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.

A215572 a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=106.

Original entry on oeis.org

2, 5, 106, 550, 6531, 44999, 435973, 3384404, 30252969, 246877464, 2135653370, 17793576423, 151867661753, 1276243154087, 10832435479322, 91356359187721, 773637352766062, 6534137016412674, 55281085635664595, 467187197014742851, 3951025667301212597, 33398969150217473532
Offset: 0

Views

Author

Roman Witula, Aug 16 2012

Keywords

Comments

The Ramanujan-type sequence number 8 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569 for the numbers: 1, 1a, 2, 2a, 3-7 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215569) connected with the following recurrence relation:
(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) are defined in comments to A215560 (see also A215569). It follows that a(n)=ct(3*n+2), at(3*n+2)=bt(3*n+2)=0, which implies the first formula below.
We note that if a(n), a(n+1) and a(n+2) are all odd for some n in N then a(n+3) is even, a(n+4) is odd, a(n+5) and a(n+6) are both even, and the numbers a(n+7), a(n+8), a(n+9) are all odd again. In consequence, this situation hold for every n of the form 7*k+4, k=0,1,..., in the other words cyclical through all sequence a(n), n=4,5,... (from n=1 whenever we start from odd-even-even sequence).

Examples

			From 4*a(1)+5*a(2)=a(3) we obtain 4*((c(1)^4/c(2))^(5/3) + (c(2)^4/c(4))^(5/3) + (c(4)^4/c(1))^(5/3)) + 5*((c(1)^4/c(2))^(8/3) + (c(2)^4/c(4))^(8/3) + (c(4)^4/c(1))^(8/3)) = (4 + 5*c(1)^4/c(2))*((c(1)^4/c(2))^(5/3) + (4 + 5*c(2)^4/c(4))*((c(2)^4/c(4))^(5/3) + (4 + 5*c(4)^4/c(1))*((c(4)^4/c(1))^(5/3) = (c(1)^4/c(2))^(11/3) + (c(2)^4/c(4))^(11/3) + (c(4)^4/c(1))^(11/3) = 550*49^(1/3).
		

References

  • R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3,46,1}, {2,5,106}, 50]
    CoefficientList[Series[(2 - x - x^2)/(1 - 3*x - 46*x^2 - x^3), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)
  • PARI
    Vec((2-x-x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

Formula

49^(1/3)*a(n) = (c(1)^4/c(2))^(n+2/3) + (c(2)^4/c(4))^(n+2/3) + (c(4)^4/c(1))^(n+2/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+2) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+2) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+2).
G.f.: (2-x-x^2)/(1-3*x-46*x^2-x^3).

Extensions

More terms from Michel Marcus, Apr 20 2016