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%I #82 Feb 16 2024 06:38:40
%S 0,2,6,44,116,798,2090,14328,37512,257114,673134,4613732,12078908,
%T 82790070,216747218,1485607536,3889371024,26658145586,69791931222,
%U 478361013020,1252365390980,8583840088782,22472785106426,154030760585064,403257766524696,2763969850442378
%N First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.
%H G. C. Greubel, <a href="/A077259/b077259.txt">Table of n, a(n) for n = 0..1000</a>
%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://ajc.maths.uq.edu.au/pdf/77/ajc_v77_p318.pdf">Counting families of generalized balancing numbers</a>, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
%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://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 Hermann Stamm-Wilbrandt, <a href="/A077259/a077259_2.svg">6 interlaced bisections</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,18,-18,-1,1).
%F Let b(n) be A007805(n). Then with a(0)=0, a(1)=2, a(2*n+2) = 2*a(2*n+1) - a(2*n) + 2*b(n), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + 2*b(n+1).
%F a(n) = (A000045(A007310(n+1))-1)/2. - _Vladeta Jovovic_, Nov 02 2002 [corrected by _R. J. Mathar_, Sep 16 2009]
%F a(0)=0, a(1)=2, a(n+2) = 4 + 9*a(n) + 2*sqrt(1 +20*a(n) +20*a(n)^2). - _Herbert Kociemba_, May 12 2008
%F a(0)=0, a(1)=2, a(2)=6, a(3)=44, a(n) = 18*a(n-2) - a(n-4) + 8. - Robert Phillips, Sep 01 2008
%F G.f.: 2*x*(1+x)^2/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
%F a(n) = a(-1-n) for all n in Z. - _Michael Somos_, Jul 15 2018
%F a(2*n) = A049651(2*n); a(2*n+1) = A110679(2*n+1). See "6 interlaced bisections" link. - _Hermann Stamm-Wilbrandt_, Apr 18 2019
%F a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5). - _Wesley Ivan Hurt_, Jul 24 2020
%F From _Vladimir Pletser_, Feb 07 2021: (Start)
%F a(n) = ((5+sqrt(5))*(2+sqrt(5))^n + (5-sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for even n;
%F a(n) = ((5+3*sqrt(5))*(2+sqrt(5))^n + (5-3*sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for odd n. (End)
%e a(3) = (2*6) - 2 + (2*17) = 12 - 2 + 34 = 44.
%e G.f. = 2*x + 6*x^2 + 44*x^3 + 116*x^4 + 798*x^5 + 2090*x^6 + 14328*x^7 + ... - _Michael Somos_, Jul 15 2018
%p f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 18*a(n - 2) - a(n - 4) + 8}, a(n), remember): map(f, [$ (0 .. 40)])[]; # _Vladimir Pletser_, Jul 24 2020
%t LinearRecurrence[{1, 18, -18, -1, 1}, {0, 2, 6, 44, 116}, 30] (* _G. C. Greubel_, Jul 15 2018 *)
%t a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ 2 x (x + 1)^2 / ((1 - x) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, m}]]; (* _Michael Somos_, Jul 15 2018 *)
%o (PARI) my(x='x+O('x^30)); concat([0], Vec(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))) \\ _G. C. Greubel_, Jul 15 2018
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))); // _G. C. Greubel_, Jul 15 2018
%Y Cf. A007805, A077260, A077261, A077262.
%Y Cf. A053141.
%K easy,nonn
%O 0,2
%A Bruce Corrigan (scentman(AT)myfamily.com), Nov 01 2002
%E More terms from _Colin Barker_, Mar 23 2014
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