P[n+ 1]= P[n]-P[n-3]/ 16(n≥4)

[] is a subscript.

Multiply both sides by 2 (n+ 1)

p[n+ 1]*2^(n+ 1)=p[n]*2^(n+ 1)-p[n-3]*2^(n-3)

Let p [n] * 2 n = a [n], so a [1] = 2; a[2]= 4； a[3]= 7； A[4]= 13

Then the above formula can be changed to

A[n+ 1]=2A[n]-A[n-3]

Which means both.

A[n+ 1]-A[n]=A[n]-A[n-3]

Iterative availability

A[n+ 1]-A[4]= A[n]+A[n- 1]+A[n-2]-A[ 1]-A[2]-A[3]

Replace known data:

A[n+ 1]= A[n]+A[n- 1]+A[n-2]

This is a series similar to Fibonacci, which is not easy to understand.

After finding the general term of A[n], reduce it with p [n] = a [n]/2 n]/2 n.

The following is not a solution, but a discussion. I write this in the hope of attracting more attention.

Discussion, the characteristic equation corresponding to A[n] should be

x^3+x^2+x= 1

Characteristic solution x = 0.5436890 13. ...

(equivalent to Fibonacci sequence y 2+y =1,y = 0.6 18 ...)

So it can be said that when n tends to infinity, a [n- 1] = a [n] * 0.55438+03. ...

Therefore, p [n]/p [n-1] =1(2 * 0.54689013 ...) = 0.91964378. ...

So P[n] tends to zero.