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 1-8 of 8 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

A332437 Decimal expansion of 2*cos(Pi/9).

Original entry on oeis.org

1, 8, 7, 9, 3, 8, 5, 2, 4, 1, 5, 7, 1, 8, 1, 6, 7, 6, 8, 1, 0, 8, 2, 1, 8, 5, 5, 4, 6, 4, 9, 4, 6, 2, 9, 3, 9, 8, 7, 2, 4, 1, 6, 2, 6, 8, 5, 2, 8, 9, 2, 9, 2, 6, 6, 1, 8, 0, 5, 7, 3, 3, 2, 5, 5, 4, 8, 4, 4, 2, 4, 2, 1, 9, 9, 1, 7, 7, 8, 9, 1, 7, 8, 9, 9, 4, 9, 1, 7, 7, 9, 6, 7, 5, 8, 9, 6, 1, 3, 4, 9
Offset: 1

Views

Author

Wolfdieter Lang, Mar 27 2020

Keywords

Comments

This algebraic number called rho(9) of degree 3 = A055034(9) has minimal polynomial C(9, x) = x^3 - 3*x - 1 (see A187360).
rho(9) gives the length ratio diagonal/side of the smallest diagonal in the regular 9-gon.
The length ratio diagonal/side of the second smallest and the third smallest (or the largest) diagonal in the regular 9-gon are rho(9)^2 - 1 = A332438 - 1 and rho(9) + 1, respectively. - Mohammed Yaseen, Oct 31 2020

Examples

			rho(9) = 1.87938524157181676810821855464946293987241626852892926618...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

Crossrefs

Programs

  • Mathematica
    RealDigits[2 * Cos[Pi/9], 10, 100][[1]] (* Amiram Eldar, Mar 27 2020 *)
  • PARI
    2*cos(Pi/9) \\ Michel Marcus, Mar 28 2020

Formula

rho(9) = 2*cos(Pi/9).
Equals (-1)^(-1/9)*((-1)^(1/9) - i)*((-1)^(1/9) + i). - Peter Luschny, Mar 27 2020
Equals 2*A019879. - Michel Marcus, Mar 28 2020
Equals sqrt(A332438). - Mohammed Yaseen, Oct 31 2020
From Peter Bala, Oct 20 2021: (Start)
The zeros of x^3 - 3*x - 1 are r_1 = -2*cos(2*Pi/9), r_2 = -2*cos(4*Pi/9) and r_3 = -2*cos(8*Pi/9) = 2*cos(Pi/9).
The polynomial x^3 - 3*x - 1 is irreducible over Q (since it is irreducible mod 2) with discriminant equal to 3^4, a square. It follows that the Galois group of the number field Q(2*cos(Pi/9)) over Q is cyclic of order 3.
The mapping r -> 2 - r^2 cyclically permutes the zeros r_1, r_2 and r_3. The inverse cyclic permutation is given by r -> r^2 - r - 2.
The first differences r_1 - r_2, r_2 - r_3 and r_3 - r_1 are the zeros of the cyclic cubic polynomial x^3 - 9*x - 9 of discriminant 3^6.
First quotient relations:
r_1/r_2 = 1 + (r_3 - r_1); r_2/r_3 = 1 + (r_1 - r_2); r_3/r_1 = 1 + (r_2 - r_3);
r_2/r_1 = (r_3 - r_2) - 2; r_3/r_2 = (r_1 - r_3) - 2; r_1/r_3 = (r_2 - r_1) - 2;
r_1/r_2 + r_2/r_3 + r_3/r_1 = 3; r_2/r_1 + r_3/r_2 + r_1/r_3 = -6.
Thus the first quotients r_1/r_2, r_2/r_3 and r_3/r_1 are the zeros of the cyclic cubic polynomial x^3 - 3*x^2 - 6*x - 1 of discriminant 3^6. See A214778.
Second quotient relations:
(r_1*r_2)/(r_3^2) = 3*r_2 - 6*r_1 - 8, with two other similar relations by cyclically permuting the 3 zeros. The three second quotients are the zeros of the cyclic cubic polynomial x^3 + 24*x^2 + 3*x - 1 of discriminant 3^10.
(r_1^2)/(r_2*r_3) = 1 - 3*(r_2 + r_3), with two other similar relations by cyclically permuting the 3 zeros. (End)
Equals i^(2/9) + i^(-2/9). - Gary W. Adamson, Jun 25 2022
Equals Re((4+4*sqrt(3)*i)^(1/3)). - Gerry Martens, Mar 19 2024
From Amiram Eldar, Nov 22 2024: (Start)
Equals Product_{k>=1} (1 - (-1)^k/A056020(k)).
Equals 1 + Product_{k>=1} (1 + (-1)^k/A156638(k)). (End)

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).

A215917 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.

Original entry on oeis.org

0, 6, -15, 45, -129, 372, -1071, 3084, -8880, 25569, -73623, 211989, -610398, 1757571, -5060724, 14571774, -41957751, 120812529, -347865813, 1001639688, -2884106535, 8304453792, -23911721688, 68851058529, -198248721795, 570834443697, -1643652272562
Offset: 0

Views

Author

Roman Witula, Aug 27 2012

Keywords

Comments

The Berndt-type sequence number 9 for the argument 2Pi/9 defined by the first relation from the section "Formula" below.
We have a(n) = 3*(-1)^(n+1)*A215448(n+1). From the recurrence formula for a(n) it follows that all a(3*n) are divisible by 9, a(3*n+1)/3 are congruent to 2 modulo 3, and a(3*n+2)/3 are congruent to 1 modulo 3. In the consequence also all sums a(n)+a(n+1)+a(n+2) are divisible by 9.
From general recurrence X(n) = -3*X(n-1) + X(n-3) the following formula can be deduced: 3*Sum_{k=2..n-1} X(k) = -X(n)-X(n-1)-X(n-2)+X(2)+X(1)+X(0). Hence, in the case of a(n) we obtain 3*Sum_{k=2..n-1} a(k) = -a(n)-a(n-1)-a(n-2)-9.
If we set X(n) = -3*X(n-1) + X(n-3), n in Z, with a(n) = X(n) for n=0,1,... then X(-n) = abs(A215666(n)) = (-1)^n*A215666(n), for every n=0,1,...
The following decomposition holds true (X - c(1)*(-c(4))^(-n))*(X - c(2)*(-c(1))^(-n))*(X - c(4)*(-c(2))^(-n)) = X^3 - a(n)*X^2 + (-1)^n*(A215665(n) - A215664(n))*X + 1.

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

  • Maple
    We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.
  • Mathematica
    LinearRecurrence[{-3,0,1}, {0,6,-15}, 50]
  • PARI
    concat(0,Vec(3*(x+2)/(1+3*x-x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

Formula

a(n) = c(1)*(-c(4))^(-n) + c(2)*(-c(1))^(-n) + c(4)*(-c(2))^(-n), where c(j) := 2*cos(2*Pi*j/9).
a(n) = (-1)^n*(A215885(n+1) - A215885(n)).
G.f.: 3*x(x+2)/(1+3*x-x^3).

A215919 a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=-3, a(2)=12.

Original entry on oeis.org

0, -3, 12, -36, 105, -303, 873, -2514, 7239, -20844, 60018, -172815, 497601, -1432785, 4125540, -11879019, 34204272, -98487276, 283582809, -816544155, 2351145189, -6769852758, 19493014119, -56127897168, 161613838746, -465348502119, 1339917609189, -3858138988821
Offset: 0

Views

Author

Roman Witula, Aug 27 2012

Keywords

Comments

The Berndt-type sequence number 10 for the argument 2Pi/9 defined by the first trigonometric relation from the section "Formula" below. The sequence a(n) is connected with sequences A215917 and A215885 - see the respective formula.
We have A035045(n)=abs(a(n+1)/3) for every n=0,1,...,5 and A035045(7) + a(7)/3 = 1, A035045(8) - a(8)/3 = 10, A035045(9) + a(9)/3 = 63, and A035045(10) - a(10)/3 = 320 - all these four results-numbers are in A069269.

Examples

			We have a(2)=-4*a(1), a(3)=-3*a(2), a(6)/a(3) = -24.25, and a(9) = 579*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[{-3, 0, 1}, {0, -3, 12}, 50]

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).
a(n) = A215917(n+1) + A215917(n) - 2*(-1)^n*A215885(n).
G.f.: -3*x*(1-x)/(1+3*x-x^3).

A274018 Number of n-bead ternary necklaces (no turning over allowed) that avoid the subsequence 110.

Original entry on oeis.org

1, 3, 6, 10, 21, 42, 103, 237, 603, 1519, 3942, 10257, 27131, 71940, 192462, 516933, 1395636, 3781356, 10283911, 28050600, 76732047, 210414811, 578330649, 1592821005, 4395261552, 12149386569, 33637309323, 93267459520, 258961863288, 719938597227, 2003881480452, 5583818718102, 15575529493713
Offset: 0

Views

Author

Marko Riedel, Jun 06 2016

Keywords

Comments

The pattern in this enumeration must be contiguous (all three values next to each other in one sequence of three letters).

Examples

			The necklace
    1--1
   /    \
  0      0
  |      |
  1      2
   \    /
    0--0
contains one instance of the subsequence starting in the upper left corner. Unlike a bracelet, the necklace is oriented.
		

Crossrefs

Formula

G.f.: 1 - Sum_{n>=1} (phi(n)/n)*log(x^(3*n) - q*x^n + 1), where q=3 is the number of symbols in the alphabet we are using. - Petros Hadjicostas, Sep 12 2017
a(n) = (1/n)*Sum_{d|n} phi(n/d)*A215885(d) for n >= 1. - Petros Hadjicostas, Sep 13 2017

A217336 a(n) = 3^(-1+floor(n/2))*A(n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.

Original entry on oeis.org

1, 1, 11, 35, 345, 1129, 11091, 36315, 356721, 1168017, 11473371, 37567443, 369023049, 1208298105, 11869049763, 38863020555, 381749439969, 1249968331809, 12278374244523, 40203278289027, 394914722339385, 1293075627640713
Offset: 0

Views

Author

Roman Witula, Oct 01 2012

Keywords

Comments

The Berndt-type sequence number 14 for the argument 2Pi/9 defined by the relation: A(n)*(-sqrt(3))^n = t(1)^n + (-t(2))^n + t(4)^n = (-sqrt(3) + 4*s(1))^n + (-sqrt(3) - 4*s(2))^n + (-sqrt(3) + 4*s(4))^n, where s(j) := sin(2*Pi*j/9) and t(j) := tan(2*Pi*j/9).
The definitions of the other Berndt-type sequences for the argument 2Pi/9 like A215945, A215948, A216034 in Crossrefs are given.
We note that all a(2*n), n=2,3,..., are divisible by 3, and it is only when n=5 that a(2*n) is divisible by 9.

Examples

			Note that A(0)=A(1)=3, a(0)=a(1)=1, A(2)=a(2)=11, A(3)=a(3)=35, A(4)=115, a(4)=345 and A(5) = 1129/3, which implies the equality  3387*sqrt(3) = -t(1)^5 + t(2)^5 - t(4)^5.
		

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

  • Magma
    /* By definition: */ i:=22; I:=[3,3,11]; A:=[m le 3 select I[m] else 3*Self(m-1)+Self(m-2)-Self(m-3)/3: m in [1..i]]; [3^(-1+Floor((n-1)/2))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012
  • Mathematica
    LinearRecurrence[{0, 33, 0, -27, 0, 3}, {1, 1, 11, 35, 345, 1129},25] (* Paolo Xausa, Feb 23 2024 *)

Formula

G.f.: (1+x-22*x^2+2*x^3+9*x^4+x^5)/(1-33*x^2+27*x^4-3*x^6). - Bruno Berselli, Oct 01 2012
Showing 1-8 of 8 results.