Tuba Fantasy (music generated from Padovan numbers)

A musical graph related to A000931 (the Padovan numbers) at the OEIS: a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0. For each integer m, I use an adaptation of an algorithm I described in 2008/9 Fibonacci Quarterly to find a solution to the equation m*a(n) = a(j)+...+a(k) such that j..k satisfy the equation m=b^j+...+b^k, where b=(1/2+sqrt(23/108))^(1/3) + (1/2-sqrt(23/108))^(1/3) (the plastic number). The numbers were a bit close together, so I used the the Padovan sequence to spread them out (the number n is mapped to the nth (offset) Padovan number.

The algorithm I used is also described in http://www.youtube.com/watch?v=eOxuQQlYG94 and the adaptation depends on the identity a(n+m) == a(n+3)*a(m) + a(n+4)*a(m-1) + a(n+2)*a(m-2). This is a variant of the third identity in "Proofs that Really Count" by Benjamin and Quinn and it can be proved combinatorially as in this book.

More info is available, send me an email if interested.

Category:

Science & Technology

Tags:

• math