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%I #81 Apr 29 2022 03:22:15
%S 0,3,105,3570,121278,4119885,139954815,4754343828,161507735340,
%T 5486508657735,186379786627653,6331426236682470,215082112260576330,
%U 7306460390622912753,248204571168918457275,8431648959352604634600,286427860046819639119128
%N Triangular numbers that are half other triangular numbers.
%C This is the sequence of 1/2 the 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)=3. - _George F. Johnson_, Aug 24 2012
%H Colin Barker, <a href="/A075528/b075528.txt">Table of n, a(n) for n = 0..653</a>
%H Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, <a href="http://www.jstor.org/stable/20871097">Misinterpretations can sometimes be a good thing</a>, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
%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 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>, 2022.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F a(n) = 3*A029546(n-1) = A029549(n)/2.
%F G.f.: 3*x/((1-x)*(1-34*x+x^2)).
%F From _George F. Johnson_, Aug 24 2012: (Start)
%F a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
%F 8*a(n)+1 = A000129(2*n+1)^2.
%F 16*a(n)+1 = A002315(n)^2.
%F 128*a(n)^2 + 24*a(n) + 1 is a perfect square.
%F a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
%F a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
%F a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
%F a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
%F a(n) = A084159(n)*A046729(n)/4 = A001652(n)*A046090(n)/4.
%F Lim_{n->infinity} a(n)/a(n-1) = 17 + 12*sqrt(2).
%F Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
%F Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
%F Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
%F (End)
%F a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - _R. J. Mathar_, Nov 07 2015
%F a(n) = A000217(A053141(n)). - _R. J. Mathar_, Aug 16 2019
%F a(n) = (a(n-1)*(a(n-1)-3))/a(n-2) for n > 2. - _Vladimir Pletser_, Apr 08 2020
%t CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* _Robert G. Wilson v_, Jun 24 2011 *)
%o (PARI) concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ _Colin Barker_, Jun 18 2015
%Y Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.
%K nonn,easy
%O 0,2
%A _Christian G. Bower_, Sep 19 2002
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