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%I A029549 #155 Sep 12 2024 07:49:41
%S A029549 0,6,210,7140,242556,8239770,279909630,9508687656,323015470680,
%T A029549 10973017315470,372759573255306,12662852473364940,430164224521152660,
%U A029549 14612920781245825506,496409142337836914550,16863297918705209269200,572855720093639278238256
%N A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.
%C A029549 Triangular numbers that are twice other triangular numbers. - _Don N. Page_
%C A029549 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 A029549 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 A029549 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 A029549 This sequence satisfy the same recurrence as A165518. - _Ant King_, Dec 13 2010
%C A029549 Intersection of A000217 and A002378.
%C A029549 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 A029549 Reinhard Zumkeller, <a href="/A029549/b029549.txt">Table of n, a(n) for n = 0..100</a>
%H A029549 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 A029549 Shyam Sunder Gupta <a href="http://www.shyamsundergupta.com/triangle.htm">Fascinating Triangular Numbers</a>
%H A029549 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 A029549 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 A029549 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 A029549 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 A029549 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 A029549 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 A029549 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 A029549 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F A029549 G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).
%F A029549 a(n) = 6*A029546(n-1) = 2*A075528(n).
%F A029549 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 A029549 From _Bill Gosper_, Feb 07 2010: (Start)
%F A029549 a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.
%F A029549 a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)
%F A029549 a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - _Mitch Harris_, Apr 19 2007, _R. J. Mathar_, Jun 26 2009
%F A029549 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 A029549 Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - _Ant King_, Dec 13 2010
%F A029549 a(n) = a(n - 1) + A001109(2n). - _Charlie Marion_, Feb 10 2011
%F A029549 a(n+2) = 34*a(n + 1) - a(n) + 6. - _Charlie Marion_, Feb 11 2011
%F A029549 From _George F. Johnson_, Aug 20 2012: (Start)
%F A029549 a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.
%F A029549 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 A029549 a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
%F A029549 a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
%F A029549 a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).
%F A029549 a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).
%F A029549 Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).
%F A029549 Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
%F A029549 Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.
%F A029549 Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)
%F A029549 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 A029549 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 A029549 a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - _Pontus von Brömssen_, Sep 11 2024
%p A029549 A029549 := proc(n)
%p A029549     option remember;
%p A029549     if n <= 1 then
%p A029549         op(n+1,[0,6]) ;
%p A029549     else
%p A029549         34*procname(n-1)-procname(n-2)+6 ;
%p A029549     end if;
%p A029549 end proc: # _R. J. Mathar_, Feb 05 2016
%t A029549 Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by _Ray Chandler_, Jul 09 2015 *)
%t A029549 CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
%t A029549 Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* _Bill Gosper_, Feb 07 2010 *)
%t A029549 LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* _Harvey P. Dale_, Jun 06 2011 *)
%t A029549 (LucasL[4Range[20] - 2, 2] -6)/32 (* _G. C. Greubel_, Jan 13 2020 *)
%o A029549 (Macsyma) (makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* _Bill Gosper_, Feb 07 2010 */
%o A029549 (Haskell)
%o A029549 a029549 n = a029549_list !! n
%o A029549 a029549_list = [0,6,210] ++
%o A029549    zipWith (+) a029549_list
%o A029549                (map (* 35) $ tail delta)
%o A029549    where delta = zipWith (-) (tail a029549_list) a029549_list
%o A029549 -- _Reinhard Zumkeller_, Sep 19 2011
%o A029549 (PARI) concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ _Charles R Greathouse IV_, Jun 13 2013
%o A029549 (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 A029549 (Scala) val triNums = (0 to 39999).map(n => (n * n + n)/2)
%o A029549 triNums.filter(_ % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte_, Jan 12 2020
%o A029549 (Sage) [(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # _G. C. Greubel_, Jan 13 2020
%o A029549 (GAP) List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # _G. C. Greubel_, Jan 13 2020
%Y A029549 Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).
%Y A029549 Cf. A000129, A001108, A002203, A009111, A245031.
%Y A029549 Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.
%K A029549 nonn,easy
%O A029549 0,2
%A A029549 _Don N. Page_
%E A029549 Additional comments from _Christian G. Bower_, Sep 19 2002; _T. D. Noe_, Nov 07 2006; and others
%E A029549 Edited by _N. J. A. Sloane_, Apr 18 2007, following suggestions from _Andrew S. Plewe_ and _Tanya Khovanova_

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