|
|
A060802
|
|
To weigh from 1 to n, make the heaviest weight as small as possible, under the condition of using fewest pieces of different, single weights; a(n) = weight of the heaviest weight.
|
|
1
|
|
|
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7, 8, 6, 6, 6, 7, 7, 8, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 10, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 18, 18, 18, 19, 19, 20, 20, 21
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Starting at n = 2^x (x > 2) you get: 3 entries of 2^floor(log_2(x)-2)+2, then 2 entries of each subsequent integer until you reach the halfway point between 2^x and 2^(x+1), then 1 entry of each subsequent integer until you reach 2^(x+1)-1. Proved (see link). - David Consiglio, Jr., Jan 08 2015
|
|
LINKS
|
|
|
FORMULA
|
After the 8th term:
If 2^x <= n <= (2^x)+2 then a(n) = 2 ^ floor(base2log(x)-2)+2 (see A052548)
If (2^x)+2 < n and n+1 < (2^x + 2^x+1)/2 then a(n) and a(n+1) = a(n-1)+1
|
|
EXAMPLE
|
a(20)=7 because every number from 1 to 20 can be obtained from {1,2,4,6,7}.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|