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

Author

Ben Paul Thurston, Aug 18 2019

Keywords

• nonn

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

Examples

			a(6) = 13 because sqrt(2)+sqrt(13) is closer to an integer than any of the previous 5 terms.
		

Programs

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 *)

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)

Extensions

More terms from Giovanni Resta, Aug 19 2019