A233075 Numbers that are midway between the nearest square and the nearest cube.

6, 26, 123, 206, 352, 498, 1012, 1350, 1746, 2203, 2724, 3428, 4977, 5804, 6874, 8050, 9335, 10732, 12244, 13874, 17500, 19782, 21928, 24519, 26948, 29860, 32946, 35829, 39254, 42862, 50639, 54814, 59184, 63752, 69045, 74036, 79234, 85224, 90863, 97340, 104076

Offset: 1

Author

Alex Ratushnyak, Dec 03 2013

Keywords

• nonn
• nice
• easy

Comments

The sequence of roots of nearest squares begins: 2, 5, 11, 14, 19, 22, 32, 37, 42, 47, 52, 59, 71, 76, 83, 90, 97, 104, 111, 118, 132, ...

The sequence of cube roots of nearest cubes begins: 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, ... (Cf. A000037)

The sequence of k-k2 (equals k3-k) begins: 2, 1, 2, 10, -9, 14, -12, -19, -18, -6, 20, -53, -64, 28, -15, -50, -74, -84, -77, -50, ...

If we allow k2=k3 then first missing terms are 0, 1, 64, 729, 4096, ... . - Zak Seidov, Dec 10 2013

Examples

			26 = 5^2 + 1 = 3^3 - 1.
352 = 19^2 - 9 = 7^3 + 9.
		

Links

• Zak Seidov and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 2108 from Seidov)

Crossrefs

Cf. A000290, A000578, A075454, A233074.

Cf. A002760 (Squares and cubes).

Cf. A001014 (Additional terms if k2=k3 were allowed).

Programs

Java

import java.math.*;
public class A233075 {
  public static void main (String[] args) {
    for (long k = 1; ; k++) { // ok for small k's
      long r2=(long)Math.sqrt(k), r3=(long)Math.cbrt(k);
      long b2=r2*r2, a2=b2+r2*2+1; //squares below and above
      long b3=r3*r3*r3, a3=b3+3*r3*(r3+1)+1; //cubes below, above
      if ((b2+a3==k*2 && k-b2<=a2-k && a3-k<=k-b3) ||
          (b3+a2==k*2 && k-b3<=a3-k && a2-k<=k-b2))
            System.out.printf("%d, ", k);
    }
  }
}

Mathematica

max = 10^6; u = Union[Range[Ceiling[Sqrt[max]]]^2,Range[Ceiling[ max^(1/3) ]]^3]; Reap[Do[x = u[[k]]; y = u[[k+1]]; If[Not[IntegerQ[Sqrt[x]] && IntegerQ[Sqrt[y]]] && Not[IntegerQ[x^(1/3)] && IntegerQ[y^(1/3)]] && IntegerQ[m = (x+y)/2], Sow[m]], {k, 1, Length[u]-2}]][[2, 1]] (* Jean-François Alcover, Dec 03 2015 *)
Module[{upto=150000,nns},nns=Union[Join[Range[Floor[Sqrt[upto]]]^2,Range[Floor[Surd[upto,3]]]^3]];Mean/@Select[Partition[nns,2,1],EvenQ[Total[#]]&]] (* Harvey P. Dale, Nov 06 2017 *)

PARI

list(lim)=my(v=List(),m=2,n=2,m2=4,n3=8,s=12); lim*=2; while(s <= lim, if(s%2==0 && m2!=n3 && abs(s/2-m2)<=abs(s/2-(m-1)^2) && abs(s/2-m2)<=abs(s/2-(m+1)^2) && abs(s/2-m2)<=abs(s/2-(n-1)^3) && abs(s/2-m2)<=abs(s/2-(n+1)^3), listput(v,s/2)); if(m2n3, n3=n++^3, m2=m++^2; n3=n++^3); s=m2+n3); Vec(v) \\ Charles R Greathouse IV, Jul 29 2016

Python

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
a=[]
for c in range(1, 10000):
    cube = c*c*c
    srB = isqrt(cube)
    srB2= srB**2
    if srB2==cube: continue
    if ((srB2^cube)&1)==0:
        n = (srB2+cube)//2
    else:
        n = (srB2+2*srB+1+cube)//2
    a.append(n)
print(a)