Integer Sequences Related To π For any given integers s0 and s1 we can construct the infinite sequence sn, n = 0, 1, 2, ...  generated by the recurrence For example, with the initial values s0 = 0, s1 = 1 we produce the sequence On the other hand, if we take the initial values s0 = 1, s1 = 0 we produce the sequence For convenience, let sk(m,n) denote sk based on the initial values s0 = m and s1 = n.  The two sequences listed above can serve as the "basis" for all sequences of this form, according to the relation The ratios of corresponding terms in these sequences approach some interesting values.  For example, we can show that for any integers m,n (positive or negative) Simply dividing two general expressions of this form, we get the more general relation for any w,x,y,z This applies to any real values of w,x,y,z, not just integers, so it gives the limit of ratios of terms of any sequence sk(x,y) to terms of the sequence sk(w,z).  To illustrate, note that the sequence sk(1,1) has the values and the sequence sk(-2,1) has the values The ratio of the 9th terms of these sequences is If we add 2 to this ratio and multiply by 4/3 we get 3.141593...  Another interesting observation concerns the growth rate of the individual sequences.  It appears that the difference in consecutive ratios of consecutive values approaches 1 + √2.  In other words, for any given initial values, we have It would be interesting to determine an explicit solution of the recurrence (1), which is a simple homogeneous linear 2nd-order recurrence, but with non-constant coefficients. Since (3) shows that all the sequences are easily related, it's enough to consider just the ratio of two sequences.  Define the integer sequences Dk and Nk such that both satisfy the recurrence relation (1) with the initial values N0 = 0, N1 = 1 and D0 = 1, D1 = 1. Thus the values of N and D are It can be shown that the ratio Nk/Dk equals the kth convergent of the continued fraction for the inverse tangent of 1: Incidentally, this continued fraction can also be inferred from the fact that The continued fraction can be converted into the following infinite series where the kth term ck is just the difference between the (k–1)th and the (k+1)th convergent of the continued fraction.  In other words As a result, the partial sums of the series are just the sums of two consecutive convergents of the continued fraction. Clearly the sequence of numerators is just the odd numbers. Is there a simple way of characterizing the denominators? In that regard, notice that if we let g denote the greatest common divisor of it follows that and where dk is the denominator of the kth term in the above series for π/2.  Now it appears that so g = [(k+1)!]2, which implies that Equation (5) is a kind of convolution of the terms of the two sequences, which may be a more natural definition of their relation than the original linear 2nd order recurrence. Actually (5) is just one of a set of formulas relating the "cross-products" of these two series.  In general we have where Q(k) is a polynomial in k of degree j–1.  The first few such polynomials are listed below The constant coefficients are just the values of the N sequence, whereas the leading coefficients 1, 2, 5, 12, 29, 70, ... are the "Pell numbers" that satisfy the recurrence pn = 2pn–1 + pn–2. Also, notice that the sums of the coefficients for the jth polynomial are 1, 5, 44, 476, 6336, 99504, ...  which are the values of the other "basis sequence" mentioned previously. In other words, these values satisfy the same recurrence as do the N and D sequences, but with the initial values 1,0,... Incidentally, using the original sequence notation, we have the identity, applicable to any linear second-order recurrence: This is discussed in the note on anti-symmetric arrays for linear recurrences. Return to MathPages Main Menu