptic functions. One starting point is Pfaffian triples and the Weierstrass property.

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Michael Somos <michael.somos@gmail.com>