cp's OEIS Frontend

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

%I A233075 #48 May 07 2021 09:10:13
%S A233075 6,26,123,206,352,498,1012,1350,1746,2203,2724,3428,4977,5804,6874,
%T A233075 8050,9335,10732,12244,13874,17500,19782,21928,24519,26948,29860,
%U A233075 32946,35829,39254,42862,50639,54814,59184,63752,69045,74036,79234,85224,90863,97340,104076
%N A233075 Numbers that are midway between the nearest square and the nearest cube.
%C A233075 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, ...
%C A233075 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)
%C A233075 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, ...
%C A233075 If we allow k2=k3 then first missing terms are 0, 1, 64, 729, 4096, ... . - _Zak Seidov_, Dec 10 2013
%H A233075 Zak Seidov and Charles R Greathouse IV, <a href="/A233075/b233075.txt">Table of n, a(n) for n = 1..10000</a> (first 2108 from Seidov)
%e A233075 26 = 5^2 + 1 = 3^3 - 1.
%e A233075 352 = 19^2 - 9 = 7^3 + 9.
%t A233075 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 *)
%t A233075 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 *)
%o A233075 (Java)
%o A233075 import java.math.*;
%o A233075 public class A233075 {
%o A233075   public static void main (String[] args) {
%o A233075     for (long k = 1; ; k++) { // ok for small k's
%o A233075       long r2=(long)Math.sqrt(k), r3=(long)Math.cbrt(k);
%o A233075       long b2=r2*r2, a2=b2+r2*2+1; //squares below and above
%o A233075       long b3=r3*r3*r3, a3=b3+3*r3*(r3+1)+1; //cubes below, above
%o A233075       if ((b2+a3==k*2 && k-b2<=a2-k && a3-k<=k-b3) ||
%o A233075           (b3+a2==k*2 && k-b3<=a3-k && a2-k<=k-b2))
%o A233075             System.out.printf("%d, ", k);
%o A233075     }
%o A233075   }
%o A233075 }
%o A233075 (Python)
%o A233075 def isqrt(a):
%o A233075     sr = 1 << (int.bit_length(int(a)) >> 1)
%o A233075     while a < sr*sr:  sr>>=1
%o A233075     b = sr>>1
%o A233075     while b:
%o A233075         s = sr + b
%o A233075         if a >= s*s:  sr = s
%o A233075         b>>=1
%o A233075     return sr
%o A233075 a=[]
%o A233075 for c in range(1, 10000):
%o A233075     cube = c*c*c
%o A233075     srB = isqrt(cube)
%o A233075     srB2= srB**2
%o A233075     if srB2==cube: continue
%o A233075     if ((srB2^cube)&1)==0:
%o A233075         n = (srB2+cube)//2
%o A233075     else:
%o A233075         n = (srB2+2*srB+1+cube)//2
%o A233075     a.append(n)
%o A233075 print(a)
%o A233075 (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(m2<n3, m2=m++^2, m2>n3, n3=n++^3, m2=m++^2; n3=n++^3); s=m2+n3); Vec(v) \\ _Charles R Greathouse IV_, Jul 29 2016
%Y A233075 Cf. A000290, A000578, A075454, A233074.
%Y A233075 Cf. A002760 (Squares and cubes).
%Y A233075 Cf. A001014 (Additional terms if k2=k3 were allowed).
%K A233075 nonn,nice,easy
%O A233075 1,1
%A A233075 _Alex Ratushnyak_, Dec 03 2013