Search: a214699 -id:a214699
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A006054
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a(n) = 2*a(n-1) + a(n-2) - a(n-3), with a(0) = a(1) = 0, a(2) = 1.
(Formerly M1396)
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+10
59
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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
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OFFSET
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0,4
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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Maximilian Fichtner, K. Voigt, 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.
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FORMULA
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G.f.: x^2/(1-2*x-x^2+x^3).
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+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
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EXAMPLE
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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
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MAPLE
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MATHEMATICA
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PROG
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(Maxima)
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
(PARI) x='x+O('x^66);
concat([0, 0], Vec(x^2/(1-2*x-x^2+x^3))) \\ Joerg Arndt, May 05 2011
(Haskell)
a006054 n = a006053_list !! n
a006054_list = 0 : 0 : 1 : zipWith (+) (map (2 *) $ drop 2 a006054_list)
(zipWith (-) (tail a006054_list) a006054_list)
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CROSSREFS
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Cf. A005578, A006053, A006356, A007583, A080937, A094790, A214683, A214699, A214779, A215112, A306334.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A215664
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a(n) = 3*a(n-2) - a(n-3), with a(0)=3, a(1)=0, and a(2)=6.
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+10
14
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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
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OFFSET
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0,1
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COMMENTS
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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.
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REFERENCES
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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).
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MATHEMATICA
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LinearRecurrence[{0, 3, -1}, {3, 0, 6}, 50].
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PROG
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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A215512
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a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.
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+10
11
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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FORMULA
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G.f.: (1-2*x-x^2)/(1-5*x+6*x^2-x^3).
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EXAMPLE
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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).
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MATHEMATICA
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LinearRecurrence[{5, -6, 1}, {1, 3, 8}, 50]
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PROG
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(PARI) 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
(Magma) 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
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CROSSREFS
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Cf.A215694, A215695, A215007, A215008, A215143, A215493, A215494, A215510, A215575, A215455, A214683, A214699.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A214779
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a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.
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+10
10
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-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
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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G.f.: -(1-x+x^2)/(1-3*x^2+x^3).
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EXAMPLE
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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).
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MATHEMATICA
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LinearRecurrence[{0, 3, -1}, {-1, 1, -4}, 40] (* T. D. Noe, Jul 30 2012 *)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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A052931
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Expansion of 1/(1 - 3*x^2 - x^3).
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+10
9
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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
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OFFSET
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0,3
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COMMENTS
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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
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)
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LINKS
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FORMULA
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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
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EXAMPLE
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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)
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(PARI) x='x+O('x^40); Vec(1/(1-3*x^2-x^3)) \\ Altug Alkan, Feb 20 2018
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/(1-3*x^2-x^3) )); // G. C. Greubel, Oct 17 2019
(Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P(1/(1-3*x^2-x^3)).list()
(GAP) 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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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encyclopedia(AT)pommard.inria.fr, Jan 25 2000
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EXTENSIONS
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STATUS
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approved
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A214778
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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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+10
9
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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
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OFFSET
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0,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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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)).
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MATHEMATICA
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LinearRecurrence[{3, 6, 1}, {3, 3, 21}, 40] (* T. D. Noe, Jul 30 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A215665
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a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.
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+10
9
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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
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OFFSET
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0,2
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COMMENTS
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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,...
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REFERENCES
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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).
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MATHEMATICA
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LinearRecurrence[{0, 3, -1}, {0, -3, -3}, 50].
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PROG
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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A215666
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a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=-3, and a(2)=6.
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+10
9
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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
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OFFSET
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0,2
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COMMENTS
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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,...
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REFERENCES
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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.
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MATHEMATICA
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LinearRecurrence[{0, 3, -1}, {0, 3, -6}, 50].
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PROG
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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A214951
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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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+10
7
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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
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OFFSET
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0,1
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COMMENTS
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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.
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REFERENCES
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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)
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LINKS
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FORMULA
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G.f.: (2-x-x^2)/(1-3*x-6*x^2-x^3).
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EXAMPLE
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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
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MATHEMATICA
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LinearRecurrence[{3, 6, 1}, {2, 5, 26}, 40] (* T. D. Noe, Jul 30 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A215560
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a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=a(1)=3, a(2)=101.
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+10
7
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3, 3, 101, 444, 5981, 38468, 390974, 2948431, 26868565, 216624495, 1888775906, 15657923053, 134074085330, 1124375492334, 9556192325235, 80523923708399, 682280993578341, 5760499663646612, 48746948619251921, 411906111379078256, 3483838470286469746, 29447943482916260935
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OFFSET
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0,1
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COMMENTS
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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.
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REFERENCES
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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
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LINKS
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FORMULA
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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).
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MATHEMATICA
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LinearRecurrence[{3, 46, 1}, {3, 3, 101}, 50]
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PROG
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(PARI) Vec((3-6*x-46*x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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