A225786 -id:A225786 - cp's OEIS frontend

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

Alex Ratushnyak, May 16 2013

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

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

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).

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)).

#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;
}

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 *)

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 */

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

More terms from Bruno Berselli, May 18 2013

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).

Original entry on oeis.org

0, 1, 324, 104329, 33593616, 10817040025, 3483053294436, 1121532343768369, 361129931640120384, 116282716455774995281, 37442673568827908360100, 12056424606446130716956921, 3882131280602085262951768464, 1250034215929265008539752488489

Offset: 0

Bruno Berselli, Jan 12 2018

16*k*a(n) provides infinitely many x-values solutions (x,y) of x*(5*x + k) = y^2.
This follows from the fact that 5*16*a(n) + 1 is a perfect square: more precisely, 80*a(n) + 1 = A023039(n)^2.
This is a divisibility sequence, that is a(n) divides a(m) if n divides m. It is the case P1 = 324, P2 = 644, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Jan 19 2018

Colin Barker, Table of n, a(n) for n = 0..300
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A023039, A049660, A225786, A253368, A298271.

P:=proc(n) trunc(evalf(((2+sqrt(5))^(4*n)+(2-sqrt(5))^(4*n)-2)/320,1000));
end: seq(P(i),i=0..13); # Paolo P. Lava, Jan 18 2018

CoefficientList[x (1 + x)/((1 - x) (1 - 322 x + x^2)) + O[x]^20, x]

makelist(coeff(taylor(x*(1+x)/((1-x)*(1-322*x+x^2)), x, 0, n), x, n), n, 0, 20);

first(n) = Vec(x*(1 + x)/((1 - x)*(1 - 322*x + x^2)) + O(x^n), -n) \\ Iain Fox, Jan 12 2018

gf = x*(1+x)/((1-x)*(1-322*x+x^2))
print(taylor(gf, x, 0, 20).list())

G.f.: x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).
a(n) = a(-n) = ((2 + sqrt(5))^(4*n) + (2 - sqrt(5))^(4*n) - 2)/320.
a(n) = A225786(n)/48. This is the case k=3 of the first comment. Example: for n = 2, 16*3*a(2) = A225786(2) = 15552 and 15552*(5*15552+3) = 34776^2.
a(n) = A049660(n)^2.
a(n)*(80*a(n) + 1) = 81*A253368(n)^2 for n>0.
a(n)*a(n-2) = (a(n-1) - 1)^2.
a(n) = 322*a(n-1) - a(n-2) + 2.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3). - Iain Fox, Jan 12 2018
a(n) = A298271(n)+A298271(n-1). - R. J. Mathar, Nov 20 2020

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

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A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).

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cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C

Mathematica

PARI

Formula

Extensions

A298101 Expansion of x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Sage

Formula

A298101 Expansion of x(1 + x)/((1 - x)(1 - 322*x + x^2)).