[seqfan] Re: Looking for an interpretation

Susanne Wienand susanne.wienand at gmail.com
Wed Dec 23 13:37:09 CET 2015


Hello seqfans,

there is a proof for Ron's recursion.

proof that x[n] fulfills a[n] = 2*a[n-1] - 2*a[n-3] + a[n-4]

1. proof that the above equation is fulfilled, if x[n-4] is a multiple of
y[n-4]

>From the assumption that
x[n-4] = f*y[n-4]

it follows that
x[n-3] = x[n-4] + y[n-4] = f*y[n-4] + y[n-4] = (f+1)*y[n-4]
y[n-3] = x[n-4]/y[n-4] + 1 = f+1

x[n-2] = x[n-3] + y[n-3] = (f+1)*y[n-4] + f+1 = (f+1)*(y[n-4]+1)
y[n-2] = x[n-3]/y[n-3] + 1 = y[n-4] + 1

x[n-1] = x[n-2] + y[n-2] = (f+1)*(y[n-4]+1) + y[n-4]+1 = (f+2)*(y[n-4]+1)
y[n-1] = x[n-2]/y[n-2] + 1 = f+2

x[n] = x[n-1] + y[n-1] = (f+2)*(y[n-4]+1) + f+2 = (f+2)*(y[n-4]+2)

x[n] = f*y[n-4] + 2*y[n-4] + 2*f + 4 = x[n-4] + 2*y[n-4] + 2*f + 4

still to show: 2*x[n-1] - 2*x[n-3] = 2*y[n-4] + 2*f + 4

2*x[n-1]-2*x[n-3] = 2*(f+2)*(y[n-4]+1)-2*(f+1)*y[n-4]

2*x[n-1]-2*x[n-3] = 2*f*y[n-4]+4*y[n-4]+2*f+4-2*f*y[n-4]-2*y[n-4]

2*x[n-1]-2*x[n-3] = 2*y[n-4] + 2*f + 4

so x[n] = 2*x[n-1] - 2*x[n-3] + x[n-4] if x[n-4] is a multiple of y[n-4]

2. proof that x[i+1] is a multiple of y[i+1] if x[i] is a multiple of y[i]

from the assumption that
x[i] = f*y[i]

it follows that
x[i+1] = x[i] + y[i] = f*y[i] + y[i] = (f+1)*y[i]
y[i+1] = x[i]/y[i] + 1 = f+1

so x[i+1] is a multiple of y[i+1] if x[i] is a multiple of y[i]

The x,y pairs start
1,2
3,1
4,4
8,2
10,5

The first pair where x is a multiple of y, is 3,1. For all subsequent
pairs, x is a multiple of y and, starting with the sixth (initially I had
said fifth, that is wrong)
pair, all x-values fulfill the equation x[n] = 2*x[n-1]-2*x[n-3]+x[n-4].

I wish you nice relaxing holidays.
Susanne

>
> Empirical a(n)=2*a(n-1)-2*a(n-3)+a(n-4)
>  rhhardin at mindspring.com rhhardin at att.net (either)
>
>


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