A232177 Least positive k such that triangular(n) + triangular(k) is a square.

1, 2, 1, 2, 3, 1, 5, 6, 7, 8, 9, 5, 2, 12, 13, 1, 15, 16, 17, 3, 5, 20, 2, 22, 23, 8, 4, 26, 12, 3, 29, 30, 1, 5, 33, 34, 4, 36, 37, 15, 6, 29, 22, 5, 43, 19, 45, 7, 15, 48, 6, 50, 11, 52, 8, 41, 22, 7, 57, 58, 59, 9, 26, 62, 8, 64, 19, 66, 10, 68, 5, 9, 71, 2

Offset: 0

Author

Alex Ratushnyak, Nov 20 2013

Keywords

• nonn

Comments

Triangular(k) = A000217(k) = k*(k+1)/2.

For n>1, a(n) <= n-1, because with k=n-1: triangular(n) + triangular(k) = n*(n+1)/2 + (n-1)*n/2 = n^2.

Crossrefs

Cf. A000217, A000290.

Cf. A082183 (least k>0 such that triangular(n) + triangular(k) is a triangular number).

Cf. A212614 (least k>1 such that triangular(n) * triangular(k) is a triangular number).

Cf. A232176 (least k>0 such that n^2 + triangular(k) is a square).

Cf. A232179 (least k>=0 such that n^2 + triangular(k) is a triangular number).

Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number).

Cf. A232178 (least k>=0 such that triangular(n) + k^2 is a square).

Programs

Mathematica

Table[k = 1; tri = n*(n + 1)/2; While[k <= n+2 && ! IntegerQ[Sqrt[tri + k*(k + 1)/2]], k++]; k, {n, 0, 100}] (* T. D. Noe, Nov 21 2013 *)

Python

import math
for n in range(77):
  tn = n*(n+1)//2
  for k in range(1, n+9):
    sum = tn + k*(k+1)//2
    r = int(math.sqrt(sum))
    if r*r == sum:
      print(str(k), end=',')
      break