A077399 Triangular numbers that are 1/7 of triangular numbers.
0, 3, 15, 780, 3828, 198135, 972315, 50325528, 246964200, 12782485995, 62727934503, 3246701117220, 15932648399580, 824649301287903, 4046829965558835, 209457675826010160, 1027878878603544528, 53201425010505292755, 261077188335334751295, 13512952494992518349628
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..832
- 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 (1,254,-254,-1,1).
Programs
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Magma
Q:=Rationals(); R
:=PowerSeriesRing(Q,30); [0] cat Coefficients(R!(-3*x*(x^2+4*x+1)/((x-1)*(x^2-16*x+1)*(x^2+16*x+1)))); // G. C. Greubel, Jan 18 2018 -
Maple
f := gfun:-rectoproc({a(-2) = 3, a(-1) = 0, a(0) = 0, a(1) = 3, a(n) = 254*a(n-2)-a(n-4)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] #Vladimir Pletser, Feb 21 2021 -
Mathematica
Select[Accumulate[Range[0,5700000]],IntegerQ[(Sqrt[56#+1]-1)/2]&] (* Harvey P. Dale, Jan 18 2013 *) LinearRecurrence[{1,254,-254,-1,1},{0,3,15,780,3828}, 30] (* G. C. Greubel, Jan 18 2018 *)
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PARI
concat(0, Vec(-3*x*(x^2+4*x+1)/((x-1)*(x^2-16*x+1)*(x^2+16*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015
Formula
Let b(n) be A077398(n) then a(n) = b(n)*(b(n)+1)/2.
G.f.: -3*x*(x^2+4*x+1) / ((x-1)*(x^2-16*x+1)*(x^2+16*x+1)). - Colin Barker, Jul 02 2013
Comment from _Vladimir Pletser, Feb 21 2021: (Start)
a(n) = 254 a(n - 2) - a (n - 4) + 18.
a(n) = a(n - 1) + 254 (a(n - 2) - a(n - 3)) - (a (n - 4) - a(n - 5)). (End)