A212614 -id:A212614 - cp's OEIS frontend

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

Alex Ratushnyak, Nov 20 2013

nonn

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.

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).

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 *)

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

A212615 Least k > 1 such that the product pen(n) * pen(k) is pentagonal, or zero if no such k exists, where pen(k) is the k-th pentagonal number.

Original entry on oeis.org

2, 39, 2231, 40, 14, 94974, 47, 212, 1071, 477, 124, 261, 15120, 5, 180, 375638, 2413, 22, 4270831, 924, 278, 18, 126, 33510, 355, 376, 9047610, 37313170, 1533015, 7315, 1687018, 520, 363155, 8827, 13514, 11701449166, 670, 3290, 2, 4, 817, 31175067

Offset: 1

T. D. Noe, Jun 07 2012

nonn

That is, pen(k) = k*(3k-1)/2.

			For n = 2, pen(n) = 5 and the first k is 39 because pen(39) = 2262 and 5*2262 = 11310 which is the 87th pentagonal number.

Cf. A188663 (pentagonal numbers that are pen(x) * pen(y) for some x,y > 1).
Cf. A212614 (similar sequence for triangular numbers).
Cf. A000326 (pentagonal numbers).

kMax = 10^7; PentagonalQ[n_] := IntegerQ[(1 + Sqrt[1 + 24*n])/6]; Table[t = n*(3*n - 1)/2; k = 2; While[t2 = k*(3*k - 1)/2; k < kMax && ! PentagonalQ[t*t2], k++]; If[k == kMax, 0, k], {n, 15}]

a(25) corrected and a(28)-a(42) from Donovan Johnson, Feb 08 2013

cp's OEIS Frontend

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

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