A214683 a(n+3) = -a(n+2) + 2a(n+1) + a(n) with a(0)=-1, a(1)=0, a(2)=-3.

-1, 0, -3, 2, -8, 9, -23, 33, -70, 113, -220, 376, -703, 1235, -2265, 4032, -7327, 13126, -23748, 42673, -77043, 138641, -250054, 450293, -811760, 1462292, -2635519, 4748343, -8557089, 15418256, -27784091, 50063514, -90213440, 162556377, -292919743

Offset: 0

Author

Roman Witula, Jul 25 2012

Keywords

• sign
• easy

Comments

Ramanujan-type sequence number 1 for the argument 2Pi/7.

The discussed sequence is associated with the sequence A006053 (with respect to the similar trigonometric formulas describing both sequences). Indeed, we have 7^(1/3)*a(n) = (c(1)/c(2))^(1/3)*(2c(1))^n + (c(2)/c(4))^(1/3)*(2c(2))^n + (c(4)/c(1))^(1/3)*(2c(4))^n = (c(1)/c(2))^(1/3)*(2c(2))^(n+1) + (c(2)/c(4))^(1/3)*(2c(4))^(n+1) + (c(4)/c(1))^(1/3)*(2c(1))^(n+1), where c(j) := Cos(2Pi*j/7), which is "almost" the copy of the respective formula for A006053. From a(0), A006053(0) and a(1), A006053(1), (and again) A006053(0) we deduce the following attractive decompositions

x^3 - 7^(1/3)*x - 1 = (x - (c(1)/c(4))^(1/3))*(x - (c(2)/c(1))^(1/3))*(x - (c(4)/c(2))^(1/3)), and

x^3 - 49^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3)*2c(1))*(x - (c(2)/c(4))^(1/3)*2c(2))*(x - (c(4)/c(1))^(1/3)*2c(4)).

From Newton-Girard formulas applied to these polynomials we generate two new sequences of real numbers S(n) := (c(1)/c(4))^(n/3) + (c(2)/c(1))^(n/3) + (c(4)/c(2))^(n/3), and T(n) := ((c(1)/c(2))^(1/3)*2c(1))^n + ((c(2)/c(4))^(1/3)*2c(2))^n + ((c(4)/c(1))^(1/3)*2c(4))^n. In first Witula's paper it is proved that S(n) = as(n) + bs(n)*7^(1/3) + cs(n)*49^(1/3), where as(n+3) = as(n) + 7cs(n+1), bs(n+3) = bs(n) + as(n+1), cs(n+3) = cs(n) + bs(n+1), as(0)=3, as(1)=as(2)=bs(0)=bs(1)=0, bs(2)=2, cs(0)=cs(1)=cs(2)=0, and T(n) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where at(n+3) = at(n) + 7bt(n+1), bt(n+3) = bt(n) + 7ct(n+1), ct(n+3) = ct(n) + at(n+1), at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2. All six sequences as(n),bs(n),...,ct(n) are created from integers and will be discussed in separate sequences .

Examples

			From values of a(k), for k=0,1,..,5 we deduce that (c(1)/c(2))^(1/3)*A + (c(2)/c(4))^(1/3)*B + (c(4)/c(1))^(1/3)*C = 0  in the following cases: A=2c(1), B=2c(2), C=2c(4) or A=-1+(2c(1))^2+(2c(1))^3, B=-1+(2c(2))^2+(2c(2))^3, C=-1+(2c(3))^2+(2c(3))^3 or  A=1+(2c(1))^4+(2c(1))^5, B=1+(2c(2))^4+(2c(2))^5, C=1+(2c(3))^4+(2c(3))^5.
		

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.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
• Roman Witula, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.
• Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.
• Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
• Index entries for linear recurrences with constant coefficients, signature (-1,2,1).

Crossrefs

Cf. A006053.

Programs

Magma

a:=[-1,0,-3]; [ n le 3 select a[n] else -Self(n-1) + 2*Self(n-2) + Self(n-3): n in [1..35]]; // Marius A. Burtea, Oct 03 2019

Mathematica

LinearRecurrence[{-1, 2, 1}, {-1, 0, -3}, 40]

SageMath

@CachedFunction
def a(n): # a = A214683
    if (n<3): return (-1,0,-3)[n]
    else: return -a(n-1) + 2*a(n-2) + a(n-3)
[a(n) for n in range(40)] # G. C. Greubel, Nov 25 2022

Formula

a(n+3) + a(n+2) - 2a(n+1) - a(n) = 0, a(0)=-1, a(1)=0, a(2)=-3.

G.f.: -(1+x+x^2)/(1+x-2*x^2-x^3).