A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k.
0, 3, 8, 35, 84, 351, 836, 3479, 8280, 34443, 81968, 340955, 811404, 3375111, 8032076, 33410159, 79509360, 330726483, 787061528, 3273854675, 7791105924, 32407820271, 77123997716, 320804348039, 763448871240, 3175635660123, 7557364714688, 31435552253195
Offset: 0
Examples
b(3)=630 so a(3) = (-1 + sqrt(8*630 + 1))/2 = (-1 + sqrt(5041))/2 = (71 - 1)/2 = 35.
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 (1,10,-10,-1,1).
Programs
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Maple
f := gfun:-rectoproc({a(-2) = -4, a(-1) = -1, a(0) = 0, a(1) = 3, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 26 2020 -
Mathematica
LinearRecurrence[{1,10,-10,-1,1},{0,3,8,35,84},30] (* Harvey P. Dale, Oct 11 2019 *) -
PARI
concat(0, Vec(x*(x^3+3*x^2-5*x-3)/((x-1)*(x^4-10*x^2+1)) + O(x^100))) \\ Colin Barker, May 15 2015
Formula
Let b(n) be A077290. Then a(n) = (-1 + sqrt(8*b(n) + 1))/2.
G.f.: x*(x^3+3*x^2-5*x-3) / ((x-1)*(x^4-10*x^2+1)). - Colin Barker, Mar 09 2012
From Vladimir Pletser, Jul 26 2020: (Start)
a(n) = 10*a(n-2) - a(n-4) + 4 with a(-2)=-4, a(-1)=-1, a(0)=0, a(1)=3.
Can be defined for negative n by setting a(-n) = - a(n-1) - 1 for all n in Z.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). (End)
Comments