Atomic (a Blondie fan, huh? [

])

I’m not sure of the significance of what you’re doing.

An irrational number (i.e. one which cannot be expressed as a ratio of two integers) has an "aperiodic" decimal expansion – i.e. if written as x.xxxxxx... then the digits after the decimal point go on forever and never repeat in any sort of pattern. (So pi, e, sqrt(2), etc, when written as decimals, have an infinitely long sequence of numbers after the decimal point that appear "random" and never settle into any sort of periodic pattern.)

But that doesn’t mean that irrational numbers can’t be generated from an (infinite) sequence with a "discernible pattern". For example:

Pi/4 = 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + 1/13 – 1/15 + …

(Pi^2)/6 = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + …

e = 1/1! + 1/2! + 1/3! + 1/4! + …

I’ve no idea whether the sum you calculated above

(0.824908067279466195032000541687011719)

is irrational or not, but I don’t recognise it as being "special" in any way (it doesn’t appear to be a square or square root or exponential or log of any integer or of any other "special" number such as pi or e).

In any case, it’s worth noting that the sequence “(2^(-n))” is just:

1, 1/2, 1/4, 1/8, 1/16, …

and the sum of all these terms = 2.

By multiplying each term by "A(n)" you’re just choosing to include only half of these terms in your sum. If you chose every odd term (i.e. if A(n) = 1 0 1 0 1 0 ...) then the sum would be 1+1/4+1/16+... = 1.3333..., and if you chose every even term (i.e. if A(n) was 0 1 0 1 0 1 ...) then the sum would be 0.6666...

For the particular sequence A(n) you’re using, you get a sum of 0.8249... which is between 0 and 2. But any other random sequence A(n) (which was half 0s and half 1s) would also give you a sum between 0 and 2. And I can’t see why 0.8249... is any more "special" than any of these other sums.

Am I helping here, or am I missing the point?

Paul.