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%I #51 Feb 16 2024 06:35:34
%S 0,1,3,14,34,143,341,1420,3380,14061,33463,139194,331254,1377883,
%T 3279081,13639640,32459560,135018521,321316523,1336545574,3180705674,
%U 13230437223,31485740221,130967826660,311676696540,1296447829381,3085281225183,12833510467154
%N First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.
%C Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - _Gerald McGarvey_, Apr 02 2005
%H G. C. Greubel, <a href="/A077288/b077288.txt">Table of n, a(n) for n = 0..1000</a>
%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/2102.13494">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 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,10,-10,-1,1).
%F Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
%F G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
%F a(n) = a(-1-n) for all n in Z. - _Michael Somos_, Jul 15 2018
%F a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - _Vladimir Pletser_, Feb 29 2020
%F a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - _Wesley Ivan Hurt_, Jul 24 2020
%F 2*a(n) + 1 = A080806(n+1). - _R. J. Mathar_, Oct 01 2021
%e a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
%e G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - _Michael Somos_, Jul 15 2018
%p f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - _Vladimir Pletser_, Jul 24 2020
%t CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}],x] (* _T. D. Noe_, Jun 04 2012 *)
%t LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* _G. C. Greubel_, Jul 15 2018 *)
%t a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* _Michael Somos_, Jul 15 2018 *)
%o (PARI) my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ _G. C. Greubel_, Jul 15 2018
%o (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // _G. C. Greubel_, Jul 15 2018
%Y The k values are in A077291
%Y Cf. A077289, A077290, A077291.
%Y Cf. A053141.
%K easy,nonn
%O 0,3
%A Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002
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