A276122 -id:A276122 - cp's OEIS frontend

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

1, 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: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 28 2005

Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is an integer. For this sequence K=4. - Andrey Vyshnevyy, Sep 18 2015
The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - N. J. A. Sloane, Sep 08 2018

Seiichi Manyama, Table of n, a(n) for n = 0..1750
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A061278, A001353, A001835, A005246, A276122, A276271.

I:=[1,1,2]; [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 18 2015

LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* Vincenzo Librandi, Sep 18 2015 *)
CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 11 2016 *)
a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* Michael Somos, Jul 09 2017 *)

M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))

{a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* Michael Somos, Jul 09 2017 */

a(n) = A101265(n), n>0. - R. J. Mathar, Aug 30 2008
a(n) = A079935(n+1) - A001571(n). - Gerry Martens, Jun 05 2015
a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - Seiichi Manyama, Aug 11 2016
From Ilya Gutkovskiy, Aug 11 2016: (Start)
G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
a(n) = 4*a(n-1) - a(n-2) - 1. - Seiichi Manyama, Aug 26 2016
From Seiichi Manyama, Sep 03 2016: (Start)
a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
a(n) = A005246(n)*A005246(n+1). (End)
From Michael Somos, Jul 09 2017: (Start)
0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)

a(26)-a(27) from Vincenzo Librandi, Sep 18 2015

A276160 A recurrence of order 3 : a(0)=a(1)=a(2)=1 ; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-1) + a(n-2) + 1)/a(n-3).

Original entry on oeis.org

1, 1, 1, 5, 33, 1153, 266337, 2149605893, 4007637093066433, 60303882185826956720761345, 1691732525726797389070758961468800814420801, 714126272449521825808382965880022542720530687818734820147878380094981

Offset: 0

Seiichi Manyama, Aug 22 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..16

Cf. A101368, A276122.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 1] + a[n - 2] + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 22 2016 *)
nxt[{a_,b_,c_}]:={b,c,(c^2+b^2+c+b+1)/a}; NestList[nxt,{1,1,1},15][[All,1]] (* Harvey P. Dale, Sep 16 2021 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(0){|s, i| s + i * i} + a[1..-1].inject(:+) + 1
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276160(n)
  A(3, n)
end

a(n) = 7*a(n-1)*a(n-2) - a(n-3) - 1.

A276271 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)+a(n-2)+a(n-3))/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 6, 46, 2206, 4870846, 3954191749561, 339905052007042640998641, 52373274877565894156748130733610185904753361, 563138297002425210235477817802336090254190075906443582099838858026136728896536841

Offset: 0

Seiichi Manyama, Aug 26 2016

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..15

Cf. A101879, A276122.

nxt[{a_,b_,c_,d_}]:={b,c,d,(b^2+c^2+d^2+b+c+d)/a}; NestList[nxt,{1,1,1,1},12][[All,1]] (* Harvey P. Dale, Mar 10 2017 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(0){|s, i| s + i * i} + a[1..-1].inject(:+)
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A276271(n)
  A(4, n)
end

a(n) = 8*a(n-1)*a(n-2)*a(n-3)-a(n-4)-1.

a(n) ~ 2^(-3/2) * c^(d^n), where c = 1.2578918597... and d = A058265 = 1.83928675521416... = (1 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3))/3. - Vaclav Kotesovec, Mar 20 2017

cp's OEIS Frontend

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

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A276160 A recurrence of order 3 : a(0)=a(1)=a(2)=1 ; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-1) + a(n-2) + 1)/a(n-3).

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A276271 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)+a(n-2)+a(n-3))/a(n-4).

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Author

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Formula

cp's OEIS Frontend

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A276160 A recurrence of order 3 : a(0)=a(1)=a(2)=1 ; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-1) + a(n-2) + 1)/a(n-3).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Ruby

Formula

A276271 a(0) = a(1) = a(2) = a(3) = 1; for n > 3, a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)+a(n-2)+a(n-3))/a(n-4).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Ruby

Formula

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).