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 1-6 of 6 results.

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

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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

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

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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

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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]

A215945 a(n) = - 3^n*A(2*n+1), 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

-3, -105, -3387, -108945, -3504051, -112702329, -3624894315, -116589061665, -3749904995427, -120609834867081, -3879226882922139, -124769271310005681, -4013008656895890963, -129072153032843014809, -4151404124161560449739
Offset: 0

Views

Author

Roman Witula, Aug 28 2012

Keywords

Comments

The Berndt-type sequence number 11 for the argument 2Pi/9 connecting with the following sequence T(n) := t(1)^n + (-t(2))^n + t(4)^n, where t(j) := tan(2*Pi*j/9). More precisely we have sqrt(3)*a(n) = T(2*n+1) = (-sqrt(3))^n*W(n;2*i/sqrt(3)), where W(n;d) := (1 + 2*i*d*s(1))^n
+ (1 - 2*i*d*s(2))^n + (1 + 2*i*d*s(4))^n, n=0,1,..., d in C. For example we have W(0;d)=W(1;d)=3, W(2;d)=3-6*d^2. The characteristic polynomial of W(n;d) has the form ((x-1)-2*i*d*s(1))*((x-1)-2*i*d*s(2))*((x-1)-2*i*d*s(4)) = (x-1)^3 + 3*d^2*(x-1) - sqrt(3)*i*d^3 = x^3 - 3*x^2 + 3*(1+d^2)*x - 1 - 3*d^2 - sqrt(3)*i*d^3, which implies the recurrence form of W(n;d) = 3*W(n-1;d) - 3*(1+d^2)*W(n-2;d) + (1+3*d^2+sqrt(3)*i*d^3)*W(n-3;d). In consequence we obtain the title recurrence relation for
A(n) := W(n;2*i/sqrt(3)). The polynomials W(n;d) are equivalent of the big omega function with index n of argument d defined and discussed in Witula-Slota's paper (Section 6) and in comments to A215794.
We note that all numbers a(n)/3 and 3^(-1+floor((n+1)/3))*A(n) = A216034(n) are integers.
The following decomposition hold true: (X - t(1)^n)*(X - (-t(2))^n)*(X - t(4)^n) = X^3 - (-sqrt(3))^n*A(n)*X^2 + A215829(n)*X - sqrt(3)^n. Moreover we have A215829(n) = (T(n)^2 - T(2*n))/2, which implies A215829(2*n+1) = (3*a(n)^2 - A215948(2*n+1))/2 and A215829(2*n) = (A215948(n)^2 - A215948(2*n))/2.

Examples

			We have 35*(t(1) - t(2) + t(3)) =  t(1)^3 - t(2)^3 + t(4)^3, t(1)^7 - t(2)^7 + t(4)^7 = -5*81*269*sqrt(3) and t(1)^9 - t(2)^9 + t(4)^9 = -9*389339*sqrt(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

  • Magma
    m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(1+x)^2/(1-33*x+27*x^2-3*x^3))); // Bruno Berselli, Aug 29 2012
    
  • Magma
    I:=[-3, -105, -3387]; [n le 3 select I[n] else 33*Self(n-1)-27*Self(n-2)+3*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Mar 19 2013
  • Mathematica
    LinearRecurrence[{33, -27, 3}, {-3, -105, -3387}, 17] (* Bruno Berselli, Aug 29 2012 *)
    CoefficientList[Series[-3 (1 + x)^2/(1 - 33 x + 27 x^2 - 3 x^3), {x, 0, 20}], x] (* Vincenzo Librandi, Mar 19 2013 *)

Formula

sqrt(3)*a(n) = t(1)^(2*n+1) - t(2)^(2*n+1) + t(4)^(2*n+1) = (-sqrt(3) + 4*s(1))^(2*n+1) + (-sqrt(3) - 4*s(2))^(2*n+1) + (-sqrt(3) + 4*s(4))^(2*n+1), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9).
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
G.f.: -3*(1+x)^2/(1-33*x+27*x^2-3*x^3). - Bruno Berselli, Aug 29 2012

A215948 a(n) = 3^n*A(2*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

3, 33, 1035, 33273, 1070163, 34420113, 1107069147, 35607149289, 1145248319907, 36835122733569, 1184744167018155, 38105444942752473, 1225602095969542131, 39419576386041628017, 1267869080483024344443, 40779027899804588036553, 1311593714249667872790339
Offset: 0

Views

Author

Roman Witula, Aug 28 2012

Keywords

Comments

The Berndt-type sequence number 12 for the argument 2*Pi/9 defined by the first trigonometric relations from the section "Formula" below (it is the complement of the sequence A215945). For more information see comments to A215945. We note that all a(n)/3 and 3^(-1 + floor((n+3)/3))*A(n) = A216034(n) are integers.

Examples

			We have t(1)^4 + t(2)^4 + t(4)^4 = 1035 = (345/11)*(t(1)^2 + t(2)^2 + t(4)^2) and (1 - 4*s(1)/sqrt(3))^4 + (1 + 4*s(2)/sqrt(3))^4 + (1 - 4*s(4)/sqrt(3))^4 = 115. Moreover we get a(2)/a(1) = 31,(36), a(3)/a(1) = 1008,(27), a(4)/a(1) = 32429,(18).
		

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[{33,-27,3}, {3,33,1035}, 50]

Formula

a(n) = t(1)^(2*n) + t(2)^(2*n) + t(4)^(2*n) = (-sqrt(3) + 4*s(1))^(2*n) + (sqrt(3) + 4*s(2))^(2*n) + (-sqrt(3) + 4*s(4))^(2*n), where t(j) := tan(2*Pi*j/9) and s(j) := sin(2*Pi*j/9). For the respective sums of odd powers - see A215945.
a(n) = 33*a(n-1) - 27*a(n-2) + 3*a(n-3).
G.f.: 3*(1-22*x+9*x^2)/(1-33*x+27*x^2-3*x^3).
a(n) = cot(Pi/18)^(2*n) + cot(5*Pi/18)^(2*n) + cot(7*Pi/18)^(2*n). - Greg Dresden, Oct 01 2020

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 1-6 of 6 results.