1,2,4,3,6,5,10,7,14,8,16,9,18,11,22,12,24...
This sequence has simple rules with a major hint in the title.
Part 1) Determine the rule to extend the sequence.
Part 2) Find, with no aids beyond a simple calculator, the 1000th term of this sequence.
(In reply to
re: An uncertain shorthand for part II by Charlie)
Thank you Charlie for the check
I was thinking doing it on my own, so I try to give form to the relation between the number of term (M) and the last single number (not doubled) of the sequence (n).
It should be like this:
M = n + d [1] being:
- d=[number of odds + number of multiples of strict even powers of 2] in the interval (n/2,n-1).
- "strict even powers of 2" (like 4, 16, 64,...) = it is only multiple of that even power of 2, (f. ex: is a multiple of 4 but not of 8, 16, etc.).
This could be reduce with good estimation to:
M= 1 + n + n/4 + n/16 + n/64 + ...
(this work perfect for the numbers 3*2^(2b), as 3, 12, 48, ... 768
To check that for M=1000, n=751 with the above formula [1]:
For n=751 interval is (376,750).
The number of odds in the interval is (750-376)/2=187
To calculate the number of multiples of strict powers of 2^(2b) in the interval (376-750) a good idea is operate in binary
750=1011101110 [strict multiples of 4= (1011101)+1=94; sm of 16=(10111)=23; sm of 64=(101)+1=6; sm of 256=(1)=1]
376=101111000 (sm of 4=(101111)=47; sm of 16=(1011)+1=12; sm of 64=(10)+1=3]
Then d=187 + (94-47) + (23-12) + (6-3) + 1 = 249
M= n + d = 751+249=1000
Edited on July 27, 2016, 2:04 pm
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Posted by armando
on 2016-07-26 16:26:50 |
re(2): An uncertain shorthand for part II
