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A253181 Numbers n such that the distance between n^3 and the nearest square is less than n.

%I A253181 #23 May 05 2021 13:40:17
%S A253181 1,2,3,4,5,9,13,15,16,17,25,32,35,36,37,40,43,46,49,52,56,63,64,65,81,
%T A253181 99,100,101,109,121,136,143,144,145,152,158,169,175,190,195,196,197,
%U A253181 225,243,255,256,257,289,312,317,323,324,325,331,336,351,356,361,366,377
%N A253181 Numbers n such that the distance between n^3 and the nearest square is less than n.
%C A253181 Distance can be zero, that is, cubes that are squares are included.
%C A253181 Numbers n such that A002938(n) < n.
%H A253181 Daniel Mondot, <a href="/A253181/b253181.txt">Table of n, a(n) for n = 1..10000</a>
%e A253181 The distance between 5^3=125 and the nearest square 11^2=121 is less than 5, so 5 is in the sequence.
%t A253181 dnsQ[n_]:=Module[{n3=n^3,sr},sr=Sqrt[n3];Min[n3-Floor[sr]^2, Ceiling[ sr]^2- n3]<n]; Select[Range[400],dnsQ] (* _Harvey P. Dale_, Dec 23 2015 *)
%o A253181 (Python)
%o A253181 def isqrt(a):
%o A253181     sr = 1 << (int.bit_length(int(a)) >> 1)
%o A253181     while a < sr*sr:  sr>>=1
%o A253181     b = sr>>1
%o A253181     while b:
%o A253181         s = sr + b
%o A253181         if a >= s*s:  sr = s
%o A253181         b>>=1
%o A253181     return sr
%o A253181 for n in range(1000):
%o A253181     cube = n*n*n
%o A253181     r = isqrt(cube)
%o A253181     sqr = r**2
%o A253181     if cube-sqr < n or sqr+2*r+1-cube < n:  print(str(n), end=',')
%Y A253181 Cf. A000290, A000578, A002760.
%Y A253181 Cf. A116885, A002938, A077119.
%Y A253181 Cf. A268509, A268510.
%K A253181 nonn
%O A253181 1,2
%A A253181 _Alex Ratushnyak_, Mar 23 2015