A214699 -id:A214699 - cp's OEIS frontend

A006054 a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

0, 0, 1, 2, 5, 11, 25, 56, 126, 283, 636, 1429, 3211, 7215, 16212, 36428, 81853, 183922, 413269, 928607, 2086561, 4688460, 10534874, 23671647, 53189708, 119516189, 268550439, 603427359, 1355888968, 3046654856, 6845771321, 15382308530, 34563733525

Offset: 0

N. J. A. Sloane

Let u(k), v(k), w(k) be defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)+w(k), v(k+1)=u(k)+v(k), w(k+1)=u(k); then {u(n)} = 1,1,3,6,14,31,... (A006356 with an extra initial 1), {v(n)} = 0,1,2,5,11,25,... (this sequence with its initial 0 deleted) and {w(n)} = {u(n)} prefixed by an extra 0 = A077998 with an extra initial 0. - Benoit Cloitre, Apr 05 2002. Also u(k)^2+v(k)^2+w(k)^2 = u(2k). - Gary W. Adamson, Dec 23 2003
Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then A006054 counts walks of length n between the vertex of degree 1 and the vertex of degree 3. - Paul Barry, Oct 02 2004
Form the digraph with matrix [1,1,0; 1,0,1; 1,1,1]. A006054(n) counts walks of length n between the vertices with loops. - Paul Barry, Oct 15 2004
Nonzero terms = INVERT transform of (1, 1, 2, 2, 3, 3, ...). Example: 56 = (1, 1, 2, 5, 11, 25) dot (3, 3, 2, 2, 1, 1) = (3 + 3 + 4 + 10 + 11 + 25). - Gary W. Adamson, Apr 20 2009
-a(n+1) appears in the formula for the nonpositive 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+1)*sigma, n >= 0, with C(n)=A077998(n), C(-1):=0. See the Steinbach reference, and a comment under A052547.
If, with the above notations, the power basis of the field Q(rho) is taken one has for nonpositive powers of rho, rho^(-n) = a(n+2)*1 + A077998(n-1)*rho - a(n+1)*rho^2. For nonnegative powers see A006053. See also the Steinbach reference. - Wolfdieter Lang, May 06 2011
a(n) appears also in the nonnegative powers of sigma,(defined in the above comment, where also the basis is given). See a comment in A106803.
The sequence b(n):=(-1)^(n+1)*a(n) forms the negative part (i.e., with nonpositive indices) of the sequence (-1)^n*A006053(n+1). In this way we obtain what we shall call the Ramanujan-type sequence number 2a for the argument 2*Pi/7 (see the comment to Witula's formula in A006053). We have b(n) = -2*b(n-1) + b(n-2) + b(n-3) and b(n) * 49^(1/3) = (c(1)/c(4))^(1/3) * (c(1))^(-n) + (c(2)/c(1))^(1/3) * (c(2))^(-n) + (c(4)/c(2))^(1/3) * (c(4))^(-n) = (c(2)/c(1))^(1/3) * (c(1))^(-n+1) + (c(4)/c(2))^(1/3) * (c(2))^(-n+1) + (c(1)/c(4))^(1/3) * (c(4))^(-n+1), where c(j) := 2*cos(2*Pi*j/7) (for the proof, see the comments to A215112). - Roman Witula, Aug 06 2012
(1, 1, 2, 5, 11, 25, 56, ...) * (1, 0, 1, 0, 1, ...) = the variant of A006356: (1, 1, 3, 6, 14, 31, ...). - Gary W. Adamson, May 15 2013
The limit of a(n+1)/a(n) for n -> infinity is, for all generic sequences with this recurrence of signature (2,1,-1), sigma = rho^2-1, approximately 2.246979603, the length ratio (largest diagonal)/side in the regular heptagon (7-gon). For rho = 2*cos(Pi/7) and sigma see a comment above, and the P. Steinbach reference. Proof: a(n+1)/a(n) = 2 + 1/(a(n)/a(n-1)) - 1/((a(n)/a(n-1))*(a(n-1)/a(n-2))), leading in the limit to sigma^3 -2*sigma^2 - sigma + 1, which is solved by sigma = rho^2-1, due to C(7, rho) = 0 , with the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho (see A187360). - Wolfdieter Lang, Nov 07 2013
Numbers of straight-chain aliphatic amino acids involving single, double or triple bonds (allowing adjacent double bonds) when cis/trans isomerism is neglected. - Stefan Schuster, Apr 19 2018
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0) = 1. Also, A(r,n)=0 for n<0. Let A_1(r,n) = Sum_{j=0..n} A(r,j). Then the expansion of 1/(1 - 2*x - x^2 + x^3) is A_1(0,n) + A_1(1,n-2) + A_1(n-4) + ... = a(n) without the initial two 0's. In general, the expansion of 1/(1 - 2*x -x^k + x^(k+1)) is equal to Sum_{j>=0} A_1(j, n-j*k). - Gregory L. Simay, May 25 2018
For n>1, a(n) is the number of ways to tile a strip of length n-1 with one color of squares and dominos, two colors of trominos and quadrominos, 3 colors of 5-minos and 6-minos, and so on. - Greg Dresden and Zhiyu Zhang, Jun 26 2025

			G.f. = x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 56*x^7 + 126*x^8 + 283*x^9 + ... - _Michael Somos_, Jun 25 2018

Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..150
C. P. de Andrade, J. P. de Oliveira Santos, E. V. P. da Silva and K. C. P. Silva, Polynomial Generalizations and Combinatorial Interpretations for Sequences Including the Fibonacci and Pell Numbers, Open Journal of Discrete Mathematics, 2013, 3, 25-32 doi:10.4236/ojdm.2013.31006. - From _N. J. A. Sloane_, Feb 20 2013
Maximilian Fichtner, K. Voigt, and S. Schuster, The tip and hidden part of the iceberg: Proteinogenic and non-proteinogenic aliphatic amino acids, Biochimica et Biophysica Acta (BBA)-General, 2016, Volume 1861, Issue 1, Part A, January 2017, Pages 3258-3269.
Brian Hopkins and Hua Wang, Restricted Color n-color Compositions, arXiv:2003.05291 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 434
S. Morier-Genoud, V. Ovsienko, and S. Tabachnikov, Introducing supersymmetric frieze patterns and linear difference operators, Math. Z. 281 (2015) 1061.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
R. Sachdeva and A. K. Agarwal, Combinatorics of certain restricted n-color composition functions, Discrete Mathematics, 340, (2017), 361-372.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Alexey Ustinov, Supercontinuants, arXiv:1503.04497 [math.NT], 2015.
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
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, D. Slota and A. Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-1).

Cf. A005578, A006053, A006356, A007583, A080937, A094790, A214683, A214699, A214779, A215112, A306334.

Row sums of A144159 and A180264.

a006054 n = a006053_list !! n
a006054_list = 0 : 0 : 1 : zipWith (+) (map (2 *) $ drop 2 a006054_list)
   (zipWith (-) (tail a006054_list) a006054_list)
-- Reinhard Zumkeller, Oct 14 2011

A006054:=z**2/(1-2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation

LinearRecurrence[{2, 1, -1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *)

a(n):=if n<2 then 0 else if n=2 then 1 else b(n-2);
b(n):=sum(sum(binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j),j,0,k),k,1,n); /* Vladimir Kruchinin, May 05 2011 */

x='x+O('x^66);
concat([0, 0], Vec(x^2/(1-2*x-x^2+x^3))) \\ Joerg Arndt, May 05 2011

G.f.: x^2/(1-2*x-x^2+x^3).
Sum_{k=0..n+2} a(k) = A077850(n). - Philippe Deléham, Sep 07 2006
Let M = the 3 X 3 matrix [1,1,0; 1,2,1; 0,1,2], then M^n*[1,0,0] = [A080937(n-1), A094790(n), A006054(n-1)]. E.g., M^3*[1,0,0] = [5,9,5] = [A080937(2), A094790(3), A006054(2)]. - Gary W. Adamson, Feb 15 2006
a(n) = round(k*A006356(n-1)), for n>1, where k = 0.3568958678... = 1/(1+2*cos(Pi/7)). - Gary W. Adamson, Jun 06 2008
a(n+1) = A187070(2n+1) = A187068(2n+3). - L. Edson Jeffery, Mar 10 2011
a(n+3) = Sum_{k=1..n} Sum_{j=0..k} binomial(j,n-3*k+2*j)*(-1)^(j-k)*binomial(k,j)*2^(-n+3*k-j); a(0)=0, a(1)=0, a(2)=1. - Vladimir Kruchinin, May 05 2011
7*a(n) = (c(2)-c(4))*(1+c(1))^n + (c(4)-c(1))*(1+c(2))^n + (c(1)-c(2))*(1+c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula et al. papers. - Roman Witula, Aug 07 2012
a(n) = -A006053(1-n) for all n in Z. - Michael Somos, Jun 25 2018

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

Roman Witula, Aug 20 2012

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.

			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).

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).

Michael De Vlieger, Table of n, a(n) for n = 0..3649
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215007, A214699, A214683.

Mathematica
```
LinearRecurrence[{0,3,-1}, {3,0,6}, 50]
```

Vec(3*(1-x^2)/(1-3*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

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).

A215512 a(n) = 5a(n-1) - 6a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

Original entry on oeis.org

1, 3, 8, 23, 70, 220, 703, 2265, 7327, 23748, 77043, 250054, 811760, 2635519, 8557089, 27784091, 90213440, 292919743, 951102166, 3088205812, 10027335807, 32558546329, 105716922615, 343260670908, 1114560365179, 3618954723062, 11750672095144, 38154192502527

Offset: 0

Roman Witula, Aug 14 2012

The Berndt-type sequence number 7 for the argument 2Pi/7 defined by the relation: sqrt(7)*a(n) = s(1)*c(4)^(2*n) + s(2)*c(1)^(2*n) + s(4)*c(2)^(2*n), where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7). If we additionally defined the following sequences:
sqrt(7)*b(n) = s(2)*c(4)^(2*n) + s(4)*c(1)^(2*n) + s(1)*c(2)^(2*n),
sqrt(7)*c(n) = s(4)*c(4)^(2*n) + s(1)*c(1)^(2*n) + s(2)*c(2)^(2*n), and
sqrt(7)*a1(n) = s(1)*c(4)^(2*n+1) + s(2)*c(1)^(2*n+1) + s(4)*c(2)^(2*n+1),
sqrt(7)*b1(n) = s(2)*c(4)^(2*n+1) + s(4)*c(1)^(2*n+1) + s(1)*c(2)^(2*n+1),
sqrt(7)*c1(n) = s(4)*c(4)^(2*n+1) + s(1)*c(1)^(2*n+1) + s(2)*c(2)^(2*n+1), then the following simple relationships between elements of these sequences hold true: a(n)=c1(n), c(n+1)=a1(n), -a(n)-b(n)=b1(n), which means that the sequences a1(n), b1(n), and c1(n) are completely and in very simple way determined by the sequences a(n), b(n) and c(n). However the last one's satisfy the following system of recurrence equations: a(n+1) = 2*a(n) + b(n), b(n+1) = a(n) + 2*b(n) - c(n), c(n+1) = c(n) - b(n). We have b(n)=A215694(n) and c(n)=A215695(n).
We note that a(n)=A000782(n) for every n=0,1,...,4 and A000782(5)-a(5)=2.
From general recurrence relation: a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), i.e. a(n) = 5*(a(n-1)-a(n-2)) + (a(n-3)-a(n-2)) the following summation formula can be easily obtained: sum{k=3,..,n} a(k) = 5*a(n-1)-a(n-2)+a(0)-5*a(1). Hence in discussed sequence it follows that: sum{k=3,..,n} a(k) = 5*a(n-1) - a(n-2) - 14.

			We have a(6) = 10*a(4)+a(1), a(5) = 11*(a(3)-a(1)), a(10)-a(4)+a(3)+a(1)+a(0) = 77*10^3, and a(11)-a(4)+a(3)-a(2)+a(0) = 25*10^4 = (5^6)*(2^4).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
Index entries for linear recurrences with constant coefficients, signature (5, -6, 1).

Cf.A215694, A215695, A215007, A215008, A215143, A215493, A215494, A215510, A215575, A215455, A214683, A214699.

I:=[1,3,8]; [n le 3 select I[n] else 5*Self(n-1) - 6*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 23 2018

Mathematica
```
LinearRecurrence[{5,-6,1}, {1,3,8}, 50]
```

x='x+O('x^30); Vec((1-2*x-x^2)/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 23 2018

G.f.: (1-2*x-x^2)/(1-5*x+6*x^2-x^3).

A214779 a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.

Original entry on oeis.org

-1, 1, -4, 4, -13, 16, -43, 61, -145, 226, -496, 823, -1714, 2965, -5965, 10609, -20860, 37792, -73189, 134236, -257359, 475897, -906313, 1685050, -3194836, 5961463, -11269558, 21079225, -39770137, 74507233, -140389636, 263291836, -495676141, 930265144

Offset: 0

Roman Witula, Jul 28 2012

Ramanujan-type sequence number 2 for argument 2Pi/9 is connected with the sequence A214699 (see also sequences A006053, A214683) - all have "similar" trigonometric description, for example in the case of a(n) the following formula hold true: 9^(1/3)*a(n) = (c(1)/c(4))^(1/3)*c(1)^n + (c(2)/c(1))^(1/3)*c(2)^n + (c(4)/c(2))^(1/3)*c(4)^n = -( (c(2)/c(1))^(1/3)*c(1)^(n+1) + (c(4)/c(2))^(1/3)*c(2)^(n+1) + (c(1)/c(4))^(1/3)*c(4)^(n+1) ), where c(j) := 2*Cos(2Pi*j/9) - for the proof see Witula et al.'s papers.
From a(0),A214699(0),a(2) and c(1)+c(2)+c(4)=0 we deduce
x^3 - 9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3))*(x - (c(2)/c(4))^(1/3))*(x - (c(4)/c(1))^(1/3)), and
x^3 - 7*9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3)*c(1)^2)*(x - (c(2)/c(4))^(1/3)*c(2)^2)*(x - (c(4)/c(1))^(1/3)*c(4)^2). We note that applying the Newton-Girard formulas to these polynomials two new sequences of real numbers can be discussed: X(n) := (c(1)/c(2))^(n/3) + (c(2)/c(4))^(n/3) + (c(4)/c(1))^(n/3), and Y(n) := ((c(1)/c(2))^(1/3)*c(1)^2)^n + ((c(2)/c(4))^(1/3)*c(2)^2)^n + ((c(4)/c(1))^(1/3)*c(4)^2)^n, where X(n)=9^(1/3)*X(n-2)+X(n-3), X(0)=3, X(1)=0, X(2)=2*9^(1/3), Y(n)=7*9^(1/3)Y(n-2)+Y(n-3), Y(0)=3, Y(1)=0, Y(2)=14*9^(1/3). It could be obtained the following decompositions: X(n) = ax(n) + 9^(1/3)*bx(n) + 81^(1/3)*cx(n), ax(0)=3, bx(0)=cx(0)=ax(1)=bx(1)=cx(1)=ax(2)=bx(2)=0, cx(2)=2, ax(n)=ax(n-3)+9*cx(n-2), bx(n)=bx(n-3)+ax(n-2), cx(n)=cx(n-3)+bx(n-2), and  Y(n) = ay(n) + 9^(1/3)*by(n) + 81^(1/3)*cy(n), ay(0)=3, by(0)=cy(0)=ay(1)=by(1)=cy(1)=ay(2)=cy(2)=0, by(2)=14, ay(n)=ay(n-3)+63*cy(n-2), by(n)=by(n-3)+7*ay(n-2), cy(n)=cy(n-3)+7*by(n-2). All these new sequence of positive integers ax(n),bx(n),...,cy(n) will be presented separately as A214778, A214951, A214954. - Roman Witula, Sep 27 2012
We note that all sums a(n+1) + a(n) are divisible by 3, which easily from recurrence formula for a(n) follows. Then it can be deduced the formula a(n+1) + a(n) = -A214699(n). - Roman Witula, Oct 06 2012

			From a(0)=-1 and A214699(0)=0 we obtain (c(1)/c(4))^(2/3) + (c(2)/c(1))^(2/3) + (c(4)/c(2))^(2/3) = 3*3^(1/3), whereas from a(1)=-1 and A214699(1)=3*3^(1/3) we get (c(1)/c(4))^(2/3)*2c(2) + (c(2)/c(1))^(2/3)*2c(4) + (c(4)/c(2))^(2/3)*2c(1) = 3*3^(1/3).

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.

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 (0,3,-1).

Cf. A214699, A214778, A214951, A214954, A217053, A217052, A217069.

LinearRecurrence[{0, 3, -1}, {-1, 1, -4}, 40] (* T. D. Noe, Jul 30 2012 *)

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

A052931 Expansion of 1/(1 - 3*x^2 - x^3).

Original entry on oeis.org

1, 0, 3, 1, 9, 6, 28, 27, 90, 109, 297, 417, 1000, 1548, 3417, 5644, 11799, 20349, 41041, 72846, 143472, 259579, 503262, 922209, 1769365, 3269889, 6230304, 11579032, 21960801, 40967400, 77461435, 144863001, 273351705, 512050438, 964918116, 1809503019

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Let A be the tridiagonal unit-primitive matrix (see [Jeffery]) A = A_{9,1} = [0,1,0,0; 1,0,1,0; 0,1,0,1; 0,0,1,1]. Then a(n)=[A^n](2,3). - _L. Edson Jeffery, Mar 19 2011
From Wolfdieter Lang, Oct 02 2013: (Start)
This sequence a(n) appears in the formula for the nonnegative powers of the algebraic number rho(9) := 2*cos(Pi/9) of degree 3, the ratio of the smallest diagonal/side in the regular 9-gon, in terms of the power basis of the algebraic number field Q(rho(9)) (see A187360, n=9).
rho(9)^n = A(n)*1 + B(n)*rho(9) + C(n)*rho(9)^2, with A(0) = 1, A(1) = 0, A(n) = B(n-2), n >= 2, B(0) = 0, B(n) = a(n-1), n >= 1, C(0) = 0, C(n) = B(n-1), n >= 1. (End)

			From _Wolfdieter Lang_, Oct 02 2013: (Start)
In the 9-gon (enneagon), powers of rho(9) = 2*cos(pi/9):
rho(9)^5 = A(5)*1 + B(5)*rho(9) + C(5)*rho(9)^2, with A(5) = B(3) = a(2) = 3, B(5) = a(4) = 9 and C(5) = B(4) = a(3) = 1:
  rho(9)^5 = 3 + 9*rho(9) + rho(9)^2. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
N. Gogin and A. Mylläri, Padovan-like sequences and Bell polynomials, Proceedings of Applications of Computer Algebra ACA, 2013.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 917
L. E. Jeffery, Unit-primitive matrices
Index entries for linear recurrences with constant coefficients, signature (0,3,1).

Cf. A214699.

a:=[1,0,3];; for n in [4..40] do a[n]:=3*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Oct 17 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/(1-3*x^2-x^3) )); // G. C. Greubel, Oct 17 2019

spec := [S,{S=Sequence(Prod(Z,Union(Z,Z,Z,Prod(Z,Z))))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
seq(coeff(series(1/(1-3*x^2-x^3), x, n+1), x, n), n = 0..40); # G. C. Greubel, Oct 17 2019

CoefficientList[Series[1/(1-3x^2-x^3),{x,0,40}],x] (* or *) LinearRecurrence[{0,3,1},{1,0,3},40] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)

x='x+O('x^40); Vec(1/(1-3*x^2-x^3)) \\ Altug Alkan, Feb 20 2018

def A052931_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(1/(1-3*x^2-x^3)).list()
A052931_list(40) # G. C. Greubel, Oct 17 2019

G.f.: 1/(1-3*x^2-x^3).
a(n) = 3*a(n-2) + a(n-3), with a(0)=1, a(1)=0, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+3*z^2+z^3)} (1/9)*(-1 +5*alpha +2*alpha^2) * alpha^(-1-n).
a(n) = Sum_{k=0..floor(n/2)} binomial(k, n-2k)3^(3k-n). - Paul Barry, Oct 04 2004
a(n) = A187497(3*(n+1)). - L. Edson Jeffery, Mar 19 2011.
3*a(n) = abs(A214699(n+1)). - Roman Witula, Oct 06 2012

More terms from James Sellers, Jun 06 2000

A214778 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 21.

Original entry on oeis.org

3, 3, 21, 84, 381, 1668, 7374, 32511, 143445, 632775, 2791506, 12314613, 54325650, 239656134, 1057236915, 4663973199, 20574997221, 90766067772, 400412159841, 1766407883376, 7792462676946, 34376247490935, 151649926417857, 668999726876127, 2951274986626458

Offset: 0

Roman Witula, Jul 28 2012

Ramanujan-type sequence number 3 for the argument 2Pi/9 is equal to the subsequence ax(3n) of the sequence ax(n), which (with its two conjugate sequences bx(n) and cx(n)) is defined in the comments to the sequence A214779 (we note that simultaneously we have bx(3n)=cx(3n)=0).
From example below follows that a(n) is equal to the sum of n-th powers of the roots of the polynomial x^3-3*x^2-6*x-1.
We note that all a(n) are divisible by 3 and a(n)/3 == 1 (mod 3). - Roman Witula, Oct 06 2012

			From a(1)=3 (after squaring) and a(2)=21 the following equality follows c(1)/c(4) + c(4)/c(2) + c(2)/c(1) = -6, which implies the decomposition x^3 - 3*x^2 - 6*x - 1 =(x - c(1)/c(2))*(x - c(2)/c(4))*(x - c(4)/c(1)).

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.

Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,6,1).

Cf. A214699, A214779.

LinearRecurrence[{3, 6, 1}, {3, 3, 21}, 40] (* T. D. Noe, Jul 30 2012 *)

Vec((3-6*x-6*x^2)/(1-3*x -6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

polsym(x^3 - 3*x^2 - 6*x - 1, 30) \\ Charles R Greathouse IV, Jul 20 2016

a(n) = (c(1)/c(2))^n + (c(2)/c(4))^n + (c(4)/c(1))^n, where c(j) := Cos(2*Pi*j/9).
G.f.: (3-6*x-6*x^2)/(1-3*x -6*x^2-x^3).
a(n+1) = A214951(n+1) - A214951(n). - Roman Witula, Oct 06 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

Roman Witula, Aug 20 2012

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,...

			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).

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).

Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

LinearRecurrence[{0,3,-1}, {0,-3,-3}, 50]

concat(0,Vec(-3*(1+x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

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

Roman Witula, Aug 20 2012

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,...

			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.

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).

Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

LinearRecurrence[{0,3,-1}, {0,-3,6}, 50]

concat(0,Vec(-3*(1-2*x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

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).

A214951 a(n) = 3a(n-1) + 6a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.

Original entry on oeis.org

2, 5, 26, 110, 491, 2159, 9533, 42044, 185489, 818264, 3609770, 15924383, 70250033, 309906167, 1367143082, 6031116281, 26606113502, 117372181274, 517784341115, 2284192224491, 10076654901437, 44452902392372, 196102828810229, 865102555686356, 3816377542312814

Offset: 0

Roman Witula, Jul 30 2012

Ramanujan-type sequence number 4 for the argument 2*Pi/9 is defined by the following relation: 9^(1/3)*a(n)=(c(1)/c(2))^(n - 1/3) + (c(2)/c(4))^(n - 1/3) + (c(4)/c(1))^(n - 1/3), where c(j) := Cos(2Pi*j/9) - for the proof see Witula et al.'s papers. We have a(n)=bx(3n-1), where the sequence bx(n) and its two conjugate sequences ax(n) and cx(n) are defined in the comments to the sequence A214779. We note that ax(3n-1)=cx(3n-1)=0. Moreover we have ax(3n)=A214778(n), bx(3n)=cx(3n)=0 and cx(3n+1)=A214954(n), ax(3n+1)=bx(3n+1)=0.

			We have 2*9^(1/3) = (c(2)/c(1))^(1/3) + (c(4)/c(2))^(1/3) + (c(1)/c(4))^(1/3), 5*9^(1/3) = (c(1)/c(2))^(2/3) + (c(2)/c(4))^(2/3) + (c(4)/c(1))^(2/3), and 110*9^(1/3)=(c(1)/c(2))^(8/3) + (c(2)/c(4))^(8/3) + (c(4)/c(1))^(8/3). Moreover we obtain a(6)-a(2)-a(1)-a(0)=9500, a(12)-a(2)-a(1)-a(0)=70250000 and a(12)-a(6)=3^3*43*a(1)*a(3)^2. - _Roman Witula_, Oct 06 2012

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. (in review)

Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,6,1).

Cf. A214778, A214779, A214954, A214699, A217053, A217052, A217069.

LinearRecurrence[{3, 6, 1}, {2, 5, 26}, 40] (* T. D. Noe, Jul 30 2012 *)

Vec((2-x-x^2)/(1-3*x-6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (2-x-x^2)/(1-3*x-6*x^2-x^3).

a(n+1) - a(n) = A214778(n+1). - Roman Witula, Oct 06 2012

A215560 a(n) = 3a(n-1) + 46a(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

Roman Witula, Aug 16 2012

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.

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

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 (3,46,1).

Cf. A214683, A215112, A006053, A006054, A215076, A215100, A120757, A214699.

LinearRecurrence[{3, 46, 1}, {3, 3, 101}, 50]

Vec((3-6*x-46*x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

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).

More terms from Michel Marcus, Apr 20 2016

cp's OEIS Frontend

A006054 a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Maxima

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A215512 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A214779 a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A052931 Expansion of 1/(1 - 3*x^2 - x^3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A214778 a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 21.

Views

Author

Keywords

Comments

Examples

References

Links

A215512 a(n) = 5a(n-1) - 6a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

A214778 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 21.

A214951 a(n) = 3a(n-1) + 6a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.

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