cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

Original entry on oeis.org

0, 5, 14, 63, 152, 637, 1518, 6319, 15040, 62565, 148894, 619343, 1473912, 6130877, 14590238, 60689439, 144428480, 600763525, 1429694574, 5946945823, 14152517272, 58868694717, 140095478158, 582740001359, 1386802264320
Offset: 1

Views

Author

Ant King, Sep 25 2009, Oct 01 2009

Keywords

Comments

Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2 + 9k + 8 = 2s^2. Hence solutions occur whenever (3k^2 + 9k + 8)/2 is a perfect square, or equivalently when s>=2 and sqrt(24s^2 - 15) is congruent to 3 mod 6. This sequence returns the index of the smallest of the 3 triangular numbers, the values of s^2 are given in A165516 and, with the exception of the first term, the values of s are in A129445.

Examples

			The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 = T(63) + T(64) + T(65). Hence a(4)=63.

Links

Crossrefs

Cf. A000290, A129445, A000217, A165516.

Programs

  • Magma
    I:=[0, 5, 14, 63, 152]; [n le 5 select I[n] else Self(n-1) + 10*Self(n-2) - 10*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018
  • Mathematica
    TriangularNumber[ n_ ]:=1/2 n (n+1);Select[ Range[ 0,10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]
CoefficientList[Series[x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)), {x,0,50}], x] (* or *) LinearRecurrence[{1,10,-10,-1,1}, {0, 5, 14, 63, 152}, 50] (* G. C. Greubel, Feb 17 2017 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)))) \\ G. C. Greubel, Feb 17 2017

Formula

a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5).
G.f.: x(x^3 + x^2 - 9x - 5)/((x-1)(x^4 - 10x^2 + 1)).
a(n) = 10*a(n-2) - a(n-4) + 12. - Zak Seidov, Sep 25 2009

Extensions

a(1) = 0 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky
More terms from Zak Seidov, Sep 25 2009

A165518 Perfect squares (A000290) that can be expressed as the sum of four consecutive triangular numbers (A000217).

Original entry on oeis.org

4, 100, 3364, 114244, 3880900, 131836324, 4478554084, 152139002500, 5168247530884, 175568277047524, 5964153172084900, 202605639573839044, 6882627592338442564, 233806732499933208100, 7942546277405390632804, 269812766699283348307204, 9165691521498228451812100, 311363698964240484013304164
Offset: 1

Views

Author

Ant King, Sep 28 2009

Keywords

Comments

As T(n) + T(n+1) = (n+1)^2 and T(n+2) + T(n+3) = (n+3)^2, it follows that the equation T(n) + T(n+1) + T(n+2) + T(n+3) = s^2 becomes (n+1)^2 + (n+3)^2 = s^2. Hence the solutions to this equation correspond to those Pythagorean triples with shorter legs that differ by two, such as 6^2 + 8^2 = 10^2.
Terms are the squares of the hypotenuses of Pythagorean triangles where other two sides are m and m+2, excepting the initial 4. See A075870. - Richard R. Forberg, Aug 15 2013

Examples

			As the third perfect square that can be expressed as the sum of four consecutive triangular numbers is 3364 = T(39) + T(40) + T(41) + T(42), we have a(3)=3364.
The first term, 4, equals T(-1) + T(0) + T(1) + T(2).

Links

Crossrefs

Cf. A000290, A000217, A165516 (squares that can be expressed as the sum of three consecutive triangular numbers), A029549, A075870.

Programs

  • Magma
    I:=[4,100,3364]; [n le 3 select I[n] else 35*Self(n-1) - 35*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 21 2018
  • Maple
    A165518:=n->(1/2)*(2+(3+2*sqrt(2))^(2*n+1)+(3-2*sqrt(2))^(2*n+1)); seq(A165518(k), k=1..20); # Wesley Ivan Hurt, Oct 24 2013
  • Mathematica
    TriangularNumber[n_]:=1/2 n (n+1); data=Select[Range[10^7],IntegerQ[Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1]+TriangularNumber[ #+2]+TriangularNumber[ #+3]]] &];2(#^2+4#+5)&/@data
t={4, 100}; Do[AppendTo[t, 34 t[[-1]] - t[[-2]] - 32], {20}]; t
LinearRecurrence[{35,-35,1},{4,100,3364},20] (* Harvey P. Dale, May 22 2012 *)
  • PARI
    x='x+O('x^50); Vec(4*x*(1-10*x+x^2)/((1-x)*(1-34*x+x^2))) \\ G. C. Greubel, Oct 21 2018

Formula

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 32.
a(n) = (2 + (3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1))/2.
a(n) = ceiling((1/2)*(2 + (3+2*sqrt(2))^(2n+1))).
G.f.: 4*x*(x^2-10*x+1)/((1-x)*(x^2-34*x+1)).
a(n) = 4*A008844(n-1). - R. J. Mathar, Dec 14 2010
a(n) = A075870(n)^2. - Richard R. Forberg, Aug 15 2013

Extensions

Extended by T. D. Noe, Dec 09 2010

A249483 Squares (A000290) which are also centered triangular numbers (A005448).

Original entry on oeis.org

1, 4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201
Offset: 1

Views

Author

Colin Barker, Jan 13 2015

Keywords

Comments

Apart from the first term the same as A165516. - R. J. Mathar, Jan 20 2015

Examples

			64 is in the sequence because the 8th square is 64, which is also the 7th centered triangular number.

Links

Crossrefs

Cf. A000290, A005448, A129444, A129445.

Programs

  • Magma
    I:=[1,4,64,361,6241]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 20 2015
  • Mathematica
    CoefficientList[Series[(x^2 - 5 x + 1) (x^2 + 8 x + 1) / ((1 - x) (x^2 - 10 x + 1) (x^2 + 10 x + 1)), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 20 2015 *)
  • PARI
    Vec(-x*(x^2-5*x+1)*(x^2+8*x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))

Formula

a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).
G.f.: -x*(x^2-5*x+1)*(x^2+8*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).
Showing 1-3 of 3 results.

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