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March 29th, 2009, 07:25 PM   #1
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Values of a*e+b*pi where a and b are natural numbers

Say one has a ascending list of all values of a*e + b*pi where a and b are natural numbers (0, e, pi, 2e, e+pi, 2pi, 3e, 2e+pi...). Then one looked at the difference of term n and term n-1. These differences have very few values: e, 2e-pi, 3e-2pi, 4e-3pi, 5e-4pi, 6e-5pi, pi-e, 7pi-8e, 7e-6pi, 22e-19pi, 13pi-15e, 37e-32pi, 89e-77pi, 45pi-52e, 103pi-119e, and 141e-122pi account for the first 28,000+ differences between values. Is there any way to predict these differences, or successive ascending values?
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March 29th, 2009, 07:38 PM   #2
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Re: Values of a*e+b*pi where a and b are natural numbers

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Originally Posted by 8Pickle
Is there any way to predict these differences, or successive ascending values?
I don't know of one.

This actually looks like an interesting and 'deep' question. Looking at the continued fraction expansion of their ratio may give some insight, I'm not sure.
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June 9th, 2009, 01:53 PM   #3
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Re: Values of a*e+b*pi where a and b are natural numbers

Quote:
Originally Posted by 8Pickle
Say one has a ascending list of all values of a*e + b*pi where a and b are natural numbers (0, e, pi, 2e, e+pi, 2pi, 3e, 2e+pi...). Then one looked at the difference of term n and term n-1. These differences have very few values: e, 2e-pi, 3e-2pi, 4e-3pi, 5e-4pi, 6e-5pi, pi-e, 7pi-8e, 7e-6pi, 22e-19pi, 13pi-15e, 37e-32pi, 89e-77pi, 45pi-52e, 103pi-119e, and 141e-122pi account for the first 28,000+ differences between values. Is there any way to predict these differences, or successive ascending values?
Do you count negative numbers also as natural numbers? From the above it seems that yes (eg. 22e - 19pi, so a=22 and b=-19). My opinion is that given arbitrary interval <m,n> m<n we can find a, b, so that a.pi + b.e falls into <m,n>. And so the distance between every consecutive numbers of the form a.pi + b.e (when ordered) is simply less than you will say.
Or maybe I don't understand ordering and generation of the set of these numbers.
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June 9th, 2009, 08:11 PM   #4
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Re: Values of a*e+b*pi where a and b are natural numbers

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Originally Posted by honzik
Quote:
Originally Posted by 8Pickle
Say one has a ascending list of all values of a*e + b*pi where a and b are natural numbers (0, e, pi, 2e, e+pi, 2pi, 3e, 2e+pi...). Then one looked at the difference of term n and term n-1. These differences have very few values: e, 2e-pi, 3e-2pi, 4e-3pi, 5e-4pi, 6e-5pi, pi-e, 7pi-8e, 7e-6pi, 22e-19pi, 13pi-15e, 37e-32pi, 89e-77pi, 45pi-52e, 103pi-119e, and 141e-122pi account for the first 28,000+ differences between values. Is there any way to predict these differences, or successive ascending values?
Do you count negative numbers also as natural numbers? From the above it seems that yes (eg. 22e - 19pi, so a=22 and b=-19). My opinion is that given arbitrary interval <m,n> m<n we can find a, b, so that a.pi + b.e falls into <m,n>. And so the distance between every consecutive numbers of the form a.pi + b.e (when ordered) is simply less than you will say.
Or maybe I don't understand ordering and generation of the set of these numbers.
No, negatives aren't natural numbers. 8P is considering the list
0, e, pi, 2e, e+pi, 2pi, 3e, 2e+pi,
taking its first difference,
e-0, pi - e, 2e - pi, (e + pi) - 2e = pi - e, 2pi - (e + pi) = pi - e, 3e - 2pi, ...,
and removing duplicates:
e, 2e-pi, 3e-2pi, 4e-3pi, 5e-4pi, 6e-5pi, pi-e, 7pi-8e, 7e-6pi, 22e-19pi, 13pi-15e, 37e-32pi, 89e-77pi, 45pi-52e, 103pi-119e, 141e-122pi, ....
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June 10th, 2009, 12:34 PM   #5
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Re: Values of a*e+b*pi where a and b are natural numbers

At the first sight - to predict the sequnce is I think almost as hard as to predict the next digit in pi expansion.
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June 10th, 2009, 06:31 PM   #6
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Re: Values of a*e+b*pi where a and b are natural numbers

Quote:
Originally Posted by honzik
At the first sight - to predict the sequnce is I think almost as hard as to predict the next digit in pi expansion.
That easy?
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