A002938 -id:A002938 - cp's OEIS frontend

A077119 a(n) = A077118(n) - n^3.

0, 0, 1, -2, 0, -4, 9, 18, 17, 0, 24, -35, 36, 12, -40, -11, 0, -13, -56, 30, -79, -45, -39, -67, 100, 0, 113, -83, -48, -53, -104, 138, -7, 163, -100, -26, 0, -28, -116, 217, 9, 248, -104, 17, 80, 79, 8, -139, 297, 0, 316, -155, 17, 119, 145, 89, -55

Offset: 0

Reinhard Zumkeller, Oct 29 2002

sign

a(n)=0 iff n = m^(6*k).
Values d=x^3-y^2 of extremal points of elliptic Mordell curves. Definition for extremal points see A200656. Each value x has only one value of distance d when coordinate x is extremal point, but for many fixed distances d, the elliptic curve has more than 1 extremal point. - Artur Jasinski, Nov 30 2011
Theorem (Artur Jasinski): If a(n)>0 then a(n)<(4n^(3/2)-1)/4 for every n. If a(n)<0 then a(n)>(-4n^(3/2)-1)/4 for every n. a(n)=0 then n is perfect square. - Artur Jasinski, Dec 08 2011

			A077118(10)=1024=32^2 is the nearest square to 10^3=1000, therefore a(10)=1024-1000=24.

Cf. A000578, A077118, A077111.

|a(n)| = A002938(n).

[Round(Sqrt(n^3))^2-n^3: n in [0..60]]; // Vincenzo Librandi, Mar 24 2015

A077119 := proc(n)
    (round( sqrt(n^3) ))^2-n^3 ;
end proc: # R. J. Mathar, Jan 18 2021

Table[Round[Sqrt[x^3]]^2 - x^3, {x, 0, 100}]  (* Artur Jasinski, Nov 30 2011 *)

from math import isqrt
def A077119(n): return ((m:=isqrt(k:=n**3))+int((k-m*(m+1)<<2)>=1))**2-k # Chai Wah Wu, Jul 29 2022

a(n) = if A077116(n) < A070929(n) then -A077116(n) else A070929(n).

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

Original entry on oeis.org

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, 99, 100, 101, 109, 121, 136, 143, 144, 145, 152, 158, 169, 175, 190, 195, 196, 197, 225, 243, 255, 256, 257, 289, 312, 317, 323, 324, 325, 331, 336, 351, 356, 361, 366, 377

Offset: 1

Alex Ratushnyak, Mar 23 2015

nonn

Distance can be zero, that is, cubes that are squares are included.

Numbers n such that A002938(n) < n.

			The distance between 5^3=125 and the nearest square 11^2=121 is less than 5, so 5 is in the sequence.

Daniel Mondot, Table of n, a(n) for n = 1..10000

Cf. A000290, A000578, A002760.
Cf. A116885, A002938, A077119.
Cf. A268509, A268510.

dnsQ[n_]:=Module[{n3=n^3,sr},sr=Sqrt[n3];Min[n3-Floor[sr]^2, Ceiling[ sr]^2- n3]Harvey P. Dale, Dec 23 2015 *)

def isqrt(a):
    sr = 1 << (int.bit_length(int(a)) >> 1)
    while a < sr*sr:  sr>>=1
    b = sr>>1
    while b:
        s = sr + b
        if a >= s*s:  sr = s
        b>>=1
    return sr
for n in range(1000):
    cube = n*n*n
    r = isqrt(cube)
    sqr = r**2
    if cube-sqr < n or sqr+2*r+1-cube < n:  print(str(n), end=',')

A073072 Minimum value of abs(n^2-x^3) x>0.

Original entry on oeis.org

0, 3, 1, 8, 2, 9, 15, 0, 17, 25, 4, 19, 44, 20, 9, 40, 54, 19, 18, 57, 71, 28, 17, 64, 104, 53, 0, 55, 112, 100, 39, 24, 89, 156, 106, 35, 38, 113, 190, 128, 47, 36, 121, 208, 172, 81, 12, 107, 204, 244, 143, 40, 65, 172, 281, 239, 126, 11, 106, 225, 346, 252, 127, 0, 129

Offset: 1

Benoit Cloitre, Aug 17 2002

Cf. A002938.

Table[Min[n^2-Floor[(n^2)^(1/3)]^3,(Floor[(n^2)^(1/3)]+1)^3-n^2],{n,100}] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)

a(n)=vecmin(vector(ceil(n^(2/3)),i,abs(n^2-i^3)))

a(n^3) = 0.

a(n) = abs(A077111(n)). - R. J. Mathar, May 01 2008

cp's OEIS Frontend

A077119 a(n) = A077118(n) - n^3.

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

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Comments

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Mathematica

Python

A073072 Minimum value of abs(n^2-x^3) x>0.

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Author

Keywords

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Programs

Mathematica

PARI

Formula