A077398 First member of the Diophantine pair (m,k) that satisfies 7*(m^2+m) = k^2+k; a(n)=m.
0, 2, 5, 39, 87, 629, 1394, 10032, 22224, 159890, 354197, 2548215, 5644935, 40611557, 89964770, 647236704, 1433791392, 10315175714, 22850697509, 164395574727, 364177368759, 2620014019925, 5803987202642, 41755828744080, 92499617873520, 665473245885362
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
- Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
- Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
- Index entries for linear recurrences with constant coefficients, signature (1,16,-16,-1,1).
Programs
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Magma
I:=[0,2,5,39,87]; [n le 5 select I[n] else Self(n-1)+16*Self(n-2) -16*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 18 2018
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Maple
f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 16*a(n - 2) - a(n - 4) + 7}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020 -
Mathematica
LinearRecurrence[{1,16,-16,-1,1},{0,2,5,39,87}, 30] (* G. C. Greubel, Jan 18 2018 *) CoefficientList[Series[x (2+3x+2x^2)/((1-x)(1-16x^2+x^4)),{x,0,40}],x] (* Harvey P. Dale, Aug 19 2022 *) -
PARI
{a(n)=if(n<0,0,polcoeff(x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)) + x*O(x^n),n))};
Formula
G.f.: x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)).
a(n) = 16*a(n-2) - a(n-4) + 7, n >= 3. [corrected by Vladimir Pletser, Feb 29 2020]
Let b(n) be A077397 then a(n+2) = 2*a(n+1) - a(n) + b(n) with a(0)=0 a(1)=2.
a(0)=0, a(1)=2; a(n+2) = (7 + 16*a(n) + 3*sqrt(1+28*a(n)+28*a(n)^2))/2. - Herbert Kociemba, May 12 2008
a(n) = a(n-1) + 16*a(n-2) - 16*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
Comments