A214779 -id:A214779 - 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

A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.

Original entry on oeis.org

0, 3, 0, 9, -3, 27, -18, 84, -81, 270, -327, 891, -1251, 3000, -4644, 10251, -16932, 35397, -61047, 123123, -218538, 430416, -778737, 1509786, -2766627, 5308095, -9809667, 18690912, -34737096, 65882403, -122902200, 232384305, -434589003, 820055115, -1536151314

Offset: 0

Roman Witula, Jul 26 2012

All a(n) are divisible by 3.
The Ramanujan-type sequence number 1 for the argument 2*Pi/9 defined by the following identity:
3^(1/3)*a(n) = (c(1)/c(2))^(1/3)*c(1)^n + (c(2)/c(4))^(1/3)*c(2)^n + (c(4)/c(1))^(1/3)*c(4)^n = -( (c(1)/c(2))^(1/3)*c(2)^(n+1) + (c(2)/c(4))^(1/3)*c(4)^(n+1) + (c(4)/c(1))^(1/3)*c(1)^(n+1) ), where c(j) := 2*cos(2*Pi*j/9).
The definitions of other Ramanujan-type sequences, for the argument of 2*Pi/9 in one's, are given in the Crossrefs section.

			We have a(2) = a(1) + a(4) = a(4) + a(7) + a(8) = -a(3) + a(5) + a(6) = 0, which implies
(c(1)/c(2))^(1/3)*c(1)^2 + (c(2)/c(4))^(1/3)*c(2)^2 + (c(4)/c(1))^(1/3)*c(4)^2 = (c(1)/c(2))^(1/3)*(c(1) + c(1)^4) + (c(2)/c(4))^(1/3)*(c(2) + c(2)^4) + (c(4)/c(1))^(1/3)*(c(4) + c(4)^4) = (c(1)/c(2))^(1/3)*(c(1)^4 + c(1)^7 + c(1)^8) + (c(2)/c(4))^(1/3)*(c(2)^4 + c(2)^7 + c(2)^8) + (c(4)/c(1))^(1/3)*(c(4)^4 + c(4)^7 + c(4)^8) = 0.
Moreover we have 3000*3^(1/3) = (c(1)/c(2))^(1/3)*c(1)^13 + (c(2)/c(4))^(1/3)*c(2)^13 + (c(4)/c(1))^(1/3)*c(4)^13. - _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

G. C. Greubel, Table of n, a(n) for n = 0..1000
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. A006053, A052931, A214683, A214778, A214779.

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

[n le 3 select 3*(1+(-1)^n)/2 else 3*Self(n-2) - Self(n-3): n in [1..40]]; // G. C. Greubel, Jan 08 2024

LinearRecurrence[{0,3,-1}, {0,3,0}, 30]
CoefficientList[Series[3*x/(1 - 3*x^2 + x^3),{x,0,34}],x] (* James C. McMahon, Jan 09 2024 *)

def a(n): # a=A214699
    if (n<3): return 3*(n%2)
    else: return 3*a(n-2) - a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Jan 08 2024

G.f.: 3*x/(1 - 3*x^2 + x^3).
From Roman Witula, Oct 06 2012: (Start)
a(n+1) = 3*(-1)^n*A052931(n), which from recurrence relations for a(n) and A052931 can easily be proved inductively.
a(n) = -A214779(n+1) - A214779(n). (End)

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

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

A214954 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

Original entry on oeis.org

0, 2, 7, 33, 143, 634, 2793, 12326, 54370, 239859, 1058123, 4667893, 20592276, 90842309, 400748476, 1767891558, 7799007839, 34405121341, 151777302615, 669561643730, 2953753868221, 13030408769658, 57483311162030, 253586139972259, 1118688695658615

Offset: 0

Roman Witula, Jul 30 2012

Ramanujan-type sequence number 5 for the argument 2*Pi/9 is defined by the following relation: 81^(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's et al. papers. We have a(n)=cx(3n+1), where the sequence cx(n) and its two conjugate sequences ax(n) and bx(n) are defined in the comments to the sequence A214779. We note that ax(3n+1)=bx(3n+1)=0. Further we have ax(3n)=A214778(n), bx(3n)=cx(3n)=0 and bx(3n-1)=A214951(n), ax(3n-1)=cx(3n-1)=0.

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. A214779, A214778, A214951, A214699.

LinearRecurrence[{3, 6, 1}, {0, 2, 7}, 40] (* T. D. Noe, Jul 30 2012 *)
CoefficientList[Series[(2x+x^2)/(1-3x-6x^2-x^3),{x,0,30}],x] (* Harvey P. Dale, Sep 13 2021 *)

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).

A217052 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

Original entry on oeis.org

1, 1, 19, 82, 703, 4096, 29242, 186733, 1266103, 8309143, 55500634, 367187437, 2441886670, 16193659132, 107553444913, 713750040577, 4738726458775, 31453733795086, 208804386436435, 1386041496850144, 9200883498819958, 61076450807299765, 405436597890428431

Offset: 0

Roman Witula, Sep 25 2012

The Ramanujan type sequence number 10 for the argument 2*Pi/9 defined by the relation a(n) = ((1/3 - c(1))^n + (1/3 - c(2))^n + (1/3 - c(4))^n)*3^(n-1), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are sequences A217053 and A217069.
  For more informations about connections a(n) with these two sequences - see comments in A217053.
The 3-valuation of the sequence a(n) is equal to (1).

			We have a(4)=37*a(2) and a(5) = 2^(12), which implies (1/3 - c(1))^4 + (1/3 - c(2))^4 + (1/3 - c(4))^4 = (37/9)*((1/3 - c(1))^2 + (1/3 - c(2))^2 + (1/3 - c(4))^2) = (37/27)*19 = 703/27, (1/3 - c(1))^5 + (1/3 - c(2))^5 + (1/3 - c(4))^5 = (8/3)^4. Moreover we have a(10) = 676837*a(3).

Roman 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, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217053, A217069, A214699, A214779, A214778, A214951, A214954, A215560, A214569, A215572.

LinearRecurrence[{3,24,1}, {1,1,19}, 30]

Vec((1-2*x-8*x^2)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (1-2*x-8*x^2)/(1-3*x-24*x^2-x^3).

A217053 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0) = 2, a(1) = 5, and a(2) = 62.

Original entry on oeis.org

2, 5, 62, 308, 2417, 14705, 102431, 662630, 4460939, 29388368, 195890270, 1297452581, 8623112591, 57204089987, 379864424726, 2521114546457, 16737293922782, 111098495308040, 737511654617345, 4895636145167777, 32498286641627651, 215727639063526946

Offset: 0

Roman Witula, Sep 25 2012

The Ramanujan type sequence number 9 for the argument 2*Pi/9 defined by the following relation a(n)*3^(-n + 1/3) = ((-1)^n)*(c(1) - 1/3)^(n + 2/3) + ((-1)^n)*(c(2) - 1/3)^(n + 2/3) + (1/3 - c(4))^(n + 2/3), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are the sequences A217052 and A217069.
In Witula's et al.'s paper the following sequence is discussed: S(n) := sum{j=0,1,2} (1/3 - c(2^j))^(n/3). It is proved that S(n+3) = (3^(1/3))*S(n+1) + S(n)/3, n=0,1,... Moreover the following decomposition holds true S(n) = x(n) + y(n)*3^(1/3) + z(n)*9^(1/3), which implies the system of recurrence relations: x(n+3) = 3*z(n+1) + x(n)/3, y(n+3) = x(n+1) + y(n)/3, z(n+3) = y(n+1) + z(n)/3, x(0)=3, y(0)=z(0)=x(1)=y(1)=z(1)=x(2)=z(2)=0, y(2)=2. Then it can be generated the relations X'(n+9) - 3*X'(n+6) - 24*X'(n+3) - X'(n) = 0, where X'(n) = X(n)*3^n for every X=x,y,z and n=0,1,..., from which we obtain that x(3*n+1)=x(3*n+2)=y(3*n)=y(3*n+1)=z(3*n)=z(3*n+2)=0 and a(n) = y(3*n+2)*3^n, A217052(n) = x(3*n)*3^(n-1), and A217069(n) = z(3*n+1)*3^(n-1).
Each a(n)+1 is divisible by 3 but which a(n)+1 are divisible by 9 - it is a question?

			We have 2*3^(1/3) = (c(1) - 1/3)^(2/3) + (c(2) - 1/3)^(2/3) + (1/3 - c(4))^(2/3), and 5*3^(-2/3) = -(c(1) - 1/3)^(5/3) - (c(2) - 1/3)^(5/3) + (1/3 - c(4))^(5/3).
Moreover we have 12*a(1) + a(0) = a(2), 5*a(2) = a(3) + a(0).

Roman 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, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217052, A217069, A214699, A214778, A214779, A214951, A214954, A215560, A215569, A215572.

LinearRecurrence[{3,24,1}, {2,5,62}, 30]

Vec((2-x-x^2)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012

G.f.: (2-x-x^2)/(1-3*x-24*x^2-x^3).

A217069 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.

Original entry on oeis.org

0, 2, 7, 69, 377, 2794, 17499, 119930, 782560, 5243499, 34631867, 230522137, 1527974718, 10151087309, 67355177296, 447219602022, 2968334148479, 19705628071261, 130804123379301, 868315777996646, 5763951923164423, 38262238564792074, 253989877628319020

Offset: 0

Roman Witula, Sep 26 2012

The Ramanujan sequence number 11 for the argument 2*Pi/9 defined by the relation a(n)*9^(1/3) = (3^(n-1))*(((-1)^(n-1))*(c(1) - 1/3)^(n + 1/3) + ((-1)^(n-1))*(c(2) - 1/3)^(n + 1/3) + (1/3 - c(4))^(n + 1/3)), where c(j) := 2*cos(2*Pi*j/9). The sequences A217052 and A217053 are conjugate with the sequence a(n). For more information on these connections - see Comments in A217053.

The 3-valuation of the sequence a(n) is equal to (0,2,1).

			We have a(4)-5*a(3)=32, 8*a(4)-a(5)=222, a(9)-a(6)=5226000. Furthermore from a(0)=0 we get (c(1) - 1/3)^( 1/3) + (c(2) - 1/3)^(1/3) = (1/3 - c(4))^(1/3), while from a(3)=69 we obtain 23*9^(-1/6) = (c(1) - 1/3)^(10/3) + (c(2) - 1/3)^(10/3) + (1/3 - c(4))^(10/3).

Roman 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, Eger, Hungary, 2012, in review.

Index entries for linear recurrences with constant coefficients, signature (3,24,1).

Cf. A217052, A217053, A214699, A214779, A214778, A214951, A214954, A215560, A215569, A215572.

Mathematica
```
LinearRecurrence[{3,24,1}, {0,2,7}, 30]
```

Vec((2+x)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 27 2012

G.f.: x*(2+x)/(1-3*x-24*x^2-x^3).

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.

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A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.

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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.

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

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A214954 a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

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A217052 a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

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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.

A214954 a(n) = 3a(n-1) + 6a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, and a(2) = 7.

A217052 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.

A217053 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0) = 2, a(1) = 5, and a(2) = 62.

A217069 a(n) = 3a(n-1) + 24a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.