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 1-2 of 2 results.

A248417 Sum of n-th powers of the roots of x^3 +25* x^2 + 31*x - 1.

Original entry on oeis.org

3, -25, 563, -13297, 314947, -7460905, 176745971, -4187046273, 99189570819, -2349764090041, 55665038509363, -1318684086371985, 31239136201419331, -740043533319442377, 17531356426655688179, -415311321997288071457, 9838570957172556010499, -233072091590971314359129, 5521391278779936334581299
Offset: 0

Views

Author

Kai Wang, Jul 01 2016

Keywords

Comments

This is the other half of A274592.
a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial
x^3 +25* x^2 + 31*x - 1.
x1 = (tan(2*Pi/7)*tan(4*Pi/7))/(tan(Pi/7))^2,
x2 = (tan(4*Pi/7)*tan(Pi/7))/(tan(2*Pi/7))^2,
x3 = (tan(Pi/7)*tan(2*Pi/7))(tan(4*Pi/7))^2.

Links

Crossrefs

Cf. A274592.

Programs

  • Mathematica
    CoefficientList[Series[(3 + 50 x + 31 x^2)/(1 + 25 x + 31 x^2 - x^3), {x, 0, 18}], x] (* Michael De Vlieger, Jul 01 2016 *)
  • PARI
    Vec((3+50*x+31*x^2)/(1+25*x+31*x^2-x^3) + O(x^20)) \\ Colin Barker, Jul 01 2016
  • PARI
    polsym(x^3 +25* x^2 + 31*x - 1, 30) \\ Charles R Greathouse IV, Jul 20 2016

Formula

a(n) = ((tan(Pi/7))^2/(tan(2*Pi/7)*tan(4*Pi/7)))^(-n)+((tan(2*Pi/7))^2/(tan(4*Pi/7)*tan(Pi/7)))^(-n)+((tan(4*Pi/7))^2/(tan(Pi/7)*tan(2*Pi/7)))^(-n).
a(n) = -25*a(n-1) - 31*a(n-2) + a(n-3).
G.f.: (3+50*x+31*x^2) / (1+25*x+31*x^2-x^3). - Colin Barker, Jul 01 2016

A274663 Sum of n-th powers of the roots of x^3 + 4*x^2 - 11*x - 1.

Original entry on oeis.org

3, -4, 38, -193, 1186, -6829, 40169, -234609, 1373466, -8034394, 47011093, -275049240, 1609284589, -9415668903, 55089756851, -322322100748, 1885860059450, -11033893589177, 64557712909910, -377717821061137, 2209972232664381, -12930227249420121
Offset: 0

Views

Author

Kai Wang, Jul 01 2016

Keywords

Comments

This is half of a two sided sequences.
The other half is A274664. - Kai Wang, Aug 02 2016
a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 + 4*x^2 - 11*x - 1.
x1 = (cos(Pi/7))^2/(cos(2*Pi/7)*cos(4*Pi/7)),
x2 = -(cos(2*Pi/7))^2/(cos(4*Pi/7)*cos(Pi/7)),
x3 = -(cos(4*Pi/7))^2/(cos(Pi/7) *cos(2*Pi/7)).

Links

Crossrefs

Cf. A215076, A248417, A274220, A274592, A274664.

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 3, a[1] == -4, a[2] == 38, a[n] == -4 a[n - 1] + 11 a[n - 2] + a[n - 3]}, a, {n, 0, 20}] (* Michael De Vlieger, Jul 02 2016 *)
LinearRecurrence[{-4,11,1},{3,-4,38},30] (* Harvey P. Dale, Dec 28 2022 *)
  • PARI
    polsym(x^3 + 4*x^2 - 11*x - 1, 21)
  • PARI
    Vec((3+8*x-11*x^2)/(1+4*x-11*x^2-x^3) + O(x^99)) \\ Altug Alkan, Jul 08 2016

Formula

a(n) = ((cos(Pi/7))^2/(cos(2*Pi/7)*cos(4*Pi/7)))^n + (-(cos(2*Pi/7))^2/(cos(4*Pi/7)*cos(Pi/7)))^n + (-(cos(4*Pi/7))^2/(cos(Pi/7)*cos(2*Pi/7)))^n.
a(n) = -4*a(n-1) + 11*a(n-2) + a(n-3) for n>2.
G.f.: (3+8*x-11*x^2)/(1+4*x-11*x^2-x^3). - Wesley Ivan Hurt, Jul 02 2016
a(n) = (-1/8)^(-n)*cos(Pi/7)^(3*n) + (-8)^n*sin(Pi/14)^(3*n) +
8^n*sin(3*Pi/14)^(3*n). - Wesley Ivan Hurt, Jul 11 2016
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.