[math-fun] curious dyadic rationals t/b, b=2^f, t=sum of few +-2^e, ContinuedFraction(t/b) has 1's and 2's only
As an example, consider b = 2^63 = 9223372036854775808(decimal) = 8000000000000000(hexadecimal) = 1000000000000000000000000000000000000000000000000000000000000000(binary) t = 5836946587753906177(dec) = 5100FFFF00000001(hex) = 101000100000000111111111111111100000000000000000000000000000001(binary) ContinuedFraction(t/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2] I have found that there are many such examples, and they are easy to find by computer search because t is a sum of few signed powers of 2. In the present example, t is a sum of 6 of them, and since binomial(64,6) * 2^5 = 2399179776 = 2.4 * 10^9 it would be feasible to consider all possibilities. Another example has t' = 101000100000000111111111111111011111111111111111111111111111111(binary) which yields a very "nearby" continued fraction to the first example: ContinuedFraction(t'/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Hum, interesting but what is t ? what type of number is it ? what is the rule for t ? Simon Plouffe Le 29/09/2013 07:43, Warren D Smith a écrit :
As an example, consider
b = 2^63 = 9223372036854775808(decimal) = 8000000000000000(hexadecimal) = 1000000000000000000000000000000000000000000000000000000000000000(binary)
t = 5836946587753906177(dec) = 5100FFFF00000001(hex) = 101000100000000111111111111111100000000000000000000000000000001(binary)
ContinuedFraction(t/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]
I have found that there are many such examples, and they are easy to find by computer search because t is a sum of few signed powers of 2. In the present example, t is a sum of 6 of them, and since binomial(64,6) * 2^5 = 2399179776 = 2.4 * 10^9 it would be feasible to consider all possibilities.
Another example has t' = 101000100000000111111111111111011111111111111111111111111111111(binary) which yields a very "nearby" continued fraction to the first example: ContinuedFraction(t'/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]
Yes, a lot of these can be generated (as I mentioned before) using the "folding theorem" of continued fractions. See, for example, my paper with van der Poorten in JNT with that title. On 9/29/13 1:43 AM, Warren D Smith wrote:
As an example, consider
b = 2^63 = 9223372036854775808(decimal) = 8000000000000000(hexadecimal) = 1000000000000000000000000000000000000000000000000000000000000000(binary)
t = 5836946587753906177(dec) = 5100FFFF00000001(hex) = 101000100000000111111111111111100000000000000000000000000000001(binary)
ContinuedFraction(t/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]
I have found that there are many such examples, and they are easy to find by computer search because t is a sum of few signed powers of 2. In the present example, t is a sum of 6 of them, and since binomial(64,6) * 2^5 = 2399179776 = 2.4 * 10^9 it would be feasible to consider all possibilities.
Another example has t' = 101000100000000111111111111111011111111111111111111111111111111(binary) which yields a very "nearby" continued fraction to the first example: ContinuedFraction(t'/b) = [0; 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2]
participants (3)
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Jeffrey Shallit -
Simon Plouffe -
Warren D Smith