A239970 Least positive k such that triangular(k) + triangular(n+k) is a triangular number (A000217).

2, 5, 3, 6, 9, 4, 15, 18, 21, 5, 27, 30, 9, 36, 6, 42, 10, 48, 51, 14, 7, 60, 63, 15, 69, 72, 19, 8, 81, 26, 20, 13, 17, 24, 99, 9, 105, 14, 111, 114, 29, 120, 123, 126, 10, 132, 135, 34, 20, 144, 147, 35, 153, 45, 11, 29, 165, 33, 17, 174, 30, 44, 183, 186, 189, 12, 18, 23

Offset: 0

Author

Alex Ratushnyak, Mar 30 2014

Keywords

• nonn

Comments

a(n) <= 3*n-3, because triangular(3*n-3) + triangular(4*n-3) = triangular(5*n-4).

In other words, smallest k>0 such that 8*k^2 + 4*(2*k + 1)*n + 4*n^2 + 8*k + 1 = m^2 has an integer solution. - Ralf Stephan, Apr 01 2014

Examples

			n=5: the smallest solution k>0 to 8*k^2 + 64*k + 225 = m^2 is k=4, so a(5)=4.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000217, A239969.

Cf. A010054.

Programs

Haskell

a239970 n = head [k | k <- [1..],
                      a010054 (a000217 k + a000217 (n + k)) == 1]
-- Reinhard Zumkeller, Apr 03 2014

Mathematica

tr[n_]:=(n(n+1))/2;lpk[n_]:=Module[{k=1},While[!OddQ[Sqrt[8(tr[k]+tr[n+k])+1]],k++];k]; Array[lpk,70,0] (* Harvey P. Dale, Nov 11 2024 *)

PARI

triangular(n) = n*(n+1)/2;
is_triangular(n) = issquare(8*n+1);
s=[]; for(n=0, 100, k=1; while(!is_triangular(triangular(k)+triangular(n+k)), k++); s=concat(s, k)); s \\ Colin Barker, Mar 31 2014

PARI

a(n)=my(k=1); while(!ispolygonal(k*(k+n+1)+(n^2+n)/2,3), k++); k \\ Charles R Greathouse IV, Apr 01 2014

Extensions

First PROG corrected by Colin Barker, Apr 04 2014