A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

-1, 1, 8, 34, 131, 493, 1844, 6886, 25703, 95929, 358016, 1336138, 4986539, 18610021, 69453548, 259204174, 967363151, 3610248433, 13473630584, 50284273906, 187663465043, 700369586269, 2613814880036, 9754889933878, 36405744855479, 135868089488041

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 14 2003

Keywords

• easy
• sign

Comments

Apart from the initial -1, these are the numbers k such that the triangular number k*(k + 1)/2 is the sum of three consecutive triangular numbers - see A129803. - Brian Nowell, Nov 03 2009

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Crossrefs

Cf. A001571, A071954.

Programs

GAP

a:=[-1,1,8];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019

Magma

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( -(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2)) )); // G. C. Greubel, Feb 25 2019

Mathematica

CoefficientList[Series[(-1+6x-2x^2)/((1-x)(1-4x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 15 2014 *)
LinearRecurrence[{5,-5,1}, {-1,1,8}, 30] (* G. C. Greubel, Feb 25 2019 *)

PARI

is(n)=ispolygonal(3/2*n*(n+1)+4,3) || n==-1 \\ Charles R Greathouse IV, Apr 14 2014

PARI

my(x='x+O('x^30)); Vec(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

Sage

(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

Formula

a(n) = A001571(n) - 1. - N. J. A. Sloane, Nov 03 2009

G.f.: -(1 -6*x +2*x^2)/((1 - x)*(1 - 4*x + x^2)).

a(n) = -3/2 + (1/12)*( (a -2*b +5)*a^n + (b -2*a +5)*b^n ), with a = 2 + sqrt(3), b = 2 - sqrt(3):.

a(n) = -3/2 + (3/4)*A003500(n) - (1/4)*A003500(n-1).

a(n) = (1/2)*(A001834(n) - 3).

E.g.f.: ((1 + sqrt(3))*exp((2 + sqrt(3))*x) + (1 - sqrt(3))*exp((2 - sqrt(3))*x) - 6*exp(x))/4. - Franck Maminirina Ramaharo, Nov 12 2018