A225785 Numbers n such that triangular(n) + triangular(2*n) is a triangular number.

0, 12, 84, 3960, 27144, 1275204, 8740380, 410611824, 2814375312, 132215732220, 906220110180, 42573055163112, 291800061102744, 13708391546789940, 93958713454973484, 4414059505011197664, 30254413932440359200, 1421313452222058857964

Offset: 1

Author

Alex Ratushnyak, May 16 2013

Keywords

• nonn
• easy

Comments

Equivalently, numbers n such that oblong(n) + oblong(2*n) is an oblong number, where oblong(n) = A002378(n) = n*(n+1).

Also, x values in the equation A147875(x) = A000217(y) - see Ralf Stephan in Program lines. - Bruno Berselli, May 18 2013

Also, numbers m such that 2*m+1 and 10*m+1 are both squares. - Bruno Berselli, Mar 03 2016

Examples

			12*13/2 + 24*25/2 = 27*28/2, so 12 is in the sequence.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..798
• Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).

Crossrefs

Cf. A000217, A002378, A082183.

Cf. A224419 (numbers n such that triangular(n) + triangular(2*n) is a square).

Cf. A011916 (numbers n such that triangular(2*n) - triangular(n) is a triangular number).

Cf. A225786 (numbers n such that oblong(2*n) + oblong(n) is a square).

Cf. A225839 (triangular numbers of the form triangular(x) + triangular(2*x)).

Programs

C

#include 
#include 
int main() {
  unsigned long long i, s, t;
  for (i = 0; i< (1ULL<<31); i++) {
    s = 2*i*(2*i+1) + i*(i+1);
    t = sqrt(s);
    if (s==t*(t+1)) printf("%llu, ", i);
  }
  return 0;
}

Mathematica

CoefficientList[Series[12 x (1 + 6 x + x^2)/((1 - x) (1 - 18 x + x^2) (1 + 18 x + x^2)), {x, 0, 20}], x] (* Bruno Berselli, May 18 2013 *)
LinearRecurrence[{1,322,-322,-1,1},{0,12,84,3960,27144},20] (* Harvey P. Dale, Apr 08 2021 *)

PARI

for(n=1,10^9,t=n*(5*n+3)/2;x=sqrtint(2*t);if(t==x*(x+1)/2,print(n))) /* Ralf Stephan, May 17 2013 */

Formula

G.f.: 12*x*(1+6*x+x^2)/((1-x)*(1-18*x+x^2)(1+18*x+x^2)). [Bruno Berselli, May 18 2013]

a(n) = (1/20)*((3+(-1)^n*sqrt(5))*(2-sqrt(5))^(4*floor(n/2))+(3-(-1)^n*sqrt(5))*(2+sqrt(5))^(4*floor(n/2))-6). [Bruno Berselli, May 18 2013]

a(2*n) = (Fibonacci(6*n-3)^2 + Lucas(6*n-3)*Fibonacci(6*n-1))/2. - Greg Dresden, Sep 24 2023

Extensions

More terms from Bruno Berselli, May 18 2013