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.

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A238600 A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/6)*Fibonacci(3*n)*Fibonacci(4*n)/Fibonacci(n).

Original entry on oeis.org

1, 28, 408, 7896, 137555, 2496144, 44599477, 801617712, 14375440584, 258018516140, 4629531440711, 83076469908768, 1490726895438793, 26750144944686436, 480010941060482040, 8613453244178393184, 154562103244937408987, 2773504708179098411952
Offset: 1

Views

Author

Peter Bala, Mar 01 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, and normalize the sequence to have the initial term 1.
For other sequences of this type see A238601, A238602 and A238603. See also A238536.

Links

Crossrefs

Cf. A000045, A215466, A238536, A238601, A238602, A238603.

Programs

  • Magma
    [(Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n))/6: n in [1..30]]; // G. C. Greubel, Aug 07 2018
  • Maple
    with(combinat):
seq(1/6*fibonacci(3*n)*fibonacci(4*n)/fibonacci(n), n = 1..20);
  • Mathematica
    Table[(1/6)*(Fibonacci[2*n] + (-1)^n*Fibonacci[4*n] + Fibonacci[6*n]), {n, 1, 500}] (* G. C. Greubel, Aug 07 2018 *)
LinearRecurrence[{14,90,-350,90,14,-1},{1,28,408,7896,137555,2496144},20] (* Harvey P. Dale, Aug 26 2020 *)
  • PARI
    vector(30, n, (fibonacci(2*n) + (-1)^n*fibonacci(4*n) + fibonacci(6*n))/6) \\ G. C. Greubel, Aug 07 2018

Formula

a(n) = (1/6)*Fibonacci(3*n)*Fibonacci(4*n)/Fibonacci(n).
a(n) = (1/6)*( Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n) ).
The sequence can be extended to negative indices when a(-n) = -a(n).
O.g.f. x*(1 + 14*x - 74*x^2 + 14*x^3 + x^4)/( (1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) ).
Recurrence equation: a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6).

A238602 A sixth-order linear divisibility sequence related to the Pell numbers: a(n) := (1/60)*Pell(3*n)*Pell(4*n)/Pell(n).

Original entry on oeis.org

1, 238, 45507, 9063516, 1792708805, 355009117386, 70287911575687, 13916722851826872, 2755438412296182921, 545562971271797876390, 108018710075587599558731, 21387159127038457710621972, 4234549485214861760195346253, 838419411023095574089504928386
Offset: 1

Views

Author

Peter Bala, Mar 06 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 2 and Q = -1, for which U(n) is the sequence of Pell numbers, A000129, and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.
For other sequences of this type see A238600, A238601 and A238603. See also A238536.

Links

Crossrefs

Cf. A000129, A238536, A238600, A238601, A238603.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 +68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1+34*x+x^2)*(1-198*x+x^2)))); // G. C. Greubel, Aug 07 2018
  • Mathematica
    Table[(1/60)*(Fibonacci[2*n, 2] + (-1)^n*Fibonacci[4*n, 2] + Fibonacci[6*n, 2]), {n, 1, 50}] (* G. C. Greubel, Aug 07 2018 *)
  • PARI
    x='x+O('x^30); Vec(x*(1+68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1 + 34*x+x^2)*(1-198*x+x^2))) \\ G. C. Greubel, Aug 07 2018

Formula

a(n) = (1/60)*( Pell(2*n) + (-1)^n*Pell(4*n) + Pell(6*n) ).
The sequence can be extended to negative indices using a(-n) = -a(n).
O.g.f. x*(1 + 68*x - 698*x^2 + 68*x^3 + x^4)/( (1 - 6*x + x^2)*(1 + 34*x + x^2)*(1 - 198*x + x^2) ).
Recurrence equation: a(n) = 170*a(n-1) + 5745*a(n-2) - 40052*a(n-3) + 5745*a(n-4) + 170*a(n-5) - a(n-6).

A238603 A sixth-order linear divisibility sequence related to A000225: a(n) := (1/105)*(2^(3*n) - 1)*(2^(4*n) - 1)/(2^n - 1).

Original entry on oeis.org

1, 51, 2847, 170391, 10555655, 664857063, 42215949223, 2691226507047, 171901443816999, 10990938133564455, 703076406514657319, 44985901769992495143, 2878746218051469266983, 184228512166784552153127, 11790264946382521291370535, 754565442462197107544125479
Offset: 1

Views

Author

Peter Bala, Mar 06 2014

Keywords

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.
It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.
Here we take p = 3 and q = 4 with P = 3 and Q = 2, for which U(n) is the sequence A000225 (sometimes called the Mersenne numbers), and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.
For other sequences of this type see A238600, A238601 and A238602. See also A238536.

Examples

			G.f. = x + 51*x^2 + 2847*x^3 + 170391*x^4 + 10555655*x^5 + 664857063*x^6 + ... - _Michael Somos_, May 07 2017

Links

Crossrefs

Cf. A000225, A238536, A238600, A238601, A238602.

Programs

  • Magma
    [(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1): n in [1..50]]; // G. C. Greubel, Aug 07 2018
  • Maple
    seq(1/105*(2^(3*n)-1)*(2^(4*n)-1)/(2^n-1), n = 1..20);
  • Mathematica
    Table[(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1), {n, 1, 50}] (* G. C. Greubel, Aug 07 2018 *)
  • PARI
    {a(n) = if( n, (8^n - 1) * (16^n - 1) / (105 * (2^n - 1)), 0)}; /* Michael Somos, May 07 2017 */

Formula

a(n) = (1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1).
O.g.f.: x*(4096*x^4 - 4352*x^3 + 1160*x^2 - 68*x + 1 )/( (1-x)*(1-2*x)(1-4*x)*(1-16*x)*(1-32*x)*(1-64*x) ).
The formula for a(n) may be used to define it for all n in Z, and then we have a(n) = -(64)^n * a(-n). - Michael Somos, May 07 2017
Showing 1-3 of 3 results.

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