|
| |
|
|
A136322
|
|
a(n) is the ceiling of 2^n * (sqrt(2)-1), i.e. a(n)-1 is the number whose binary representation gives the n first bits of sqrt(2)-1.
|
|
1
|
|
|
|
1, 2, 4, 7, 14, 27, 54, 107, 213, 425, 849, 1697, 3394, 6787, 13573, 27146, 54292, 108584, 217168, 434335, 868669, 1737338, 3474676, 6949351, 13898701, 27797402, 55594804, 111189607, 222379213, 444758426, 889516852, 1779033704, 3558067408
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
LINKS
|
Paul Stoeber, Table of n, a(n) for n = 1..96 [from (pstoeber(AT)uni-potsdam.de), Mar 25 2008]
|
|
|
EXAMPLE
|
Alternative definition: Call an integer m >= 1 plentiful iff the square of m has twice as many bits (floor log_2 plus one) as m. For each n >= 1 there is a unique k >= 0 so that an m with 2^n <= m < 2^(n+1) is plentiful iff m >= 2^n+k. a(n) is this k.
a(8)=107 because 256+107,...,511 are plentiful and 256,...,256+107-1 are not.
|
|
|
PROG
|
{- Haskell (replace leading dots with spaces!) -}
bits :: Integer -> Integer
bits n = if n==1 then 1 else 1+bits (n`div`2)
plentiful :: Integer -> Bool
plentiful n = bits (n*n) == 2*bits n
bisect a b = let c = (a+b)`div`2 in
.. if c==a then b else if plentiful c then bisect a c else bisect c b
a :: Integer -> Integer
a n = bisect (2^n) (2^(n+1)-1) - 2^n
main = print [a n | n<-[1..]]
|
|
|
CROSSREFS
|
Sequence in context: A190822 A107949 A155099 * A160113 A171231 A094057
Adjacent sequences: A136319 A136320 A136321 * A136323 A136324 A136325
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Paul Stoeber (pstoeber(AT)uni-potsdam.de), Mar 25 2008
|
|
|
EXTENSIONS
|
I simplified the definition and mentioned the old definition as an example. - Vincent Nesme, Nov 04 2008
|
|
|
STATUS
|
approved
|
| |
|
|
