Let me see.
Let the effective 4 monthly interest rate be r
Now I need to find the effective 3 yearly interest rate equivalent.
There are 9 four monthly periods in 3 years so the effective 3 yearly rate is (1+r)9 -1
Consider any perpetuity.
Let the amount at the beginning of a period be A and the period interest rate be i then the interest earned in any period is Ai. Now if this is to pay out into perpetuity then the regular period payment must equal the period interest.
That is A+Ai-R=A
Ai=R (I have used A instead of PV)
now I want to look at the 3 year perpetuity.
Initial investment = €32
after 3 years = €32[ (1+r)9 -1+1] = €32[ (1+r)9 ] this is where the perpetuity actually begins. so this is A
A= €32 (1+r)9
i = (1+r)9 -1
$$\begin{array}{rll}
Ai&=&R\\\\
\left[32(1+r)^{9}\right]\left[(1+r)^{9}-1\right]&=&10 \\\\
32(1+r)^{18}-32(1+r)^{9}-10&=&0\\\\
16(1+r)^{18}-16(1+r)^{9}-5&=&0\\\\
\mbox{let } x=(1+r)^{9}&&\\\\
16x^2-16x-5&=&0 \mbox{ where } x > 0\\\\
\mbox{the quadratic formula gives me }x&=&1.25\\\\
(1+r)^{9}&=&1.25\\\\
1+r&=&1.25^{1/9}\\\\
r&=&1.25^{1/9}-1\approx 0.025103648\\\\
\end{array}$$
Okay now lets look at the 4 monthly perpetuity
Ai=R, A=unknown, i=r=etc, R=€1 (is that what you mean by 1 unit?)
$$\begin{array}{rll}
A&=&\frac{R}{i}\\\\
A&=&\frac{1}{1.25^{1/9}-1}\\\\
A&=&39.83484719\\\\
A&=& 39.83
\end{array}$$
PV=€39.83
I think that is right but I haven't done any checking.
Can you do that reinout? I might but it is 1:15am here. You might be a little less tired than I.
LIFE-the condition that distinguishes animals and plants from inorganic matter, including the capacity for growth, 
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