A029549 - OEIS

%I #155 Sep 12 2024 07:49:41

%S 0,6,210,7140,242556,8239770,279909630,9508687656,323015470680,

%T 10973017315470,372759573255306,12662852473364940,430164224521152660,

%U 14612920781245825506,496409142337836914550,16863297918705209269200,572855720093639278238256

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

%C Triangular numbers that are twice other triangular numbers. - _Don N. Page_

%C 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

%C 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

%C 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

%C This sequence satisfy the same recurrence as A165518. - _Ant King_, Dec 13 2010

%C Intersection of A000217 and A002378.

%C 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

%H Reinhard Zumkeller, <a href="/A029549/b029549.txt">Table of n, a(n) for n = 0..100</a>

%H H. J. Hindin, <a href="/A006062/a006062.pdf">Stars, hexes, triangular numbers and Pythagorean triples</a>, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)

%H Shyam Sunder Gupta <a href="http://www.shyamsundergupta.com/triangle.htm">Fascinating Triangular Numbers</a>

%H Roger B. Nelson, <a href="http://www.jstor.org/stable/10.4169/math.mag.89.3.159">Multi-Polygonal Numbers</a>, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.

%H Vladimir Pletser, <a href="https://arxiv.org/abs/2101.00998">Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers</a>, arXiv:2101.00998 [math.NT], 2021.

%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.12392">Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2102.12392 [math.GM], 2021.

%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.13494">Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations</a>, arXiv:2102.13494 [math.NT], 2021.

%H Vladimir Pletser, <a href="https://arxiv.org/abs/2103.03019">Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2103.03019 [math.GM], 2021.

%H Vladimir Pletser, <a href="https://doi.org/10.13140/RG.2.2.35428.91527">Searching for multiple of triangular numbers being triangular numbers</a>, 2021.

%H Vladimir Pletser, <a href="https://www.researchgate.net/profile/Vladimir-Pletser/publication/359808848_USING_PELL_EQUATION_SOLUTIONS_TO_FIND_ALL_TRIANGULAR_NUMBERS_MULTIPLE_OF_OTHER_TRIANGULAR_NUMBERS/">Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers</a>, 2021.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).

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

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

%F 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

%F From _Bill Gosper_, Feb 07 2010: (Start)

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

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

%F a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - _Mitch Harris_, Apr 19 2007, _R. J. Mathar_, Jun 26 2009

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

%F Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - _Ant King_, Dec 13 2010

%F a(n) = a(n - 1) + A001109(2n). - _Charlie Marion_, Feb 10 2011

%F a(n+2) = 34*a(n + 1) - a(n) + 6. - _Charlie Marion_, Feb 11 2011

%F From _George F. Johnson_, Aug 20 2012: (Start)

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

%F 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.

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

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

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

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

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

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

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

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

%F 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

%F 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

%F a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - _Pontus von Brömssen_, Sep 11 2024

%p A029549 := proc(n)

%p     option remember;

%p     if n <= 1 then

%p         op(n+1,[0,6]) ;

%p     else

%p         34*procname(n-1)-procname(n-2)+6 ;

%p     end if;

%p end proc: # _R. J. Mathar_, Feb 05 2016

%t Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by _Ray Chandler_, Jul 09 2015 *)

%t CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]

%t Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* _Bill Gosper_, Feb 07 2010 *)

%t LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* _Harvey P. Dale_, Jun 06 2011 *)

%t (LucasL[4Range[20] - 2, 2] -6)/32 (* _G. C. Greubel_, Jan 13 2020 *)

%o (Macsyma) (makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* _Bill Gosper_, Feb 07 2010 */

%o (Haskell)

%o a029549 n = a029549_list !! n

%o a029549_list = [0,6,210] ++

%o    zipWith (+) a029549_list

%o                (map (* 35) $ tail delta)

%o    where delta = zipWith (-) (tail a029549_list) a029549_list

%o -- _Reinhard Zumkeller_, Sep 19 2011

%o (PARI) concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ _Charles R Greathouse IV_, Jun 13 2013

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // _G. C. Greubel_, Jul 15 2018

%o (Scala) val triNums = (0 to 39999).map(n => (n * n + n)/2)

%o triNums.filter(_ % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte_, Jan 12 2020

%o (Sage) [(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # _G. C. Greubel_, Jan 13 2020

%o (GAP) List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # _G. C. Greubel_, Jan 13 2020

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

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

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

%K nonn,easy

%O 0,2

%A _Don N. Page_

%E Additional comments from _Christian G. Bower_, Sep 19 2002; _T. D. Noe_, Nov 07 2006; and others

%E Edited by _N. J. A. Sloane_, Apr 18 2007, following suggestions from _Andrew S. Plewe_ and _Tanya Khovanova_