A069017 -id:A069017 - cp's OEIS frontend

A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

0, 1, 4, 9, 26, 55, 154, 323, 900, 1885, 5248, 10989, 30590, 64051, 178294, 373319, 1039176, 2175865, 6056764, 12681873, 35301410, 73915375, 205751698, 430810379, 1199208780, 2510946901, 6989500984, 14634871029, 40737797126, 85298279275, 237437281774

Offset: 0

Raphie Frank, Sep 01 2012

Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - Alex Ratushnyak, Apr 18 2013

For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Wikipedia, Pell numbers
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A124174, A000129, A001333, A006451, A006452, A124124, A079496.

Cf. A000217, A069017 (triangular numbers of the form k^2 + k + 1).

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 4, 9, 26}, 40] (* T. D. Noe, Sep 03 2012 *)

Vec( x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 13 2014

isok(n) = ispolygonal(n*(n+1) + 1, 3); \\ Michel Marcus, Aug 13 2014

G.f.: x*(1+3*x-x^2-x^3)/((1-x)*(1+2*x-x^2)*(1-2*x-x^2)). - R. J. Mathar, Sep 08 2012
sqrt(2) = lim_{k->infinity} ((a(2k+1) + a(2k) + 1)/2)/(a(2k+1) - a(2k)) = lim_{k->infinity} A001333(2k + 1)/A000129(2k + 1).
1 + (sqrt 2) = lim_{k->infinity} (a(2k + 1) - a(2k))/(a(2k + 1) - 2*a(2k) +  a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A000129(2k).
1 + 1/(sqrt 2) = lim_{k->infinity} (a(2k+1) - a(2k))/(a(2k) - a(2k - 1)) = lim_{k->infinity} A000129(2k + 1)/A001333(2k).
a(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013
From Raphie Frank, Jan 04 2013: (Start)
A124174(n) = a(n)*(a(n) + 1)/2.
A079496(n) = a(n + 1) - a(n).
A000129(2n) = a(2n) - 2*a(2n - 1) +  a(2n - 2).
A000129(2n) = a(2n + 1) - 2*a(2n) +  a(2n - 1).
A000129(2n + 1) = a(2n + 1) - a(2n).
A001333(2n) = a(2n) - a(2n - 1).
A001333(2n + 1) = (a(2n + 1) + a(2n) + 1)/2.
A006451(n + 1) = (a(n + 2) + a(n))/2.
A006452(n + 2) = (a(n + 2) - a(n))/2.
A124124(n + 2) = (a(n + 2) + a(n))/2 + (a(n + 2) - a(n)).
(End)
a(n + 2) = sqrt(8*a(n)^2 + 8*a(n) + 9) + 3*a(n) + 1; a(0) = 0, a(1) = 1. - Raphie Frank, Feb 02 2013
a(n) = (3/8 + sqrt(2)/4)*(1 + sqrt(2))^n + (-1/8 - sqrt(2)/8)*(-1 + sqrt(2))^n + (3/8 - sqrt(2)/4)*(1 - sqrt(2))^n + (-1/8 + sqrt(2)/8)*(-1 - sqrt(2))^n - 1/2. - Robert Israel, Aug 13 2014
E.g.f.: (1/4)*(-2*cosh(x) - 2*sinh(x) + 2*cosh(sqrt(2)*x)*(cosh(x) + 2*sinh(x)) + sqrt(2)*(cosh(x) + 3*sinh(x))*sinh(sqrt(2)*x)). - Stefano Spezia, Dec 10 2019

A214838 Triangular numbers of the form k^2 + 2.

Original entry on oeis.org

3, 6, 66, 171, 2211, 5778, 75078, 196251, 2550411, 6666726, 86638866, 226472403, 2943171003, 7693394946, 99981175206, 261348955731, 3396416785971, 8878171099878, 115378189547778, 301596468440091, 3919462027838451, 10245401755863186, 133146330756959526, 348042063230908203

Offset: 1

Alex Ratushnyak, Mar 07 2013

Corresponding k values are in A077241.

Except 3, all terms are in A089982: in fact, a(2) = 3+3 and a(n) = (k-2)*(k-1)/2+(k+1)*(k+2)/2, where k = sqrt(a(n)-2) > 2 for n > 2. [Bruno Berselli, Mar 08 2013]

			2211 is in the sequence because 2211 = 47^2 + 2.

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000217, A001110, A001219, A006454, A029549, A069017, A077241, A089982, A164055, A165892.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)))); // Bruno Berselli, Mar 08 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {3, 6, 66, 171, 2211}, 25] (* Bruno Berselli, Mar 08 2013 *)

t[n]:=((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4$
makelist(expand(t[n]*(t[n]+1)/2), n, 1, 25); /* Bruno Berselli, Mar 08 2013 */

for(n=1, 10^9, t=n*(n+1)/2; if(issquare(t-2), print1(t,", "))); \\ Joerg Arndt, Mar 08 2013

import math
for i in range(2, 1<<32):
      t = i*(i+1)//2 - 2
      sr = int(math.sqrt(t))
      if sr*sr == t:
          print(f'{sr:10} {i:10} {t+2}')

G.f.: -3*x*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)). - Joerg Arndt, Mar 08 2013

a(n) = A000217(t), where t = ((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4. - Bruno Berselli, Mar 08 2013

A076049 Numbers k such that the sum of the k-th triangular number and (k+2)-nd triangular number is a triangular number.

Original entry on oeis.org

0, 3, 8, 25, 54, 153, 322, 899, 1884, 5247, 10988, 30589, 64050, 178293, 373318, 1039175, 2175864, 6056763, 12681872, 35301409, 73915374, 205751697, 430810378, 1199208779, 2510946900, 6989500983, 14634871028, 40737797125

Offset: 1

Bruce Corrigan (scentman(AT)myfamily.com), Oct 29 2002

T(a(n)) + T(a(n)+2) = A069017(n+1) where T(k) = k*(k+1)/2.

Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A001109, A069017.

Let b(n) = A001109(n). Then we have a pair of recursion formulas:
a(2n+2) = 2*a(2n+1) - a(2n) + 2*b(n+1);
a(2n+3) = 2*a(2n+2) - a(2n+1) + 2*b(n+2).
G.f.: x*(3 + 5*x - x^2 - x^3)/((1-x)*(1 - 6*x^2 + x^4)).
a(n) = -3 + (1/8)*(-1^n)((7 + 5*sqrt(2))*(-1 - sqrt(2))^n + (7 - 5*sqrt(2))*(-1 + sqrt(2))^n - (1 + sqrt(2))^n - (1 - sqrt(2))^n).

Edited by Jon E. Schoenfield, Sep 02 2019

A226500 Triangular numbers representable as 3 * x^2.

Original entry on oeis.org

0, 3, 300, 29403, 2881200, 282328203, 27665282700, 2710915376403, 265642041604800, 26030209161894003, 2550694855824007500, 249942065661590841003, 24491771739980078410800, 2399943688452386093417403, 235169989696593857076494700, 23044259046577745607403063203

Offset: 1

Alex Ratushnyak, Jun 09 2013

Colin Barker, Table of n, a(n) for n = 1..503
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A029549 (triangular numbers representable as x^2 + x).

Cf. A000217, A001219, A165892, A069017.

#include 
#include 
typedef unsigned long long U64;
U64 isTriangular(U64 a) {   // input must be < 1ULL<<63
    U64 r = sqrt(a*2);
    return (r*(r+1) == a*2);
}
int main() {
  for (U64 j, i = 0; (j=i*i*3) < (1ULL<<63); i++)
      if (isTriangular(j)) printf("%llu, ", j);
  return 0;
}

a[1]=0; a[2]=3; a[3]=300; a[n_] := a[n] = 99*(a[n-1] - a[n-2]) + a[n-3]; Array[a, 10] (* Giovanni Resta, Jun 09 2013 *)
Rest@ CoefficientList[Series[3 x^2 (1 + x)/((1 - x) (1 - 98 x + x^2)), {x, 0, 16}], x] (* or *)
3 LinearRecurrence[{99, -99, 1}, {0, 1, 100}, 16] (* Michael De Vlieger, Mar 03 2016, latter after Vincenzo Librandi at A108741 *)

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3), for n > 3. a(n) = floor((49 + 20*sqrt(6))^(n-1)/32). - Giovanni Resta, Jun 09 2013
G.f.: 3*x^2*(1+x)/((1-x)*(1-98*x+x^2)); a(n)=3*A108741(n-1). - Joerg Arndt, Jun 10 2013
a(n) = (49+20*sqrt(6))^(-n)*(49+20*sqrt(6)-2*(49+20*sqrt(6))^n+(49-20*sqrt(6))*(49+20*sqrt(6))^(2*n))/32. - Colin Barker, Mar 03 2016

a(12)-a(15) from Giovanni Resta, Jun 09 2013

A217758 Triangular numbers of the form k^2 + k - 1.

Original entry on oeis.org

1, 55, 1891, 64261, 2183005, 74157931, 2519186671, 85578188905, 2907139236121, 98757155839231, 3354836159297755, 113965672260284461, 3871478020690373941, 131516287031212429555, 4467682281040532230951, 151769681268346883422801

Offset: 1

Alex Ratushnyak, Mar 23 2013

Triangular numbers belonging to A028387. - Bruno Berselli, Apr 18 2018

Since this sequence lists triangular numbers t such that 4*t + 5 = u^2 and 8*t + 1 = v^2 by definition of triangular numbers, 2*u^2 - 9 = 2*(4*t + 5) - 9 = 8*t + 1 = v^2, that is, sqrt(4*a(n) + 5) is in A075841. - Altug Alkan, Apr 18 2018

			Triangular(10) = 55 = 7^2 + 7 - 1, so 55 is in the sequence.

Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A028387, A069017, A075841.

#include 
typedef unsigned long long U64;
U64 rootPronic(U64 a) {
    U64 sr = 1L<<32, s, b;
    while (a < sr*(sr-1))  sr>>=1;
    for (b = sr>>1; b; b>>=1) {
        s = sr+b;
        if (a >= s*(s-1))  sr = s;
    }
    return sr;
}
int main() {
  U64 a, i, t;
  for (i=0; i < 1L<<32; ++i) {
      a = i*(i+1)/2 + 1;
      t = rootPronic(a);
      if (a == t*(t-1))  printf("%llu %llu %llu\n", i, t, a-1);
  }
  return 0;
}

a[1]=1; a[2]=55; a[3]=1891; a[n_] := a[n] = 35*a[n-1] - 35*a[n-2] + a[n-3]; Array[a,20] (* Giovanni Resta, Mar 24 2013 *)
Table[Floor@(9 (17 + Sqrt@ 288)^n*(3 - Sqrt@ 8)/32), {n, 0, 16}] (* or *)
CoefficientList[Series[-x (1 + 20 x + x^2)/((x - 1) (x^2 - 34 x + 1)), {x, 0, 16}], x] (* Michael De Vlieger, Oct 08 2016 *)

Vec(x*(1+20*x+x^2)/(1-35*x+35*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Mar 25 2013

From Giovanni Resta, Mar 24 2013: (Start)
G.f.: -x*(1 + 20*x + x^2) / ( (x - 1)*(x^2 - 34*x + 1) ).
a(n) = floor(9 * (17 + sqrt(288))^n * (3 - sqrt(8))/32). (End)
a(n) = (A075841(n)^2 - 5)/4. - Altug Alkan, Apr 18 2018

cp's OEIS Frontend

A216134 Numbers k such that 2 * A000217(k) + 1 is triangular.

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A214838 Triangular numbers of the form k^2 + 2.

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A076049 Numbers k such that the sum of the k-th triangular number and (k+2)-nd triangular number is a triangular number.

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A226500 Triangular numbers representable as 3 * x^2.

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A217758 Triangular numbers of the form k^2 + k - 1.

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