A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 3.

1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 1

Author

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 25 2005

Keywords

• nonn
• easy

Comments

Let M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1 ]; then [1,0,0]*M^n = [a(n), A001353(n), A061278(n-1)] for n > 1. Further, A001353 consists of the first differences of {a(n)}, and since a(n) = A061278(n) + 1, A001353 is also the first differences of A061278. Let v(n) = [1,0,0]*M^n; then, for n >= 0, sum(v_i(n)) = A001075(n) and v_1(n) + v_3(n) = A001835(n). The characteristic polynomial of M is x^3 - 5x^2 + 5x - 1. a(n)/a(n-1) tends to 2 + sqrt(3) = 3.732.... (see A019973) (a root of the polynomial and an eigenvalue of the matrix).

Numbers k such that the RootMeanSquare([1..6*k-5]) is an integer. - Ctibor O. Zizka, Dec 17 2008

Place a(n) blue and b(n) red balls in an urn. Draw 3 balls without replacement. Then Probability(3 red balls) = Probability(1 red and 2 blue balls); binomial(b(n),3) = binomial(b(n),1)*binomial(a(n),2); b(n) = A179167(n). - Paul Weisenhorn, Jul 01 2010

Conjecture: consecutive terms of this sequence and consecutive terms of A032908 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015

Conjecture is true: see StackExchange link. - Robert Israel, Sep 06 2015

Values of a unitary Y-frieze pattern associated to the linearly oriented quiver K3 (i.e., the quiver whose underlying graph is the complete graph on the vertices {1,2,3}, oriented such that i -> j whenever i < j). - Antoine de Saint Germain, Dec 30 2024

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• R. Israel, W. Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Math StackExchange, Sep 1 2015.
• Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
• Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Crossrefs

Cf. A001075, A001353, A001835, A005246, A019973, A032908, A061278, A211955.

Programs

GAP

a:=[1,2,6];; for n in [4..20] do a[n]:=5a[n-1]-5*a[n-2]+a[n-3]; od; a; #
G. C. Greubel, Dec 23 2019

Haskell

a101265 n = a101265_list !! (n-1)
a101265_list = 1 : 2 : 6 : zipWith (+) a101265_list
    (map (* 5) $ tail $ zipWith (-) (tail a101265_list) a101265_list)
-- Reinhard Zumkeller, May 18 2014

Magma

I:=[1,2,6]; [n le 3 select I[n] else 5*Self(n-1) - 5*Self(n-2) + Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 07 2015

Maple

r:=sqrt(3): for n from 1 to 100 do a[n]:=(6+(3+r)*(2+r)^(n-1)+(3-r)*(2-r)^(n-1))/12: end do: # Paul Weisenhorn, Jul 01 2010
r:=sqrt(3): a[n]:=round((6+(3+r)*(2+r)^(n-1))/12): # Paul Weisenhorn, Jul 01 2010
f:= proc(n)
  option remember; local x;
  x:= procname(n-1);
  2*x + (sqrt(12*x^2 - 12*x + 1) - 1)/2
end proc:
f(1):= 1:
map(f, [$1..30]); # Robert Israel, Aug 26 2015
seq( simplify((ChebyshevU(n,2) - Chebyshev(n-1,2) + 1)/2), n=0..20); # G. C. Greubel, Dec 23 2019

Mathematica

LinearRecurrence[{5,-5,1},{1,2,6},25] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[(1-3x+x^2)/((1-x)(1-4x+x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 07 2015 *)
Table[(ChebyshevU[n, 2] - ChebyshevU[n-1, 2] + 1)/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)

PARI

M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1,30,print1(([1,0,0]*M^i)[1],","))

PARI

{a(n)=polcoeff(x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)+x*O(x^n)),n)}

PARI

{a(n)=if(n==0,1,if(n==1,1,a(n-1)*(a(n-1)+1)/a(n-2)))} /* Paul D. Hanna, Apr 08 2012 */

PARI

vector(21, n, (polchebyshev(n, 2, 2) - polchebyshev(n-1, 2, 2) + 1)/2 ) \\ G. C. Greubel, Dec 23 2019

Sage

[(chebyshev_U(n,2) - chebyshev_U(n-1,2) + 1)/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019

Formula

a(n) = A005246(n)*A005246(n+1). a(n+1) = a(n)*(a(n)+1)/a(n-1). - Franklin T. Adams-Watters, Apr 24 2006

a(n) = (A001835(n) + 1) / 2. - Ralf Stephan, May 16 2007

O.g.f.: x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)). - R. J. Mathar, Aug 22 2008

a(n) = 1 + A061278(n). - Ctibor O. Zizka, Dec 17 2008

a(n) = 4*a(n-1) - a(n-2) - 1. - N. Sato, Jan 21 2010

a(n) = (6+(3+r)*(2+r)^(n-1) + (3-r)*(2-r)^(n-1))/12; r=sqrt(3). - Paul Weisenhorn, Jul 01 2010

a(n+1) = a(n) * (a(n) + 1) / a(n-1) for n>1 with a(0)=1, a(1)=1. - Paul D. Hanna, Apr 08 2012

From Peter Bala, May 01 2012: (Start)

a(n+1) = 1 + Sum {k = 1..n} 2^(k-1)*binomial(n+k,2*k).

Row sums of A211955.

a(n) = T(n,u)*T(n+1,u)/u with u = sqrt(3) and T(n,x) denotes the Chebyshev polynomial of the first kind.

Sum_{n >= 0} 1/a(n) = sqrt(3). In fact, 3 - (Sum_{n = 0..2*N} 1/a(n))^2 = 2/(A001835(N+1))^2 and 3 - (Sum_{n = 0..2*N+1} 1/a(n))^2 = 3/(A001075(N+1))^2. (End)

From Robert Israel, Aug 26 2015: (Start)

(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 6 * a(n) * a(n+1).

a(n+1) = 2*a(n) + (sqrt(12*a(n)^2 - 12*a(n) + 1) - 1)/2. (End)

a(n) = (ChebyshevU(n, 2) - ChebyshevU(n-1, 2) + 1)/2 = (ChebyshevT(n, 2) + ChebyshevU(n, 2) + 2)/4. - G. C. Greubel, Dec 23 2019

a(n) = (1+a(n-1))*(1+a(n-2))/a(n-3) for n > 3. - Antoine de Saint Germain, Dec 30 2024

Extensions

a(26)-a(27) from Vincenzo Librandi, Sep 07 2015