How automatically double the number of intervals - fortran

Question

I want to double a number till 256, and when the program gets 256 make stop but the program that I created not giving me a double from 1 till 256
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Fortran code
program exercise1
implicit none
real, external :: f
real :: h,a,b,prod,integration,summ,p
integer :: i,j,n
print*, 'Welcome, This program using Composite trapezoid method to calculate the integer'
print*, ' -------- '
prod=1
summ=0
print*,'Number of intervals ',' The integer value'
do i=1,258
n=(prod*i*i) ! Getting number of interval till 256
a=-1 ! The Lower limit of the integration
b=1 ! The Upper limit of the integration
h=(b-a)/n ! Calculate the delta X
p=(h/2.)*(f(a)+f(b)) ! Here as we have the whole equation is (h/2)*[f(a)+f(b)+2*sum(Xi) ! So we calculate the first part (h/2)*[f(a)+f(b) and then calculate the anoter part
summ=summ+h*f(a+i*h) !h/2 *2* sum[f(Xi)
integration = p+summ !Here the sum the both parts
if(n == 256) then !put a limit for the number of interval
Stop
end if
print*,n,' -', integration
enddo
end
real function f(x) !the function of the integer
f=sin(x+1)
end`
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the output
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1 2.72789216
4 2.46665454
9 2.47777867
16 2.49350500
25 2.50419927
36 2.51126313
49 2.51606560
64 2.51944780
81 2.52190781
100 2.52374840
121 2.52515912
144 2.52626276
169 2.52714205
196 2.52785373
225 2.52843738

Answer 1

I think you want powers of: 2**i not squares: i**2. For example:
do i = 0,8
print*, 2**i
enddo
gives
1
2
4
8
16
32
64
128
256

Answer 2

I was originally voting to close as a typo, but your code maybe indeed does something different.
Your codes makes n to be a sequence if integer squares. 1^2, 2^2, 3^2 ... 16^2=256
But you wanted doubling the number, in that case just multiply n by 2.
n = 1
do i = 1, 10
do what you need
print *, n
n = n * 2
end do