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%I #37 Jul 30 2023 08:48:21
%S 2,5,23,32,47,52,65,86,140,161,170,193,203,228,266,312,356,389,403,
%T 438,453,490,545,610,671,716,735,782,802,851,1007,1085,1142,1166,1250,
%U 1311,1503,1598,1667,1696,1767,1870,2098,2177,2210,2291,2325,2408,2528,2792,2883
%N Numbers that are exactly midway between the nearest square and the nearest triangular number.
%C 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
%C 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, ...
%C 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, ...
%C 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, ...
%H Robert Israel, <a href="/A233074/b233074.txt">Table of n, a(n) for n = 1..10000</a>
%e 5 is in the sequence because 6 and 4 are the triangular number and square nearest to 5, and 5 = (6+4)/2.
%e 23 is in the sequence because 21 and 25 are the triangular number and square nearest to 23, and 23 = (21+25)/2.
%p f:= proc(y) local t,x,s,r,R;
%p t:= y*(y+1)/2;
%p R:= NULL;
%p for x from ceil(sqrt(t))-1 to floor(sqrt(t))+1 do
%p s:= x^2;
%p if s = t then next
%p elif s < t then if t-y > s then next fi
%p else if t+y+1 < s then next fi
%p fi;
%p r:= (s+t)/2;
%p if r::integer then R:= R, r fi
%p od;
%p R
%p end proc:
%p map(f, [$1..200]; # _Robert Israel_, Oct 06 2019
%t f[y_] := Module[{t, x, s, r, R = Nothing},
%t t = y(y+1)/2;
%t For[x = Ceiling[Sqrt[t]]-1, x <= Floor[Sqrt[t]]+1, x++,
%t s = x^2;
%t Which[s == t, Continue[], s < t,
%t If[t - y > s, Continue[]], True,
%t If[t + y + 1 < s, Continue[]]];
%t r = (s + t)/2;
%t If[IntegerQ[r], R = r]
%t ];
%t R];
%t Map[f, Range[200]] (* _Jean-François Alcover_, Jul 30 2023, after _Robert Israel_ *)
%o (Java)
%o import java.math.*;
%o public class A233074 {
%o public static void main (String[] args) {
%o for (long n = 1; ; n++) { // ok for small n
%o long r2 = (long)Math.sqrt(n), b2 = r2*r2, a2 = (r2+1)*(r2+1);
%o long t = (long)Math.sqrt(2*n), b3 = t*(t+1)/2, a3 = b3 + t + 1;
%o if (b3 > n) {
%o a3 = b3;
%o b3 = t*(t-1)/2;
%o }
%o if ((b2+a3 == n*2 && n - b2 <= a2 - n && a3 - n <= n - b3) ||
%o (b3+a2 == n*2 && n - b3 <= a3 - n && a2 - n <= n - b2))
%o System.out.printf("%d, ", n);
%o }
%o }
%o }
%Y Cf. A000217, A000290, A233075.
%K nonn,easy
%O 1,1
%A _Alex Ratushnyak_, Dec 03 2013
%E Corrected by _Alex Ratushnyak_, Jun 08 2014
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