In a comment on Rachel Thomas's recent article "The Fibonacci sequence: a brief introduction" I mention a companion sequence to Fibonacci's rabbits which I now know to be called "Narayana's cows" (OEIS A000930). This is defined by OEIS as
and goes 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41 for the first twelve terms.
This sequence differs from the more famous Fibonacci by starting with three 1's instead of two, which means there’s a delay of 3 years (or months or whatever you want the unit period to be) instead of 2 before these animals produce offspring. The delay, or D number for Narayana’s cows is 3,as given by the final term in that definition: a(n-3). For the Fibonacci sequence however the defining equation is a(n) = a(n-1) + a(n-2), so the Fibonacci D number is 2.
In a sequence where D is 1 then we get a(n) = a(n-1) + a(n-1), leading to a(n) = 2(a(n-1)). This simply means the sequence starts with a single 1 and doubles each time. A lot faster.
What if D were 0? Would that make the reproduction rate inconceivably, perhaps ifinitely, fast? Rabbits or cows giving birth to more rabbits or cows, which as soon as they’re born immediately and simultaneously produce yet more rabbits or cows and so on?
Yet the equation for this sequence, a(n) = a(n-1) + a(n-0), suggests the opposite. It solves as a(n-1) = 0, which if true for any a(n) in the sequence, means the sequence consists solely of 1’s. This in turn seems to characterise an infinite D such that no offspring is produced at all, bearing in mind that the first reproduction can only occur with summing of the first and last of the 1’s, as becomes apparent in the Narayana case.
Maybe both answers amount to the same thing, or maybe to the two different kinds of infinity – actual and potential. All suggestions gratefully received.
In a comment on Rachel Thomas's recent article "The Fibonacci sequence: a brief introduction" I mention a companion sequence to Fibonacci's rabbits which I now know to be called "Narayana's cows" (OEIS A000930). This is defined by OEIS as
a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3)
and goes 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41 for the first twelve terms.
This sequence differs from the more famous Fibonacci by starting with three 1's instead of two, which means there’s a delay of 3 years (or months or whatever you want the unit period to be) instead of 2 before these animals produce offspring. The delay, or D number for Narayana’s cows is 3,as given by the final term in that definition: a(n-3). For the Fibonacci sequence however the defining equation is a(n) = a(n-1) + a(n-2), so the Fibonacci D number is 2.
In a sequence where D is 1 then we get a(n) = a(n-1) + a(n-1), leading to a(n) = 2(a(n-1)). This simply means the sequence starts with a single 1 and doubles each time. A lot faster.
What if D were 0? Would that make the reproduction rate inconceivably, perhaps ifinitely, fast? Rabbits or cows giving birth to more rabbits or cows, which as soon as they’re born immediately and simultaneously produce yet more rabbits or cows and so on?
Yet the equation for this sequence, a(n) = a(n-1) + a(n-0), suggests the opposite. It solves as a(n-1) = 0, which if true for any a(n) in the sequence, means the sequence consists solely of 1’s. This in turn seems to characterise an infinite D such that no offspring is produced at all, bearing in mind that the first reproduction can only occur with summing of the first and last of the 1’s, as becomes apparent in the Narayana case.
Maybe both answers amount to the same thing, or maybe to the two different kinds of infinity – actual and potential. All suggestions gratefully received.
Chris G