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 A309815 a(n) is the smallest positive integer x such that sqrt(2) + sqrt(x) is closer to an integer than any other value already in the sequence. 0
 1, 2, 3, 6, 7, 13, 21, 112, 243, 275, 466, 761, 1128, 4704, 9523, 10730, 17579, 28085, 41041, 165312, 331299, 372815, 607754, 967441, 1410360, 5648160, 11300259, 12713402, 20707831, 32942845, 48005301, 192060400, 384143763, 432165299, 703818922, 1119543881, 1631318640 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS If b(n) = round(sqrt(2) + sqrt(a(n))), then (b(n)^2 + 2 - a(n))/(2*b(n)) is an approximation for sqrt(2).  Conjecture: all convergents of the continued fraction of sqrt(2) except 1 arise in this way. - Robert Israel, Aug 18 2019 LINKS EXAMPLE a(6) = 13 because sqrt(2)+sqrt(13) is closer to an integer than any of the previous 5 terms. MAPLE R:= 1: delta:= sqrt(2)-1: for r from 2 to 10000 do    x0:= ceil((r - sqrt(2)-delta)^2);    x1:= floor((r-sqrt(2)+delta)^2);    for x from x0 to x1 do      dx:= abs(sqrt(2)+sqrt(x)-r);      if is(dx < delta) then        delta:= dx;        R:= R, x;      fi    od od: R; # Robert Israel, Aug 18 2019 MATHEMATICA d[x_] := Abs[x - Round[x]]; dm = 1; s = {}; Do[If[(d1 = d[Sqrt[2] + Sqrt[n]]) < dm, dm = d1; AppendTo[s, n]], {n, 1, 10^5}]; s (* Amiram Eldar, Aug 18 2019 *) PROG (Python) import math a = 2**(1/2) l = [] closest = 1.0 for i in range(1, 100000000):     b = i**(1/2)     c = abs(a+b - round(a+b))     if c < closest:         print(i, c)         closest = c         l.append(i) print(l) CROSSREFS Sequence in context: A294916 A233423 A328024 * A256800 A172105 A092482 Adjacent sequences:  A309812 A309813 A309814 * A309816 A309817 A309818 KEYWORD nonn AUTHOR Ben Paul Thurston, Aug 18 2019 EXTENSIONS More terms from Giovanni Resta, Aug 19 2019 STATUS approved

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Last modified February 29 05:25 EST 2020. Contains 332353 sequences. (Running on oeis4.)