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.

A006053 a(n) = a(n-1) + 2*a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 9, 14, 28, 47, 89, 155, 286, 507, 924, 1652, 2993, 5373, 9707, 17460, 31501, 56714, 102256, 184183, 331981, 598091, 1077870, 1942071, 3499720, 6305992, 11363361, 20475625, 36896355, 66484244, 119801329, 215873462, 388991876, 700937471
Offset: 0

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Author

Keywords

Comments

a(n+1) = S(n) for n>=1, where S(n) is the number of 01-words of length n, having first letter 1, in which all runlengths of 1's are odd. Example: S(4) counts 1000, 1001, 1010, 1110. See A077865. - Clark Kimberling, Jun 26 2004
For n>=1, number of compositions of n into floor(j/2) kinds of j's (see g.f.). - Joerg Arndt, Jul 06 2011
Counts walks of length n between the first and second nodes of P_3, to which a loop has been added at the end. Let A be the adjacency matrix of the graph P_3 with a loop added at the end. A is a 'reverse Jordan matrix' [0,0,1; 0,1,1; 1,1,0]. a(n) is obtained by taking the (1,2) element of A^n. - Paul Barry, Jul 16 2004
Interleaves A094790 and A094789. - Paul Barry, Oct 30 2004
a(n) appears in the formula for the nonnegative powers of rho:= 2*cos(Pi/7), the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^n = C(n)*1 + C(n+1)*rho + a(n)*sigma, n>=0, with C(n) = A052547(n-2). See the Steinbach reference, and a comment under A052547. - Wolfdieter Lang, Nov 25 2010
If with the above notations the power basis <1,rho,rho^2> of Q(rho) is used, nonnegative powers of rho are given by rho^n = -a(n-1)*1 + A052547(n-1)*rho + a(n)*rho^2. For negative powers see A006054. - Wolfdieter Lang, May 06 2011
-a(n-1) also appears in the formula for the nonpositive powers of sigma (see the above comment for the definition, and the Steinbach basis <1,rho,sigma>) as follows: sigma^(-n) = A(n)*1 -a(n+1)*rho -A(n-1)*sigma, with A(n) = A052547(n), A(-1):=0. - Wolfdieter Lang, Nov 25 2010

Examples

			G.f. = x^2 + x^3 + 3*x^4 + 4*x^5 + 9*x^6 + 14*x^7 + 28*x^8 + 47*x^9 + ...
Regarding the description "number of compositions of n into floor(j/2) kinds of j's," the a(6)=9 compositions of 6 are (2a, 2a, 2a), (3a, 3a), (2a, 4a), (2a, 4b), (4a, 2a), (4b, 2a), (6a), (6b), (6c). - _Bridget Tenner_, Feb 25 2022
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Witula, E. Hetmaniok and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications (2012).

Crossrefs

Programs

  • Haskell
    a006053 n = a006053_list !! n
    a006053_list = 0 : 0 : 1 : zipWith (+) (drop 2 a006053_list)
       (zipWith (-) (map (2 *) $ tail a006053_list) a006053_list)
    -- Reinhard Zumkeller, Oct 14 2011
    
  • Magma
    [ n eq 1 select 0 else n eq 2 select 0 else n eq 3 select 1 else Self(n-1) +2*Self(n-2) -Self(n-3): n in [1..40] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=0: a[1]:=0: a[2]:=1: for n from 3 to 40 do a[n]:=a[n-1]+2*a[n-2]-a[n-3] od:seq(a[n], n=0..40); # Emeric Deutsch
    A006053:=z**2/(1-z-2*z**2+z**3); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    LinearRecurrence[{1,2,-1}, {0,0,1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
  • PARI
    {a(n) = if( n<0, n = -1-n; polcoeff( -1 / (1 - 2*x - x^2 + x^3) + x * O(x^n), n), polcoeff( x^2 / (1 - x - 2*x^2 + x^3) + x * O(x^n), n))}; /* Michael Somos, Nov 30 2014 */
    
  • SageMath
    @CachedFunction
    def a(n): # a = A006053
        if (n<3): return (n//2)
        else: return a(n-1) + 2*a(n-2) - a(n-3)
    [a(n) for n in range(41)] # G. C. Greubel, Feb 12 2023

Formula

G.f.: x^2/(1 - x - 2*x^2 + x^3). - Emeric Deutsch, Dec 14 2004
a(n) = c^(n-2) - a(n-1)*(c-1) + (1/c)*a(n-2) for n > 3 where c = 2*cos(Pi/7). Example: a(7) = 14 = c^5 - 9*(c-1) + 4/c = 18.997607... - 7.21743962... + 2.219832528... - Gary W. Adamson, Jan 24 2010
G.f.: -1 + 1/(1 - Sum_{j>=1} floor(j/2)*x^j). - Joerg Arndt, Jul 06 2011
a(n+2) = A094790(n/2+1)*(1+(-1)^n)/2 + A094789((n+1)/2)*(1-(-1)^n)/2. - Paul Barry, Oct 30 2004
First differences of A028495. - Floor van Lamoen, Nov 02 2005
a(n) = A187065(2*n+1); a(n+1) = A187066(2*n+1) = A187067(2*n). - L. Edson Jeffery, Mar 16 2011
a(n) = 2^n*(c(1)^(n-1)*(c(1)+c(2)) + c(3)^(n-1)*(c(3)+c(6)) + c(5)^(n-1)*(c(5)+c(4)) )/7, with c(j):=cos(Pi*j/7). - Herbert Kociemba, Dec 18 2011
a(n+1)*(-1)^n*49^(1/3) = (c(1)/c(4))^(1/3)*(2*c(1))^n + (c(2)/c(1))^(1/3)*(2*c(2))^n + (c(4)/c(2))^(1/3)*(2c(4))^n = (c(2)/c(1))^(1/3)*(2*c(1))^(n+1) + (c(4)/c(2))^(1/3)*(c(2))^(n+1) + (c(1)/c(4))^(1/3)*(2*c(4))^(n+1), where c(j) := cos(2Pi*j/7); for the proof, see Witula et al.'s papers. - Roman Witula, Jul 21 2012
The previous formula connects the sequence a(n) with A214683, A215076, A215100, A120757. We may call a(n) the Ramanujan-type sequence number 2 for the argument 2*Pi/7. - Roman Witula, Aug 02 2012
a(n) = -A006054(1-n) for all n in Z. - Michael Somos, Nov 30 2014
G.f.: x^2 / (1 - x / (1 - 2*x / (1 + 5*x / (2 - x / (5 - 2*x))))). - Michael Somos, Jan 20 2017
a(n) ~ r*c^n, where r=0.241717... is one of the roots of 49*x^3-7*x+1, and c=2*cos(Pi/7) (as in Gary W. Adamson's formula). - Daniel Checa, Nov 04 2022
a(2n-1) = 2*a(n+1)*a(n) - a(n)^2 - a(n-1)^2. - Richard Peterson, May 25 2023

Extensions

More terms from Emeric Deutsch, Dec 14 2004
Typo in definition fixed by Reinhard Zumkeller, Oct 14 2011

A120757 Expansion of x^2*(2+x)/(1-3*x-4*x^2-x^3).

Original entry on oeis.org

0, 2, 7, 29, 117, 474, 1919, 7770, 31460, 127379, 515747, 2088217, 8455018, 34233669, 138609296, 561217582, 2272323599, 9200450421, 37251863241, 150829715006, 610697048403, 2472661868474, 10011603514040, 40536155064419
Offset: 1

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Author

Keywords

Comments

The (1,1)-entry of the matrix M^n, where M is the 3 X 3 matrix [0,1,1; 1,1,2; 1,2,2].
a(n)/a(n-1) tends to 4.0489173...an eigenvalue of M and a root to the characteristic polynomial x^3 - 3x^2 - 4x - 1.
C(n):=a(n), with a(0):=1 (hence the o.g.f. for C(n) is (1-3*x-2*x^2)/(1-3*x-4*x^2-x^3)), appears in the following formula for the nonnegative powers of rho*sigma, where rho:=2*cos(Pi/7) and sigma:=sin(3*Pi/7)/sin(Pi/7) = rho^2-1 are the ratios of the smaller and larger diagonal length to the side length in a regular 7-gon (heptagon). See the Steinbach reference where the basis <1,rho,sigma> is used in an extension of the rational field. (rho*sigma)^n = C(n) + B(n)*rho + A(n)*sigma,n>=0, with B(n)= |A122600(n-1)|, B(0)=0, and A(n)= A181879(n). For the nonpositive powers see A085810(n)*(-1)^n, A181880(n-2)*(-1)^n and A116423(n+1)*(-1)^(n+1), respectively. See also a comment under A052547.
We have a(n)=cs(3n+1), where the sequence cs(n) and its two conjugate sequences as(n) and bs(n) are defined in the comments to the sequence A214683 (see also A215076, A215100, A006053). We call the sequence a(n) the Ramanujan-type sequence number 5 for the argument 2Pi/7. Since as(3n+1)=bs(3n+1)=0, we obtain the following relation: 49^(1/3)*a(n) = (c(1)/c(4))^(n + 1/3) + (c(4)/c(2))^(n + 1/3) + (c(2)/c(1))^(n + 1/3), where c(j) := Cos(2Pi/7) (for more details and proofs see Witula et al.'s papers). - Roman Witula, Aug 02 2012

Examples

			a(7)=1919 because M^7= [1919,3458,4312;3458,6231,7770;4312,7770,9689].
		

References

  • R. Witula, E. Hetmaniok and 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.

Crossrefs

Programs

  • Magma
    a:=[0,2,7]; [ n le 3 select a[n] else 3*Self(n-1) + 4*Self(n-2) + Self(n-3): n in [1..25]]; // Marius A. Burtea, Oct 03 2019
    
  • Maple
    with(linalg): M[1]:=matrix(3,3,[0,1,1,1,1,2,1,2,2]): for n from 2 to 25 do M[n]:=multiply(M[1],M[n-1]) od: seq(M[n][1,1],n=1..25);
  • Mathematica
    LinearRecurrence[{3,4,1},{0,2,7},40] (* Roman Witula, Aug 02 2012 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,4,3]^(n-1)*[0;2;7])[1,1] \\ Charles R Greathouse IV, Jun 22 2016
    
  • SageMath
    @CachedFunction
    def a(n): # a = A120757
        if (n<3): return (0,2,7)[n]
        else: return 3*a(n-1) + 4*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-1) + 4*a(n-2) + a(n-3) (follows from the minimal polynomial of the matrix M). See also the o.g.f. given in the name.

Extensions

Edited by N. J. A. Sloane, Dec 03 2006
New name, old name as comment; o.g.f.; reference.

A215560 a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=a(1)=3, a(2)=101.

Original entry on oeis.org

3, 3, 101, 444, 5981, 38468, 390974, 2948431, 26868565, 216624495, 1888775906, 15657923053, 134074085330, 1124375492334, 9556192325235, 80523923708399, 682280993578341, 5760499663646612, 48746948619251921, 411906111379078256, 3483838470286469746, 29447943482916260935
Offset: 0

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Author

Roman Witula, Aug 16 2012

Keywords

Comments

The Ramanujan-type sequence number 6 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757 for the numbers: 1, 1a, 2, 2a, 3, 4 and 5 respectively).
The sequence a(n) is one of the three special sequences (the remaining two are A215569 and A215572) connected with the following recurrence relation: T(n):=49^(1/3)*T(n-2)+T(n-3), with T(0)=3, T(1)=0, and T(2)=2*49^(1/3) - see the comments to A214683.
It can be proved that
T(n) = (c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3), where c(j):=2*cos(2*Pi*j/7), and the following decomposition hold true:
T(n) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),
bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the first Witula reference.
It follows that a(n)=at(3*n), bt(3*n)=ct(3*n)=0.
Every difference of the form a(n)-a(n-2)-a(n-3) is divisible by 3. Because the difference a(n+1)-a(n) is congruent to the difference a(n-4)-a(n-2) modulo 3 we easily deduce that a(6)-a(5) and a(7)-a(6)-2 are both divisible by 3.

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

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, 46, 1}, {3, 3, 101}, 50]
  • PARI
    Vec((3-6*x-46*x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

Formula

a(n) = (c(1)^4/c(2))^n + (c(2)^4/c(4))^n + (c(4)^4/c(1))^n, where c(j) = 2*cos(2*Pi*j/7).
G.f.: (3-6*x-46*x^2)/(1-3*x-46*x^2-x^3).

Extensions

More terms from Michel Marcus, Apr 20 2016

A215572 a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=106.

Original entry on oeis.org

2, 5, 106, 550, 6531, 44999, 435973, 3384404, 30252969, 246877464, 2135653370, 17793576423, 151867661753, 1276243154087, 10832435479322, 91356359187721, 773637352766062, 6534137016412674, 55281085635664595, 467187197014742851, 3951025667301212597, 33398969150217473532
Offset: 0

Views

Author

Roman Witula, Aug 16 2012

Keywords

Comments

The Ramanujan-type sequence number 8 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569 for the numbers: 1, 1a, 2, 2a, 3-7 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215569) connected with the following recurrence relation:
(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) are defined in comments to A215560 (see also A215569). It follows that a(n)=ct(3*n+2), at(3*n+2)=bt(3*n+2)=0, which implies the first formula below.
We note that if a(n), a(n+1) and a(n+2) are all odd for some n in N then a(n+3) is even, a(n+4) is odd, a(n+5) and a(n+6) are both even, and the numbers a(n+7), a(n+8), a(n+9) are all odd again. In consequence, this situation hold for every n of the form 7*k+4, k=0,1,..., in the other words cyclical through all sequence a(n), n=4,5,... (from n=1 whenever we start from odd-even-even sequence).

Examples

			From 4*a(1)+5*a(2)=a(3) we obtain 4*((c(1)^4/c(2))^(5/3) + (c(2)^4/c(4))^(5/3) + (c(4)^4/c(1))^(5/3)) + 5*((c(1)^4/c(2))^(8/3) + (c(2)^4/c(4))^(8/3) + (c(4)^4/c(1))^(8/3)) = (4 + 5*c(1)^4/c(2))*((c(1)^4/c(2))^(5/3) + (4 + 5*c(2)^4/c(4))*((c(2)^4/c(4))^(5/3) + (4 + 5*c(4)^4/c(1))*((c(4)^4/c(1))^(5/3) = (c(1)^4/c(2))^(11/3) + (c(2)^4/c(4))^(11/3) + (c(4)^4/c(1))^(11/3) = 550*49^(1/3).
		

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.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3,46,1}, {2,5,106}, 50]
    CoefficientList[Series[(2 - x - x^2)/(1 - 3*x - 46*x^2 - x^3), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)
  • PARI
    Vec((2-x-x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

Formula

49^(1/3)*a(n) = (c(1)^4/c(2))^(n+2/3) + (c(2)^4/c(4))^(n+2/3) + (c(4)^4/c(1))^(n+2/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+2) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+2) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+2).
G.f.: (2-x-x^2)/(1-3*x-46*x^2-x^3).

Extensions

More terms from Michel Marcus, Apr 20 2016

A215569 a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=0, a(1)=14, a(2)=49.

Original entry on oeis.org

0, 14, 49, 791, 4641, 50358, 365351, 3417162, 27107990, 238878773, 1967021021, 16916594611, 141471629572, 1204545261843, 10138247340452, 85965295695706, 725459810009753, 6140921279372187, 51879880394260905, 438847479843913070, 3709157858947113027
Offset: 0

Views

Author

Roman Witula, Aug 16 2012

Keywords

Comments

The Ramanujan-type sequence number 7 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560 for the numbers: 1, 1a, 2, 2a, 3-6 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215572) connected with the following recurrence relation:
(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),
bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the Witula's first paper (see also A215560). It follows that a(n)=bt(3*n+1), at(3*n+1)=ct(3*n+1)=0, which implies the first formula below.
We note that all numbers a(n) are divided by 7.

Examples

			We have  (c(1)^4/c(2))^(4/3) + (c(2)^4/c(4))^(4/3) + (c(4)^4/c(1))^(4/3) = (2/7)*(c(1)^4/c(2))^(7/3) + (c(2)^4/c(4))^(7/3) + (c(4)^4/c(1))^(7/3)).
		

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.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3,46,1},{0,14,49},30] (* Harvey P. Dale, Jan 12 2015 *)

Formula

7^(1/3)*a(n) = (c(1)^4/c(2))^(n+1/3) + (c(2)^4/c(4))^(n+1/3) + (c(4)^4/c(1))^(n+1/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+1) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+1) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+1).
G.f.: (14*x+7*x^2)/(1-3*x-46*x^2-x^3).

Extensions

More terms from Harvey P. Dale, Jan 12 2015

A215139 a(n) = (a(n-1) - a(n-3))*7^((1+(-1)^n)/2) with a(6)=5, a(7)=4, a(8)=22.

Original entry on oeis.org

5, 4, 22, 17, 91, 69, 364, 273, 1428, 1064, 5537, 4109, 21315, 15778, 81683, 60368, 312130, 230447, 1190553, 878423, 4535832, 3345279, 17267992, 12732160, 65708167, 48440175, 249956105, 184247938, 950654341, 700698236, 3615152086, 2664497745, 13746596563, 10131444477
Offset: 6

Views

Author

Roman Witula, Aug 04 2012

Keywords

Comments

The Ramanujan-type sequence the number 9 for the argument 2*Pi/7. The sequence is connecting with the following decomposition: (s(4)/s(1))^(1/3)*s(1)^n + (s(1)/s(2))^(1/3)*s(2)^n + (s(2)/s(4))^(1/3)*s(4)^n = x(n)*(4-3*7^(1/3))^(1/3) + y(n)*(11-3*49^(1/3))^(1/3), where s(j) := sin(2*Pi*j/7), x(0)=1, x(1)=-7^(1/6)/2, x(2)=y(0)=y(1)=0, y(2)=7^(1/3)/4 and X(n)=sqrt(7)*(X(n-1)-X(n-3)) for every n=3,4,..., and X=x or X=y. It could be deduced the formula 4*y(n) = a(n)*7^(1/3 + (3+(-1)^n)/4), which implies a(0)=0, a(1)= 0, a(2)= 1/7, a(3)=1/7, a(4)=1, a(5)=6/7, i.e., A163260(n)=7*a(n) for every n=0,1,...,5. The sequence a(n) is discussed in third Witula paper.

Examples

			From values of x(2),y(2) and the identity 2*sin(t)^2=1-cos(2*t) we obtain (s(4)/s(1))^(1/3)*c(1) + (s(1)/s(2))^(1/3)*c(4) + (s(2)/s(4))^(1/3)*c(1) = (4-3*7^(1/3))^(1/3) - (1/2)*(7*(11-3*49^(1/3)))^(1/3), where c(j):=cos(2*Pi*j/7). Further, from values of x(1),x(3),y(1),y(3) and the identity 4*sin(t)^3=3*sin(t)-sin(3*t) we obtain (s(4)/s(1))^(1/3)*s(4) + (s(1)/s(2))^(1/3)*s(1) + (s(2)/s(4))^(1/3)*s(2) = (-3*7^(1/6)/2 +4*7^(1/2))*(4-3*7^(1/3))^(1/3) - 7^(5/6)*(11-3*49^(1/3))^(1/3).
		

References

  • R. Witula, E. Hetmaniok and 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.

Crossrefs

Programs

  • Magma
    I:=[5,4,22,17,91,69]; [n le 6 select I[n] else 7*Self(n-2) - 14*Self(n-4) + 7*Self(n-6): n in [1..30]]; // G. C. Greubel, Apr 19 2018
  • Mathematica
    LinearRecurrence[{0,7,0,-14,0,7}, {5,4,22,17,91,69}, {1,50}] (* G. C. Greubel, Apr 19 2018 *)
  • PARI
    Vec(-x*(1+x)*(6*x^4+x^3-12*x^2-x+5)/(-1+7*x^2-14*x^4+7*x^6) + O(x^50)) \\ Michel Marcus, Apr 20 2016
    

Formula

G.f.: -x*(1+x)*(6*x^4+x^3-12*x^2-x+5) / ( -1+7*x^2-14*x^4+7*x^6 ). - R. J. Mathar, Sep 14 2012

Extensions

More terms from Michel Marcus, Apr 20 2016

A218664 Coefficients of cubic polynomials p(x+n), where p(x) = x^3 + x^2 - 2*x - 1.

Original entry on oeis.org

1, 1, -2, -1, 1, 4, 3, -1, 1, 7, 14, 7, 1, 10, 31, 29, 1, 13, 54, 71, 1, 16, 83, 139, 1, 19, 118, 239, 1, 22, 159, 377, 1, 25, 206, 559, 1, 28, 259, 791, 1, 31, 318, 1079, 1, 34, 383, 1429, 1, 37, 454, 1847, 1, 40, 531, 2339, 1, 43, 614, 2911, 1, 46, 703, 3569, 1, 49, 798, 4319
Offset: 0

Views

Author

Roman Witula, Nov 04 2012

Keywords

Comments

We have p(x) = (x - c(1))*(x - c(2))*(x - c(4)), where c(j) := 2*cos(2*Pi*j/7). We note that c(4) = c(3) = -c(1/2), c(1) = s(3) and c(2) = -s(1), where s(j) := 2*sin(Pi*j/14). Moreover we obtain -p(-x) = x^3 - x^2 - 2*x + 1 = (x + c(1))*(x + c(2))*(x + c(4)), q(x) := -x^3*p(1/x) = x^3 + 2*x^2 + x - 1 = (x - c(1)^(-1))*(x - c(2)^(-1))*(x - c(4)^(-1)), and -q(-x) = x^3 - 2*x^2 + x + 1 = (x + c(1)^(-1))*(x + c(2)^(-1))*(x + c(4)^(-1)).
We also have p(x+2) = x^3 + 7*x^2 + 14*x + 7 = (x + s(2)^2)*(x + s(4)^2)*(x + s(6)^2). The polynomial -p(-x-2) = x^3 - 7*x^2 + 14*x - 7 = (x - s(2)^2)*(x - s(4)^2)*(x - s(6)^2) is known as Johannes Kepler's cubic polynomial (see Witula's book).
Let us set r(x) := p(x+1). It can be verified that -x^3*r(1/x) = x^3 - 3*x^2 - 4*x - 1 = (x - c(1)/c(4))*(x - c(4)/c(2))*(x - c(2)/c(1)); for example, we have c(1)^3 + c(1)^2 - 2*c(1) - 1 = 0 which implies that c(1)^2 + 2*c(1) = 1/(c(1) - 1), and then c(1)^2 + 2*c(1) = c(4)/c(2) since c(4)/c(2) = (c(1)^4 - 4*c(1)^2 + 2)/(c(1)^2 - 2).
The polynomials p(x+n) and the ones obtained as above (i.e., after simple algebraic transformations) are the characteristic polynomials of many sequences in the OEIS; see crossrefs.

References

  • R. Witula, Complex Numbers, Polynomials and Partial Fraction Decomposition, Part 3, Wydawnictwo Politechniki Slaskiej, Gliwice 2010 (Silesian Technical University publishers).

Crossrefs

Formula

We have a(4*k) = 1, a(4*k + 1) = 3*k + 1, a(4*k + 2) = 3*k^2 + 2*k - 2, a(4*k + 3) = k^3 + k^2 - 2*k - 1. Further, the following relations hold true: b(k+1) = b(k) + 3, c(k+1) = 2*b(k) -2*c(k) + 3, d(k+1) = b(k) - 2*c(k) - d(k) + 1, where p(x + k) = x^3 + b(k)*x^2 + c(k)*x + d(k).
Empirical g.f.: -(x^15 - x^14 - 2*x^13 + x^12 - 5*x^11 + 10*x^10 + 3*x^9 - 3*x^8 - 3*x^7 - 11*x^6 + 3*x^4 + x^3 + 2*x^2 - x - 1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013
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