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 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 first n 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 COMMENTS Previous 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. This sequence has a geometric interpretation. Let 2^(n+1) be the length of the fixed base of all integer isosceles triangles whose equal sides are shorter than its base. Then a(n) is the number of these isosceles triangles whose height above the base is < 2^n. It includes the degenerate isosceles triangle of height 0.  - Frank M Jackson, Sep 16 2017 LINKS Paul Stoeber, Table of n, a(n) for n = 1..95 [corrected by Andrew Howroyd, Feb 28 2018] EXAMPLE a(8)=107 because 256 + 107, ..., 511 are plentiful and 256, ..., 256 + 107 - 1 are not. MATHEMATICA hcount[c_] := Module[{n=0, m}, Do[If[Sqrt[m(c+m)] 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,base AUTHOR Paul Stoeber (pstoeber(AT)uni-potsdam.de), Mar 25 2008 EXTENSIONS New simplified definition by Vincent Nesme, Nov 04 2008 STATUS approved

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Last modified January 19 00:40 EST 2020. Contains 331030 sequences. (Running on oeis4.)