cp's OEIS Frontend

The limit of a(n+1)/a(n) is equal to 2+sqrt(7) as n approaches infinity.

This is the Lucas sequence U(8,-12).

For any three-term recurrence S(n) = S(n-1)*x1 + S(n-2)*x2*1, with S(-1) = 0 and S(0) = 1, with n-independent coefficients (like here x1=8 and x2=12) one can use the standard Morse code with a dot of length 1 standing for x1 and a dash of length 2 standing for x2. The Morse code polynomial S(x1,x2;n) is then obtained by summing over all codes of length n. E.g., S(x1,x2;3) = x1^3 + 2*x1*x2 from dot dot dot, dot dash and dash dot. Here x1=8 and x2=12 (labeled dots and dashes). For example, S(3) = 8*(8^2 + 2*12) = 704 = b(3) = a(4), because in a the offset differs from the one for S. See the Graham et al. book, on Morse code polynomials (Euler's continuants), p 302. This comment was motivated by an earlier one from the author of this sequence. - Wolfdieter Lang, Mar 27 2014

a(n-1) (for n>=1) is the number of compositions of n into 8 kinds of parts 1 and 12 kinds of parts 2. - Joerg Arndt, Mar 26 2014

Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.

The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2=(k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1|< 1. Thus, a(n+1)/a(n) converges to r1 as n approaches infinity.

Note that limit a(n+1)/a(n) = 15 with k=12 and t=45. Thus, we use this as the name of the sequence.

If n > 25, then |a(n+1)/a(n) - 15| < 10^(-16).

a(n+2) is the number of strings of length n containing the 12-ary digits 0,1,...,9,A,B or any of the 45 two-consecutive digits C0,C1,...,C9,CA,...,CZ,Ca,...,Ci where A corresponds to 10, B to 11, ..., Z to 35, a to 36, ..., i to 44.

Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.

The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2 = (k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1| < 1. Thus, a(n+1)/a(n) converges to r1 as n approaches infinity.

Note that limit a(n+1)/a(n) = 15 with k=13 and t=30.

If n > 20, then |a(n+1)/a(n) - 15| < 10^(-16).

Let b(n) be the number of strings of length n containing the 13-ary digits: 0,...,9,A,B,C or the 30 two-consecutive digits D0,D1,...,D9,DA,...,DT where A corresponds to 10, ..., T corresponds to 29. Then b(0)=1=a(2) and b(1)=13=a(3). The strings q_1q_2...q_n of length n can be partitioned into 2 groups A and B where A contains the strings where q_1=0,1,...,9,A,B,C and B contains the strings where q_1=D. Thus, |A|=13*b(n-1) and |B|=30*b(n-2). Hence, b(n) = 13*b(n-1) + 30*b(n-2) for n>1. Since b(0)=a(2) and b(1)=a(3), we can show that b(n) = a(n+2).

Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.

The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2 =(k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1|< 1. Thus, the ratio a(n+1)/a(n) converges to r as n approaches infinity.

Note that limit a(n+1)/a(n) = 15 as n approaches infinity with k=14 and t=15.

If n > 15 then | a(n+1)/a(n) - 15 | < 10^(-16).

The number of walks of length n between any two distinct vertices of the complete graph K_16. - Peter Bala, May 30 2024