A233074 Numbers that are exactly midway between the nearest square and the nearest triangular number.

2, 5, 23, 32, 47, 52, 65, 86, 140, 161, 170, 193, 203, 228, 266, 312, 356, 389, 403, 438, 453, 490, 545, 610, 671, 716, 735, 782, 802, 851, 1007, 1085, 1142, 1166, 1250, 1311, 1503, 1598, 1667, 1696, 1767, 1870, 2098, 2177, 2210, 2291, 2325, 2408, 2528, 2792, 2883

Offset: 1

Author

Alex Ratushnyak, Dec 03 2013

Keywords

• nonn
• easy

Comments

Numbers k such that k = (s+t)/2, where s is the square nearest to k, t is the triangular number nearest to k, and s != t. If there are two nearest triangular numbers, either of them is acceptable. - Edited by Robert Israel, Oct 07 2019

The sequence of roots of nearest squares begins: 1, 2, 5, 6, 7, 7, 8, 9, 12, 13, 13, 14, 14, 15, 16, 18, 19, 20, 20, 21, 21, ...

The sequence of roots of nearest triangular numbers begins: 2, 3, 6, 7, 9, 10, 11, 13, 16, 17, 18, 19, 20, 21, 23, 24, 26, 27, 28, 29, ...

The sequence of k-t (equals s-k) begins: -1, -1, 2, 4, 2, -3, -1, -5, 4, 8, -1, 3, -7, -3, -10, 12, 5, 11, -3, 3, -12, -6, ...

Examples

			5 is in the sequence because 6 and 4 are the triangular number and square nearest to 5, and 5 = (6+4)/2.
23 is in the sequence because 21 and 25 are the triangular number and square nearest to 23, and 23 = (21+25)/2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000290, A233075.

Programs

Java

import java.math.*;
public class A233074 {
  public static void main (String[] args) {
    for (long n = 1; ; n++) { // ok for small n
      long r2 = (long)Math.sqrt(n), b2 = r2*r2, a2 = (r2+1)*(r2+1);
      long t = (long)Math.sqrt(2*n), b3 = t*(t+1)/2, a3 = b3 + t + 1;
      if (b3 > n) {
        a3 = b3;
        b3 = t*(t-1)/2;
      }
      if ((b2+a3 == n*2 && n - b2 <= a2 - n && a3 - n <= n - b3) ||
          (b3+a2 == n*2 && n - b3 <= a3 - n && a2 - n <= n - b2))
            System.out.printf("%d, ", n);
    }
  }
}

Maple

f:= proc(y) local t,x,s,r,R;
   t:= y*(y+1)/2;
   R:= NULL;
   for x from ceil(sqrt(t))-1 to floor(sqrt(t))+1 do
     s:= x^2;
     if s = t then next
     elif s < t then if t-y > s then next fi
     else if t+y+1 < s then next fi
     fi;
     r:= (s+t)/2;
     if r::integer then R:= R, r fi
   od;
   R
end proc:
map(f, [$1..200]; # Robert Israel, Oct 06 2019

Mathematica

f[y_] := Module[{t, x, s, r, R = Nothing},
  t = y(y+1)/2;
  For[x = Ceiling[Sqrt[t]]-1, x <= Floor[Sqrt[t]]+1, x++,
    s = x^2;
    Which[s == t, Continue[], s < t,
    If[t - y > s, Continue[]], True,
    If[t + y + 1 < s, Continue[]]];
    r = (s + t)/2;
    If[IntegerQ[r], R = r]
  ];
R];
Map[f, Range[200]] (* Jean-François Alcover, Jul 30 2023, after Robert Israel *)

Extensions

Corrected by Alex Ratushnyak, Jun 08 2014