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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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14
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
COMMENTS
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.
REFERENCES
D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).
FORMULA
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).
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{0, 3, -1}, {3, 0, 6}, 50].
a(n) = - 6*a(n-1) - 9*a(n-2) - 3*a(n-3) with a(0)=3, a(1)=-6, a(2)=18.
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11
3, -6, 18, -63, 234, -891, 3429, -13257, 51354, -199098, 772173, -2995218, 11619045, -45073827, 174857211, -678335958, 2631522330, -10208681991, 39603398850, -153636822171, 596016389349, -2312177133105, 8969825761002
COMMENTS
The Berndt-type sequence number 2 for the argument 2Pi/9 . Similarly like the respective sequence number 1 -- see A215455 -- the sequence a(n) is connected with the following general recurrence relation: X(n+3) + 6*X(n+2) + 9*X(n+1) + ((2*cos(3*g))^2)*X(n) = 0, X(0)=3, X(1)=-6, X(2)=18. The Binet formula for this one has the form: X(n) = (-4)^n*((cos(g))^(2*n) + cos(g+Pi/3))^(2*n) + cos(g-Pi/3))^(2*n)) - for details see Witula-Slota's reference and comments to A215455. The characteristic polynomial of a(n) has the form x^3 + 6*x^2 + 9*x + 3 = (x + (2*cos(Pi/18))^2)*(x+(2*cos(5*Pi/18))^2)*(x+(2*cos(7*Pi/18))^2). We note that (2*cos(Pi/18))^2 = 2 - c(4), (2*cos(5*Pi/18))^2 = 2 - c(2), and (2*cos(7*Pi/18))^2 = 2 - c(1), where c(j) = 2*cos(2*Pi*j/9) - see trigonometric relations for A215455. Furthermore all numbers a(n)*3^(-ceiling((n+1)/3)) are integers.
REFERENCES
R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).
FORMULA
a(n) = (-4)^n*((cos(Pi/18))^(2*n) + (cos(5*Pi/18))^(2*n) + (cos(7*Pi/18))^(2*n)).
G.f.: (3 + 12*x + 9*x^2)/(1 + 6*x + 9*x^2 + 3*x^3).
a(n)*(-1)^n = s(1)^(2*n) + s(2)^(2*n) + s(4)^(2*n), where s(j) := 2*sin(2*Pi*j/9) -- for the proof see Witula's book. The respective sums with odd powers of sines in A216757 are given. - Roman Witula, Sep 15 2012
MATHEMATICA
LinearRecurrence[{-6, -9, -3}, {3, -6, 18}, 50]
CoefficientList[Series[(3 + 12 x + 9 x^2)/(1 + 6 x + 9 x^2 + 3 x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 30 2017 *)
PROG
(Magma) I:=[3, -6, 18]; [n le 3 select I[n] else -6*Self(n-1)-9*Self(n-2)-3*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Aug 30 2017
CROSSREFS
Cf. A215455, A215635, A215636, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.
a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.
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9
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
COMMENTS
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,...
REFERENCES
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).
FORMULA
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).
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{0, 3, -1}, {0, -3, -3}, 50].
a(n) = 3*a(n-1) - a(n-3), with a(0) = 3, a(1) = 3, and a(2) = 9.
+10
9
3, 3, 9, 24, 69, 198, 570, 1641, 4725, 13605, 39174, 112797, 324786, 935184, 2692755, 7753479, 22325253, 64283004, 185095533, 532961346, 1534601034, 4418707569, 12723161361, 36634883049, 105485941578, 303734663373, 874569107070, 2518221379632, 7250929475523
COMMENTS
The Berndt-type sequence number 5a for the argument 2Pi/9 defined by the first relation from the section "Formula". We see that a(n) is equal to the sum of the n-th negative powers of the c(j) := 2*cos(2*Pi*j/9), j=1,2,4 (the A215664(n) is equal to the respective n-th positive powers, further both sequences can be obtained from the two-sided recurrence relation: X(n+3) = 3*X(n+1) - X(n), n in Z, with X(-1) = X(0) = 3, and X(1) = 0). From the last formula in Witula's comments to A215664 it follows that 2*(-1)^n*a(n) = A215664(n)^2 - A215664(2*n). The following decomposition holds true: (X - c(1)^(-n))*(X - c(2)^(-n))*(X - c(4)^(-n)) = X^3 - a(n)*X^2 - (-1)^n* A215664(n)*X - (-1)^n. For n >= 1, a(n) is the number of cyclic (0,1,2)-compositions of n that avoid the pattern 110 provided the positions of the parts of the composition on the circle are fixed. (Similar comments hold for the pattern 012 and for the pattern 001.) - Petros Hadjicostas, Sep 13 2017 See the Maple program by Edlin and Zeilberger for counting the q-ary cyclic compositions of n that avoid one or more patterns provided the positions of the parts of the composition are fixed on the circle. The program is located at D. Zeilberger's personal website (see links). For the sequence here, q=3 and the pattern is A=110. - Petros Hadjicostas, Sep 13 2017
FORMULA
a(n) = c(1)^(-n) + c(2)^(-n) + c(4)^(-n) = (-c(1)*c(2))^n + (-c(1)*c(4))^n + (-c(2)*c(4))^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: Sum_{n>=0} a(n)*x^n = 3-3*x*(x^2-1)/(1-3*x+x^3) = 3*(1-2*x)/(1-3*x+x^3).
G.f. of Edlin and Zeilberger (2000): 1+Sum_{n>=1} a(n)*x^n = 1-3*x*(x^2-1)/(1-3*x+x^3) = (1-2*x^3)/(1-3*x+x^3). - Petros Hadjicostas, Sep 13 2017 a(n) = ceiling(r^n) for n >= 1, where r = 1/ A130880 is the largest root of x^3 - 3*x^2 + 1. - Tamas Lengyel, Feb 20 2022
EXAMPLE
For n=3, we have a(3) = 3^3 - 3 = 24 ternary cyclic compositions of n=3 (with fixed positions on the circle for the parts) that avoid 110 because we have to exclude 110, 101, and 011. - Petros Hadjicostas, Sep 13 2017
MATHEMATICA
LinearRecurrence[{3, 0, -1}, {3, 3, 9}, 50].
PROG
(PARI) x='x+O('x^99); Vec(3*(1-2*x)/(1-3*x+x^3)) \\ Altug Alkan, Sep 13 2017
a(n) = - 12*a(n-1) - 54*a(n-2) - 112*a(n-3) - 105*a(n-4) - 36*a(n-5) - 2*a(n-6) with a(0)=a(1)=a(2)=0, a(3)=-3, a(4)=24, a(5)=-135.
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7
0, 0, 0, -3, 24, -135, 660, -3003, 13104, -55689, 232500, -958617, 3916440, -15890355, 64127700, -257698347, 1032023136, -4121456625, 16421256420, -65301500577, 259259758056, -1027901275131, 4070632899300, -16104283594083, 63657906293520, -251447560563465, 992593021410900
COMMENTS
The Berndt-type sequence number 4 for the argument 2Pi/9 defined by the relation: X(n) = b(n) + a(n)*sqrt(2), where X(n) := ((cos(Pi/24))^(2*n) + (cos(7*Pi/24))^(2*n) + ((cos(3*Pi/8))^(2*n))*(-4)^n. We have b(n)= A215635(n) (see also section "Example" below). For more details - see comments to A215635, A215634 and Witula-Slota's reference.
FORMULA
G.f.: (-3*x^3-12*x^4-9*x^5)/(1+12*x+54*x^2+112*x^3+105*x^4+36*x^5+2*x^6).
EXAMPLE
We have X(1)=-6, X(2)=18 and X(3)=-60-3*sqrt(2), which implies the equality: (cos(Pi/24))^6 + (cos(7*Pi/24))^6 + ((cos(3*Pi/8))^6 = (60+3*sqrt(2))/64.
MATHEMATICA
LinearRecurrence[{-12, -54, -112, -105, -36, -2}, {0, 0, 0, -3, 24, -135}, 50]
CROSSREFS
Cf. A215455, A215634, A215635, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.
a(n) = -3*a(n-1) + 9*a(n-2) + 3*a(n-3), with a(0)=3, a(1)=-3, a(2)=27.
+10
7
3, -3, 27, -99, 531, -2403, 11691, -55107, 263331, -1250883, 5957307, -28339875, 134882739, -641835171, 3054430539, -14535159939, 69169849155, -329162695299, 1566411248475, -7454188455651, 35472778517331, -168806797907427, 803312835011307
COMMENTS
The Berndt-type sequence number 8 for the argument 2*Pi/9 defined by the trigonometric relations from the Formula section below.
From the general recurrence relation: b(n) = -3*b(n-1) + 9*b(n-2) + 3*b(n-3), i.e., b(n) - b(n-2) = 8*b(n-2) + 3(b(n-3) - b(n-1)) the following summation formulas can be easily deduced: b(2*n+1) + 3*b(2*n) - 3*b(0) - b(1) = 8*Sum_{k=1..n} b(2*k-1) and b(2*n+2) + 3*b(2*n+1) - b(2) - 3*b(1) = 8*Sum_{k=1..n} b(2*k). Hence it follows that (a(2*n+1) + 3*a(2*n))/2 are all integers congruent to 3 modulo 4, and (a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.
We note that all numbers 3^(-1-floor(n/3))*a(n) = A215831(n) and 3^(-n-2)*a(3*n+2) are integers. The following decomposition holds true: (X - k(1)^n)*(X - (-k(2))^n)*(X - k(3)^n) = X^3 - sqrt(3)^(-n)*a(n)*X^2 + sqrt(3)^(-n)*T(n) - sqrt(3)^(-n), where T(2*n+1) = sqrt(3)* A215945(n) and T(2*n) = A215948(n). [ Roman Witula, Aug 30 2012]
REFERENCES
D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).
FORMULA
a(n) = (k(1)^n + (-k(2))^n + k(4)^n)*(sqrt(3))^n = (-1+4*c(1))^n + (-1+4*c(2))^n + (-1+4*c(4))^n, where k(j) := cot(2*Pi*j/9) and c(j) := cos(2*Pi*j/9).
G.f.: (3 + 6*x - 9*x^2)/(1 + 3*x - 9*x^2 - 3*x^3). [corrected by Georg Fischer, May 10 2019]
EXAMPLE
We have k(1)^3 - k(2)^3 + k(4)^3 = -11*sqrt(3).
MATHEMATICA
LinearRecurrence[{-3, 9, 3}, {3, -3, 27}, 50]
CROSSREFS
Cf. A215945, A215948, A216034, A215455, A215634, A215635, A215636, A215664, A215665, A215666, A215831, A215575, A108716, A215794, A215828.
a(n) = -3*a(n-1) + a(n-3), with a(0)=0, a(1)=6, and a(2)=-15.
+10
7
0, 6, -15, 45, -129, 372, -1071, 3084, -8880, 25569, -73623, 211989, -610398, 1757571, -5060724, 14571774, -41957751, 120812529, -347865813, 1001639688, -2884106535, 8304453792, -23911721688, 68851058529, -198248721795, 570834443697, -1643652272562
COMMENTS
The Berndt-type sequence number 9 for the argument 2Pi/9 defined by the first relation from the section "Formula" below.
We have a(n) = 3*(-1)^(n+1)* A215448(n+1). From the recurrence formula for a(n) it follows that all a(3*n) are divisible by 9, a(3*n+1)/3 are congruent to 2 modulo 3, and a(3*n+2)/3 are congruent to 1 modulo 3. In the consequence also all sums a(n)+a(n+1)+a(n+2) are divisible by 9. From general recurrence X(n) = -3*X(n-1) + X(n-3) the following formula can be deduced: 3*sum{k=2,..,n-1} X(k) = -X(n)-X(n-1)-X(n-2)+X(2)+X(1)+X(0). Hence, in the case of a(n) we obtain 3*sum{k=2,..,n-1} a(k) = -a(n)-a(n-1)-a(n-2)-9.
If we set X(n) = -3*X(n-1) + X(n-3), n in Z, with a(n) = X(n) for n=0,1,... then X(-n) = abs( A215666(n)) = (-1)^n* A215666(n), for every n=0,1,... The following decomposition holds true (X - c(1)*(-c(4))^(-n))*(X - c(2)*(-c(1))^(-n))*(X - c(4)*(-c(2))^(-n)) = X^3 - a(n)*X^2 + (-1)^n*( A215665(n) - A215664(n))*X + 1.
REFERENCES
D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).
FORMULA
a(n) = c(1)*(-c(4))^(-n) + c(2)*(-c(1))^(-n) + c(4)*(-c(2))^(-n), where c(j) := 2*cos(2*Pi*j/9).
G.f.: 3*x(x+2)/(1+3*x-x^3).
MAPLE
We have a(3) + 3*a(2) = 0, a(8) + 24*a(5) = 48 = a(3) + a(1)/2.
MATHEMATICA
LinearRecurrence[{-3, 0, 1}, {0, 6, -15}, 50].
a(n) = 3^(-1+floor(n/2))*A(n), where A(n) = 3*A(n-1) + A(n-2) - A(n-3)/3 with A(0)=A(1)=3, A(2)=11.
+10
1
1, 1, 11, 35, 345, 1129, 11091, 36315, 356721, 1168017, 11473371, 37567443, 369023049, 1208298105, 11869049763, 38863020555, 381749439969, 1249968331809, 12278374244523, 40203278289027, 394914722339385, 1293075627640713
COMMENTS
The Berndt-type sequence number 14 for the argument 2Pi/9 defined by the relation: A(n)*(-sqrt(3))^n = t(1)^n + (-t(2))^n + t(4)^n = (-sqrt(3) + 4*s(1))^n + (-sqrt(3) - 4*s(2))^n + (-sqrt(3) + 4*s(4))^n, where s(j) := sin(2*Pi*j/9) and t(j) := tan(2*Pi*j/9).
The definitions of the other Berndt-type sequences for the argument 2Pi/9 like A215945, A215948, A216034 in Crossrefs are given. We note that all a(2*n), n=2,3,..., are divisible by 3, and it is only when n=5 that a(2*n) is divisible by 9.
REFERENCES
D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the ninth order, (submitted, 2012).
R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math. (in press, 2012).
FORMULA
G.f.: (1+x-22*x^2+2*x^3+9*x^4+x^5)/(1-33*x^2+27*x^4-3*x^6). - Bruno Berselli, Oct 01 2012
EXAMPLE
Note that A(0)=A(1)=3, a(0)=a(1)=1, A(2)=a(2)=11, A(3)=a(3)=35, A(4)=115, a(4)=345 and A(5) = 1129/3, which implies the equality 3387*sqrt(3) = -t(1)^5 + t(2)^5 - t(4)^5.
MATHEMATICA
LinearRecurrence[{0, 33, 0, -27, 0, 3}, {1, 1, 11, 35, 345, 1129}, 25] (* Paolo Xausa, Feb 23 2024 *)
PROG
(Magma) /* By definition: */ i:=22; I:=[3, 3, 11]; A:=[m le 3 select I[m] else 3*Self(m-1)+Self(m-2)-Self(m-3)/3: m in [1..i]]; [3^(-1+Floor((n-1)/2))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012
CROSSREFS
Cf. A215455, A215634- A215636, A215664, A215885, A215665, A215666, A215829, A215831, A215917, A215919, A215945, A216034, A215948, A216757.
The sequence of coefficients of cubic polynomials p(x+n), where p(x) = x^3 - 3*x + 1.
+10
1
1, 0, -3, 1, 1, 3, 0, -1, 1, 6, 9, 3, 1, 9, 24, 19, 1, 12, 45, 53, 1, 15, 72, 111, 1, 18, 105, 199, 1, 21, 144, 323, 1, 24, 189, 489, 1, 27, 240, 703, 1, 30, 297, 971, 1, 33, 360, 1299, 1, 36, 429, 1693, 1, 39, 504, 2159, 1, 42, 585, 2703, 1, 45, 672, 3331
COMMENTS
We note that p(x) = (x - s(1))*(x + c(1))*(x - c(2)),
p(x+1) = x^3 + 3*x^2 -1 = (x + s(1)*c(1))*(x - s(1)*c(2))*(x + c(1)*c(2)), p(x+2) = x^3 + 6*x^2 + 9*x + 3 = (x + c(1/2)^2)*(x + s(2)^2)*(x + s(4)^2), and p(x + n) = (x + n - 2 + c(1/2)^2)*(x + n - 2 + s(2)^2)*(x + n - 2 + s(4)^2), n = 2,3,..., where c(j) := 2*cos(Pi*j/9) and s(j) := 2*sin(Pi*j/18). These one's are characteristic polynomials many sequences A... - see crossrefs.
A218332 is the sequence of coefficients of polynomials p(x-n).
FORMULA
We have a(4*k) = 1, a(4*k + 1) = 3*k, a(4*k + 2) = 3*k^2 - 3, and a(4*k + 3) = k^3 - 3*k + 1. Moreover we obtain
b(k+1) = b(k) + 3, c(k+1) = 2*b(k) + c(k) + 3, d(k+1) = b(k) + c(k) + d(k) + 1, where p(x + k) = x^3 + b(k)*x^2 + c(k)*x + d(k).
Empirical g.f.: -(3*x^15-3*x^13+x^12-13*x^11+9*x^10+6*x^9-3*x^8+5*x^7-12*x^6-3*x^5+3*x^4-x^3+3*x^2-1) / ((x-1)^4*(x+1)^4*(x^2+1)^4). - Colin Barker, May 17 2013
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