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.

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

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

Table of n, a(n) for n=1..37.

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: A233423 A328024 A376988 * 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