cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A238536 A fourth-order linear divisibility sequence related to the Fibonacci numbers: a(n) = (1/2)*Fibonacci(3*n)*Lucas(n).

Original entry on oeis.org

1, 12, 68, 504, 3355, 23256, 158717, 1089648, 7463884, 51170460, 350695511, 2403786672, 16475579353, 112925875764, 774003961940, 5305106018016, 36361727272627, 249227013404808, 1708227291909269, 11708364225400920, 80250321774226396, 550043889533755332, 3770056901455017263
Offset: 1

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Author

Peter Bala, Feb 28 2014

Keywords

Comments

Let P and Q be integers. The Lucas sequences U(n) and V(n) (which depend on P and Q) are a pair of integer sequences that satisfy the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1 and V(0) = 2, V(1) = P, respectively. The sequence {U(n)} n >= 1 is a linear divisibility sequence of order 2, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0. In general, V(n) is not a divisibility sequence. However, it can be shown that if p >= 3 is an odd integer then the sequence {U(p*n)*V(n)} n >= 1 is a linear divisibility sequence of order 4. For a proof and a generalization of this result see the Bala link. Here we take p = 3 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, V(n) is the sequence of Lucas numbers, A000032, and normalize the sequence to have the initial term 1. For other sequences of this type see A238537 and A238538.

References

  • S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80.

Crossrefs

Programs

  • Magma
    I:=[12, 68, 504, 3355]; [1] cat [n le 4 select I[n] else 4*Self(n-1) + 19*Self(n-2) + 4*Self(n-3) - Self(n-4): n in [1..30]]; // G. C. Greubel, Dec 25 2017
  • Maple
    with(combinat): lucas:= n->fibonacci(n+1)+ fibonacci(n-1):
    seq(1/2*lucas(n)*fibonacci(3*n), n = 1..24);
  • Mathematica
    Table[Fibonacci(3*n)*Lucas(n)/2, {n,1,30}] (* or *) Join[{1}, LinearRecurrence[{4,19,4,-1}, {12, 68, 504, 3355}, 30]] (* G. C. Greubel, Dec 25 2017 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,4,19,4]^(n-1)*[1;12;68;504])[1,1] \\ Charles R Greathouse IV, Oct 07 2016
    

Formula

a(n) = (1/2)*Fibonacci(3*n)*Lucas(n) = (1/2)*A000045(3*n)*A000032(n).
a(n) = (1/2)*Fibonacci(2*n)*Fibonacci(3*n)/Fibonacci(n).
a(n) = (1/(2*sqrt(5)))*( ((7 + 3*sqrt(5))/2)^n - ((7 - 3*sqrt(5))/2)^n + (-1)^n*((3 + sqrt(5))/2)^n - (-1)^n*((3 - sqrt(5))/2)^n ).
The sequence can be extended to negative indices by setting a(-n) = -a(n).
O.g.f. x*(1 + 8*x + x^2)/( (1 + 3*x + x^2)*(1 - 7*x + x^2) ).
Recurrence equation: a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/2) * (Fibonacci(4*n) + (-1)^n*Fibonacci(2*n)). - Ralf Stephan, Mar 01 2014

A238537 A fourth-order linear divisibility sequence related to the Pell numbers.

Original entry on oeis.org

1, 42, 1379, 47124, 1599205, 54335358, 1845747527, 62701403688, 2130000094537, 72357312787410, 2458018570699691, 83500274463891516, 2836551311028252973, 96359244313163973414, 3273377755262716618895, 111198484435049515150416, 3777475093033912744231057
Offset: 1

Views

Author

Peter Bala, Feb 28 2014

Keywords

Comments

Let P and Q be integers. The Lucas sequences U(n) and V(n) (which depend on P and Q) are a pair of integer sequences that satisfy the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1 and V(0) = 2, V(1) = P, respectively. The sequence {U(n)}n>=1 is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0. In general the sequence V(n) is not a divisibility sequence. However, it can be shown that if p >= 3 is an odd integer then the sequence {U(p*n)*V(n)}n>=1 is a divisibility sequence satisfying a linear recurrence of order 4. For a proof and a generalization of this result see the Bala link. Here we take p = 3 with P = 2 and Q = -1, for which U(n) is the sequence of Pell numbers A000129, and consider the normalized divisibility sequence with initial term equal to 1. For other sequences of this type see A238536 and A238538

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{28, 202, 28, -1}, {1, 42, 1379, 47124}, 17] (* Jean-François Alcover, Nov 02 2019 *)

Formula

a(n) = (1/5)*A000129(3*n)*A001333(n).
a(n) = (1/(20*sqrt(2)))*((1 + sqrt(2))^(3*n) - (1 - sqrt(2))^(3*n))*( (1 + sqrt(2))^n + (1 - sqrt(2))^n ).
O.g.f.: x*(1 + 14*x + x^2)/( (1 + 6*x + x^2)*(1 - 34*x + x^2) ).
Recurrence equation: a(n) = 28*a(n-1) + 202*a(n-2) + 28*a(n-3) - a(n-4).
a(n) = (1/10) * (Pell(4n) + (-1)^n*Pell(2n)), with Pell(n) = A000129(n). - Ralf Stephan, Mar 01 2014

A238540 A fourth-order linear divisibility sequence: a(n) := (3^n + 1)*(3^(3*n) - 1)/( (3 + 1)*(3^3 - 1)).

Original entry on oeis.org

1, 70, 5299, 419020, 33664741, 2719393810, 220069738519, 17820217484440, 1443290970139081, 116902609136432350, 9469004435040169339, 766986472802959676260, 62125826363286791503021, 5032189831214900660779690, 407607319514701058318401759, 33016191346720726553176114480
Offset: 1

Views

Author

Peter Bala, Mar 01 2014

Keywords

Comments

This is a divisibility sequence, that is, if n | m then a(n) | a(m). More generally, the polynomials P(n,x) := (x^n + 1)*(x^(3*n) - 1) form a sequence of divisibility polynomials in the polynomial ring Z[x]; that is, if n divides m then P(n,x) divides P(m,x) in Z[x]. See the Bala link for a proof and generalization. Here we consider the integer sequence coming from the normalized polynomials P(n,x)/P(n,1) at x = 3.
The sequence satisfies a homogeneous linear recurrence of the fourth order. However, it does not belong to the family of linear divisibility sequences of the fourth order discovered by Williams and Guy, which have o.g.f.s of the form x*(1 - q*x^2)/Q(x), Q(x) a quartic polynomial and q an integer parameter.
For sequences of a similar type see A238536 through A238541.

Crossrefs

Programs

  • Maple
    #A238540
    seq(1/104*(3^n + 1)*(3^(3*n) - 1), n = 1..20);
  • Mathematica
    LinearRecurrence[{112, -2622, 9072, -6561}, {1, 70, 5299, 419020}, 16] (* Jean-François Alcover, Nov 14 2019 *)

Formula

a(n) = (1/104)*(3^n + 1)*(3^(3*n) - 1) = (1/104)*(9^n - 1)*(27^n - 1)/(3^n - 1).
O.g.f.: x*(1 - 42*x + 81*x^2)/((1 - x)*(1 - 3*x)*(1 - 27*x)*(1 - 81*x)).
Recurrence equation: a(n) = 112*a(n-1) - 2622*a(n-2) + 9072*a(n-3) - 6561*a(n-4).

A238541 A fourth-order linear divisibility sequence: a(n) := A(n)/A(1) where A(n) := ( (3^n + 2^n)*(3^(3*n) - 2^(3*n)) ).

Original entry on oeis.org

1, 91, 7063, 538447, 41441455, 3231753343, 254851186927, 20265345051679, 1621012954550479, 130194036583465855, 10485834936321976111, 846117830539227426271, 68360837263665964839823, 5527792975131721247371327, 447241733557623755497669615
Offset: 1

Views

Author

Peter Bala, Mar 01 2014

Keywords

Comments

The bivariate polynomials P(n,x,y) := (x^n + y^n)*(x^(3*n) - y^(3*n)) form a sequence of divisibility polynomials in the polynomial ring Z[x,y]; that is, if n divides m then P(n,x,y) divides P(m,x,y) in Z[x,y] (see the Bala link). Here we consider the integer sequence coming from the normalized polynomials P(n,x,y)/P(1,x,y) when x = 3 and y = 2. Other cases include A238538(x = 2, y = 1), A238539(x = 2, y = -1) and A238540(x = 3, y = 1). See also A238536, A238537 and A215466.

Crossrefs

Programs

  • Maple
    #A238541
    seq(1/95*(3^n + 2^n)*(3^(3*n) - 2^(2*n)), n = 1..20);
  • Mathematica
    LinearRecurrence[{175,-10158,226800,-1679616},{1,91,7063,538447},20] (* Harvey P. Dale, Apr 12 2018 *)

Formula

a(n) = (1/95)*(3^n + 2^n)*(3^(3*n) - 2^(3*n)).
a(n) = (1/95)*(9^n - 4^n)*(27^n - 8^n)/(3^n - 2^n).
O.g.f.: x*(1 - 84*x + 1296*x^2)/((1 - 16*x)*(1 - 24*x)*(1 - 54*x)*(1 - 81*x)).
Recurrence equation: a(n) = 175*a(n-1) - 10158*a(n-2) + 226800*a(n-4) - 1679616*a(n-4).

A238539 A fourth-order linear divisibility sequence: a(n) := (1/9)*(2^n + (-1)^n)*(2^(3*n) - (-1)^n).

Original entry on oeis.org

1, 35, 399, 7735, 112871, 1893255, 29593159, 479082695, 7620584391, 122287263175, 1953732901319, 31282632909255, 500338874618311, 8006888009380295, 128098480026087879, 2049669505409577415, 32793961486615474631, 524709388585350492615, 8395302178969583120839
Offset: 1

Views

Author

Peter Bala, Mar 01 2014

Keywords

Comments

This is a divisibility sequence, that is, if n | m then a(n) | a(m). This is a consequence of the following more general result: The polynomials P(n,x,y) := (x^n + y^n)*(x^(3*n) - y^(3*n)) form a divisibility sequence in the polynomial ring Z[x,y]. See the Bala link.
The sequence satisfies a homogeneous linear recurrence of the fourth order. However, it does not belong to the family of linear divisibility sequences of the fourth order studied by Williams and Guy, which have o.g.f.s of the form x*(1 - q*x^2)/Q(x), Q(x) a quartic polynomial and q an integer.
For sequences of a similar type see A238536 through A238541.

Crossrefs

Programs

  • Maple
    seq(1/9*(2^n + (-1)^n)*(2^(3*n) - (-1)^n), n = 1..20);

Formula

a(n) = (1/9)*(2^n + (-1)^n)*(2^(3*n) - (-1)^n) = (1/9)*(4^n - 1)*(8^n - (-1)^n)/(2^n - (-1)^n).
O.g.f.: x*(1 + 28*x + 16*x^2)/((1 - x)*(1 + 2*x)*(1 + 8*x)*(1 - 16*x)).
Recurrence equation: a(n) = 7*a(n-1) + 138*a(n-2) + 112*a(n-4) - 256*a(n-4).
Showing 1-5 of 5 results.