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 1-7 of 7 results.

A215664 a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.

Original entry on oeis.org

3, 0, 6, -3, 18, -15, 57, -63, 186, -246, 621, -924, 2109, -3393, 7251, -12288, 25146, -44115, 87726, -157491, 307293, -560199, 1079370, -1987890, 3798309, -7043040, 13382817, -24927429, 47191491, -88165104, 166501902, -311686803, 587670810, -1101562311
Offset: 0

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Author

Roman Witula, Aug 20 2012

Keywords

Comments

The Berndt-type sequence number 5 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. The respective sums with negative powers of the cosines form the sequence A215885. Additionally if we set b(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and c(n) = c(4)*c(2)^n + c(1)*c(4)^n + c(2)*c(1)^n, where c(j):=2*cos(2*Pi*j/9), then the following system of recurrence equations holds true: b(n) - b(n+1) = a(n), a(n+1) - a(n) = c(n+1), a(n+2) - 2*a(n)=c(n). All three sequences satisfy the same recurrence relation: X(n+3) - 3*X(n+1) + X(n) = 0. Moreover we have a(n+1) + A215665(n) + A215666(n) = 0 since c(1) + c(2) + c(4) = 0, b(n)=A215665(n) and c(n)=A215666(n).
If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = A215885(n) for every n=0,1,...
From initial values and the recurrence formula we deduce that a(n)/3 and a(3n+1)/9 are all integers. We have a(n)=3*(-1)^n *A188048(n) and a(2n)=A215455(n). Furthermore the following decomposition holds: (X - c(1)^n)*(X - c(2)^n)*(X - c(4)^n) = X^3 - a(n)*X^2 + ((a(n)^2 - a(2*n))/2)*X + (-1)^(n+1), which implies the relation (c(1)*c(2))^n + (c(1)*c(4))^n + (c(2)*c(4))^n = (-c(1))^(-n) + (-c(2))^(-n) + (-c(4))^(-n) = (a(n)^2 - a(2*n))/2.

Examples

			We have c(1)^2 + c(2)^2 + c(4)^2 + 2*(c(1)^3 + c(2)^3 + c(4)^3) = 0 and 3*a(7) + a(8) = a(3).
		

References

  • D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
  • R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,3,-1}, {3,0,6}, 50]
  • PARI
    Vec(3*(1-x^2)/(1-3*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

Formula

a(n) = c(1)^n + c(2)^n + c(4)^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: 3*(1-x^2)/(1-3*x^2+x^3).

A215634 a(n) = - 6*a(n-1) - 9*a(n-2) - 3*a(n-3) with a(0)=3, a(1)=-6, a(2)=18.

Original entry on oeis.org

3, -6, 18, -63, 234, -891, 3429, -13257, 51354, -199098, 772173, -2995218, 11619045, -45073827, 174857211, -678335958, 2631522330, -10208681991, 39603398850, -153636822171, 596016389349, -2312177133105, 8969825761002
Offset: 0

Views

Author

Roman Witula, Aug 18 2012

Keywords

Comments

The Berndt-type sequence number 2 for the argument 2Pi/9 . Similarly like the respective sequence number 1 -- see A215455 -- the sequence a(n) is connected with the following general recurrence relation: X(n+3) + 6*X(n+2) + 9*X(n+1) + ((2*cos(3*g))^2)*X(n) = 0, X(0)=3, X(1)=-6, X(2)=18. The Binet formula for this one has the form: X(n) = (-4)^n*((cos(g))^(2*n) + cos(g+Pi/3))^(2*n) + cos(g-Pi/3))^(2*n)) - for details see Witula-Slota's reference and comments to A215455.
The characteristic polynomial of a(n) has the form x^3 + 6*x^2 + 9*x + 3 = (x + (2*cos(Pi/18))^2)*(x+(2*cos(5*Pi/18))^2)*(x+(2*cos(7*Pi/18))^2). We note that (2*cos(Pi/18))^2 = 2 - c(4), (2*cos(5*Pi/18))^2 = 2 - c(2), and (2*cos(7*Pi/18))^2 = 2 - c(1), where c(j) = 2*cos(2*Pi*j/9) - see trigonometric relations for A215455. Furthermore all numbers a(n)*3^(-ceiling((n+1)/3)) are integers.

References

  • R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Crossrefs

Programs

  • Magma
    I:=[3,-6,18]; [n le 3 select I[n] else -6*Self(n-1)-9*Self(n-2)-3*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Aug 30 2017
  • Mathematica
    LinearRecurrence[{-6,-9,-3}, {3,-6,18}, 50]
    CoefficientList[Series[(3 + 12 x + 9 x^2)/(1 + 6 x + 9 x^2 + 3 x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 30 2017 *)
  • PARI
    Vec((3+12*x+9*x^2)/(1+6*x+9*x^2+3*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012
    

Formula

a(n) = (-4)^n*((cos(Pi/18))^(2*n) + (cos(5*Pi/18))^(2*n) + (cos(7*Pi/18))^(2*n)).
G.f.: (3 + 12*x + 9*x^2)/(1 + 6*x + 9*x^2 + 3*x^3).
a(n)*(-1)^n = s(1)^(2*n) + s(2)^(2*n) + s(4)^(2*n), where s(j) := 2*sin(2*Pi*j/9) -- for the proof see Witula's book. The respective sums with odd powers of sines in A216757 are given. - Roman Witula, Sep 15 2012

A215665 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.

Original entry on oeis.org

0, -3, -3, -9, -6, -24, -9, -66, -3, -189, 57, -564, 360, -1749, 1644, -5607, 6681, -18465, 25650, -62076, 95415, -211878, 348321, -731049, 1256841, -2541468, 4501572, -8881245, 16046184, -31145307, 57019797, -109482105, 202204698, -385466112, 716096199
Offset: 0

Views

Author

Roman Witula, Aug 20 2012

Keywords

Comments

The Berndt-type sequence number 6 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 7) are discussed in A215664 and A215666 - for more details see comments to A215664 and Witula's reference. We have a(n) - a(n+1) = A215664(n).
From initial values and the recurrence formula we deduce that a(n)/3 are all integers.
We note that a(10) is the first element of a(n) which is positive integer and all (-1)^n*a(n+10) are positive integer, which can be obtained from the title recurrence relation.
The following decomposition holds (X - c(1)*c(2)^n)*(X - c(2)*c(4)^n)*(X - c(4)*c(1)^n) = X^3 - a(n)*X^2 - A215917(n-1)*X + (-1)^n.
If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = abs(A215919(n)) = (-1)^n*A215919(n) for every n=0,1,...

Examples

			We have a(1)=a(2)=a(8)=-3, a(3)=a(6)=-9, a(4)+a(11)=-10*a(10), and 47*a(5)=2*a(11).
		

References

  • R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math., (in press, 2012).
  • D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,3,-1}, {0,-3,-3}, 50]
  • PARI
    concat(0,Vec(-3*(1+x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

Formula

a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: -3*x*(1+x)/(1-3*x^2+x^3).

A215666 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=-3, and a(2)=6.

Original entry on oeis.org

0, -3, 6, -9, 21, -33, 72, -120, 249, -432, 867, -1545, 3033, -5502, 10644, -19539, 37434, -69261, 131841, -245217, 464784, -867492, 1639569, -3067260, 5786199, -10841349, 20425857, -38310246, 72118920, -135356595, 254667006, -478188705, 899357613
Offset: 0

Views

Author

Roman Witula, Aug 20 2012

Keywords

Comments

The Berndt-type sequence number 7 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 6) are discussed in A215664 and A215665 - for more details see comments to A215664 and Witula's reference. We have a(n) = A215664(n+2) - 2*A215664(n) and a(n+1) = A215664(n+1) - A215664(n).
From initial values and the title recurrence formula we deduce that a(n)/3 and a(3*n)/9 are all integers.
If we set X(n) = 3*X(n-2) - X(n-3), n in Z, with a(n) = X(n), for every n=0,1,..., then X(-n) = -abs(A215917(n)) = (-1)^n*A215917(n), for every n=0,1,...

Examples

			We have 8*a(3)+a(6)=5*a(6)+3*a(7)=0, a(5) + a(12) = 3000, and (a(30)-1000*a(10)-a(2))/10^5 is an integer. Further we obtain  c(4)*cos(4*Pi/7)^7 + c(1)*cos(8*Pi/7)^7 + c(2)*c(2*Pi/7)^7 = -15/16.
		

References

  • R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math., (in press, 2012).
  • D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,3,-1}, {0,-3,6}, 50]
  • PARI
    concat(0,Vec(-3*(1-2*x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

Formula

a(n) = c(4)*c(2)^n + c(1)*c(4)^n + c(2)*c(1)^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: -3*x*(1-2*x)/(1-3*x^2+x^3).

A215636 a(n) = - 12*a(n-1) - 54*a(n-2) - 112*a(n-3) - 105*a(n-4) - 36*a(n-5) - 2*a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.

Original entry on oeis.org

0, 0, 0, -3, 24, -135, 660, -3003, 13104, -55689, 232500, -958617, 3916440, -15890355, 64127700, -257698347, 1032023136, -4121456625, 16421256420, -65301500577, 259259758056, -1027901275131, 4070632899300, -16104283594083, 63657906293520, -251447560563465, 992593021410900
Offset: 0

Views

Author

Roman Witula, Aug 18 2012

Keywords

Comments

The Berndt-type sequence number 4 for the argument 2*Pi/9 defined by the relation: X(n) = b(n) + a(n)*sqrt(2), where X(n) := ((cos(Pi/24))^(2*n) + (cos(7*Pi/24))^(2*n) + (cos(3*Pi/8))^(2*n))*(-4)^n. We have b(n) = A215635(n) (see also section "Example" below). For more details - see comments to A215635, A215634 and Witula-Slota's reference.

Examples

			We have X(1)=-6, X(2)=18 and X(3)=-60-3*sqrt(2), which implies the equality: (cos(Pi/24))^6 + (cos(7*Pi/24))^6 + (cos(3*Pi/8))^6 = (60+3*sqrt(2))/64.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{-12,-54,-112,-105,-36,-2}, {0,0,0,-3,24,-135}, 50]

Formula

G.f.: (-3*x^3-12*x^4-9*x^5)/(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).

A215829 a(n) = -3*a(n-1) + 9*a(n-2) + 3*a(n-3), with a(0)=3, a(1)=-3, a(2)=27.

Original entry on oeis.org

3, -3, 27, -99, 531, -2403, 11691, -55107, 263331, -1250883, 5957307, -28339875, 134882739, -641835171, 3054430539, -14535159939, 69169849155, -329162695299, 1566411248475, -7454188455651, 35472778517331, -168806797907427, 803312835011307
Offset: 0

Views

Author

Roman Witula, Aug 24 2012

Keywords

Comments

The Berndt-type sequence number 8 for the argument 2*Pi/9 defined by the trigonometric relations from the Formula section below.
From the general recurrence relation: b(n) = -3*b(n-1) + 9*b(n-2) + 3*b(n-3), i.e., b(n) - b(n-2) = 8*b(n-2) + 3(b(n-3) - b(n-1)) the following summation formulas can be easily deduced: b(2*n+1) + 3*b(2*n) - 3*b(0) - b(1) = 8*Sum_{k=1..n} b(2*k-1) and b(2*n+2) + 3*b(2*n+1) - b(2) - 3*b(1) = 8*Sum_{k=1..n} b(2*k). Hence it follows that (a(2*n+1) + 3*a(2*n))/2 are all integers congruent to 3 modulo 4, and (a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.
We note that all numbers 3^(-1-floor(n/3))*a(n) = A215831(n) and 3^(-n-2)*a(3*n+2) are integers.
The following decomposition holds true: (X - k(1)^n)*(X - (-k(2))^n)*(X - k(3)^n) = X^3 - sqrt(3)^(-n)*a(n)*X^2 + sqrt(3)^(-n)*T(n) - sqrt(3)^(-n), where T(2*n+1) = sqrt(3)*A215945(n) and T(2*n) = A215948(n). [Roman Witula, Aug 30 2012]

Examples

			We have k(1)^3 - k(2)^3 + k(4)^3 = -11*sqrt(3).
		

References

  • D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{-3, 9, 3}, {3, -3, 27}, 50]

Formula

a(n) = (k(1)^n + (-k(2))^n + k(4)^n)*(sqrt(3))^n = (-1+4*c(1))^n + (-1+4*c(2))^n + (-1+4*c(4))^n, where k(j) := cot(2*Pi*j/9) and c(j) := cos(2*Pi*j/9).
G.f.: (3 + 6*x - 9*x^2)/(1 + 3*x - 9*x^2 - 3*x^3). [corrected by Georg Fischer, May 10 2019]

A218332 The sequence of coefficients of cubic polynomials p(x-n), where p(x) = x^3 - 3*x + 1.

Original entry on oeis.org

1, 0, -3, 1, 1, -3, 0, 3, 1, -6, 9, -1, 1, -9, 24, -17, 1, -12, 45, -51, 1, -15, 72, -109, 1, -18, 105, -197, 1, -21, 144, -321, 1, -24, 189, -487, 1, -27, 240, -701, 1, -30, 297, -969, 1, -33, 360, -1297, 1, -36, 429, -1691, 1, -39, 504, -2157, 1, -42, 585, -2701
Offset: 0

Views

Author

Roman Witula, Nov 02 2012

Keywords

Comments

We note that p(x) = (x - s(1))*(x + c(1))*(x - c(2)),
p(x-1) = x^3 - 3*x^2 + 3 = (x - s(2)*c(1/2))*(x - s(4)*c(1/2))*(x + s(2)*s(4)), p(x-2) = x^3 - 6*x^2 + 9*x - 1 = (x - c(1)^2)*(x - c(2)^2)*(x - c(4)^2), and p(x - n - 2) = (x - n - c(1)^2)*(x - n -c(2)^2)*(x - n - c(4)^2), n = 1,2,..., where c(j) := 2*cos(Pi*j/9) and s(j) = 2*sin(Pi*j/18). These one's are characteristic polynomials many sequences A... - see crossrefs.
A218489 is the sequence of coefficients of polynomials p(x+n).

Crossrefs

Formula

We have a(4*k) = 1, a(4*k+1) = -3*k, a(4*k+2) = 3*k^2 - 3, a(4*k+3) = -k^3 + 3*k + 1. Moreover we obtain the relations b(k+1) = b(k) - 3, c(k+1) = c(k) - 2*b(k) + 3, b(k) - c(k) + d(k) - 1, whenever p(x-k) = x^3 + b(k)*x^2 + c(k)*x + d(k).
Empirical g.f.: (x^15-3*x^13-x^12-7*x^11-9*x^10+6*x^9+3*x^8-x^7+12*x^6-3*x^5-3*x^4+x^3-3*x^2+1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013
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