A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

Offset: 0

Author

Don N. Page

Keywords

• nonn
• easy

Comments

Triangular numbers that are twice other triangular numbers. - Don N. Page

Triangular numbers that are also pronic numbers. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002

In other words, triangular numbers which are products of two consecutive numbers. E.g., a(2) = 210: 210 is a triangular number which is the product of two consecutive numbers: 14 * 15. - Shyam Sunder Gupta, Oct 26 2002

Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2; 1, 4, 1], [2; 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1, 4, 1] and so forth. - Gene Ward Smith, Sep 30 2006

This sequence satisfy the same recurrence as A165518. - Ant King, Dec 13 2010

Intersection of A000217 and A002378.

This is the sequence of areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + 1,z(n)) with x(0) = 0, y(0) = 1, z(0) = 1, a(0) = 0 and x(1) = 3, y(1) = 4, z(1) = 5, a(1) = 6. - George F. Johnson, Aug 20 2012

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..100
• H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
• Shyam Sunder Gupta Fascinating Triangular Numbers
• Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
• Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
• Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
• Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
• Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
• Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
• Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).

Cf. A000129, A001108, A002203, A009111, A245031.

Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

Programs

GAP

List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # G. C. Greubel, Jan 13 2020

Haskell

a029549 n = a029549_list !! n
a029549_list = [0,6,210] ++
   zipWith (+) a029549_list
               (map (* 35) $ tail delta)
   where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011

Macsyma

(makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* Bill Gosper, Feb 07 2010 */

Magma

R:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // G. C. Greubel, Jul 15 2018

Maple

A029549 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,6]) ;
    else
        34*procname(n-1)-procname(n-2)+6 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Mathematica

Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by Ray Chandler, Jul 09 2015 *)
CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* Bill Gosper, Feb 07 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* Harvey P. Dale, Jun 06 2011 *)
(LucasL[4Range[20] - 2, 2] -6)/32 (* G. C. Greubel, Jan 13 2020 *)

PARI

concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ Charles R Greathouse IV, Jun 13 2013

Sage

[(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # G. C. Greubel, Jan 13 2020

Scala

val triNums = (0 to 39999).map(n => (n * n + n)/2)
triNums.filter( % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte, Jan 12 2020

Formula

G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).

a(n) = 6*A029546(n-1) = 2*A075528(n).

a(n) = -3/16 + ((3+2*sqrt(2))/32) *(17 + 12*sqrt(2))^n + ((3-2*sqrt(2))/32) *(17 - 12*sqrt(2))^n. - Gene Ward Smith, Sep 30 2006

From Bill Gosper, Feb 07 2010: (Start)

a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.

a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)

a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009

a(n) = ceiling((3 + 2*sqrt(2))^(2n + 1) - 6)/32 = floor((1/32) (1+sqrt(2))^(4n+2)). - Ant King, Dec 13 2010

Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - Ant King, Dec 13 2010

a(n) = a(n - 1) + A001109(2n). - Charlie Marion, Feb 10 2011

a(n+2) = 34*a(n + 1) - a(n) + 6. - Charlie Marion, Feb 11 2011

From George F. Johnson, Aug 20 2012: (Start)

a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.

8*a(n) + 1 = (A002315(n))^2, 4*a(n) + 1 = (A000129(2*n + 1))^2, 32*a(n)^2 + 12*a(n) + 1 are perfect squares.

a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).

a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).

a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).

a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).

Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).

Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).

Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.

Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)

a(n) = 3 * T( b(n) ) + (2*b(n) + 1)*sqrt( T( b(n) ) ) where b(n) = A001108(n) (indices of the square triangular numbers), T(n) = A000217(n) (the n-th triangular number). - Dimitri Papadopoulos, Jul 07 2017

a(n) = (Pell(2*n + 1)^2 - 1)/4 = (Q(4*n + 2) - 6)/32, where Q(n) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Jan 13 2020

a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

Extensions

Additional comments from Christian G. Bower, Sep 19 2002; T. D. Noe, Nov 07 2006; and others

Edited by N. J. A. Sloane, Apr 18 2007, following suggestions from Andrew S. Plewe and Tanya Khovanova