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a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.
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19
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
COMMENTS
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.
REFERENCES
R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012
FORMULA
G.f.: 3*x/(1 - 3*x^2 + x^3).
a(n+1) = 3*(-1)^n* A052931(n), which from recurrence relations for a(n) and A052931 can easily be proved inductively.
EXAMPLE
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
MATHEMATICA
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 *)
PROG
(Magma) [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 (SageMath)
if (n<3): return 3*(n%2)
else: return 3*a(n-2) - a(n-3)
Decimal expansion of sine of 70 degrees.
+10
11
9, 3, 9, 6, 9, 2, 6, 2, 0, 7, 8, 5, 9, 0, 8, 3, 8, 4, 0, 5, 4, 1, 0, 9, 2, 7, 7, 3, 2, 4, 7, 3, 1, 4, 6, 9, 9, 3, 6, 2, 0, 8, 1, 3, 4, 2, 6, 4, 4, 6, 4, 6, 3, 3, 0, 9, 0, 2, 8, 6, 6, 6, 2, 7, 7, 4, 2, 2, 1, 2, 1, 0, 9, 9, 5, 8, 8, 9, 4, 5, 8, 9, 4, 9, 7, 4, 5, 8, 8, 9, 8, 3, 7, 9, 4, 8, 0, 6, 7
COMMENTS
It is well known that the length sin 70° (cos 20°) is not constructible with ruler and compass, since it is a root of the irreducible polynomial 8x^3 - 6x - 1 and 3 fails to divide any power of 2. - Jean-François Alcover, Aug 10 2014 [cf. the Maxfield ref.] The minimal polynomial of cos(Pi/9) is 8*x^3 - 6*x - 1 with discriminant (2^6)*(3^4), a square: hence the Galois group of the algebraic number field Q(sin(70°) over Q is the cyclic group of order 3.
The two other zeros of the minimal polynomial are cos(5*Pi/9) = - A019819 and cos(7*Pi/9) = - A019859. The mapping z -> 1 - 2*z^2 cyclically permutes the three zeros. The inverse permutation is given by the mapping z -> 2*z^2 - z - 1. (End)
REFERENCES
J. E. Maxfield and M. W. Maxfield, Abstract Algebra and Solution by Radicals, Dover Publications ISBN 0-486-67121-6, (1992), p. 197.
MATHEMATICA
RealDigits[Sin[70 Degree], 10, 120][[1]] (* Harvey P. Dale, Aug 17 2012 *)
a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.
+10
10
-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
COMMENTS
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
REFERENCES
R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.
FORMULA
G.f.: -(1-x+x^2)/(1-3*x^2+x^3).
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{0, 3, -1}, {-1, 1, -4}, 40] (* T. D. Noe, Jul 30 2012 *)
Decimal expansion of 2*cos(Pi/9).
+10
9
1, 8, 7, 9, 3, 8, 5, 2, 4, 1, 5, 7, 1, 8, 1, 6, 7, 6, 8, 1, 0, 8, 2, 1, 8, 5, 5, 4, 6, 4, 9, 4, 6, 2, 9, 3, 9, 8, 7, 2, 4, 1, 6, 2, 6, 8, 5, 2, 8, 9, 2, 9, 2, 6, 6, 1, 8, 0, 5, 7, 3, 3, 2, 5, 5, 4, 8, 4, 4, 2, 4, 2, 1, 9, 9, 1, 7, 7, 8, 9, 1, 7, 8, 9, 9, 4, 9, 1, 7, 7, 9, 6, 7, 5, 8, 9, 6, 1, 3, 4, 9
COMMENTS
This algebraic number called rho(9) of degree 3 = A055034(9) has minimal polynomial C(9, x) = x^3 - 3*x - 1 (see A187360). rho(9) gives the length ratio diagonal/side of the smallest diagonal in the regular 9-gon.
The length ratio diagonal/side of the second smallest and the third smallest (or the largest) diagonal in the regular 9-gon are rho(9)^2 - 1 = A332438 - 1 and rho(9) + 1, respectively. - Mohammed Yaseen, Oct 31 2020
FORMULA
rho(9) = 2*cos(Pi/9).
Equals (-1)^(-1/9)*((-1)^(1/9) - i)*((-1)^(1/9) + i). - Peter Luschny, Mar 27 2020 The zeros of x^3 - 3*x - 1 are r_1 = -2*cos(2*Pi/9), r_2 = -2*cos(4*Pi/9) and r_3 = -2*cos(8*Pi/9) = 2*cos(Pi/9).
The polynomial x^3 - 3*x - 1 is irreducible over Q (since it is irreducible mod 2) with discriminant equal to 3^4, a square. It follows that the Galois group of the number field Q(2*cos(Pi/9)) over Q is cyclic of order 3.
The mapping r -> 2 - r^2 cyclically permutes the zeros r_1, r_2 and r_3. The inverse cyclic permutation is given by r -> r^2 - r - 2.
The first differences r_1 - r_2, r_2 - r_3 and r_3 - r_1 are the zeros of the cyclic cubic polynomial x^3 - 9*x - 9 of discriminant 3^6.
First quotient relations:
r_1/r_2 = 1 + (r_3 - r_1); r_2/r_3 = 1 + (r_1 - r_2); r_3/r_1 = 1 + (r_2 - r_3);
r_2/r_1 = (r_3 - r_2) - 2; r_3/r_2 = (r_1 - r_3) - 2; r_1/r_3 = (r_2 - r_1) - 2;
r_1/r_2 + r_2/r_3 + r_3/r_1 = 3; r_2/r_1 + r_3/r_2 + r_1/r_3 = -6.
Thus the first quotients r_1/r_2, r_2/r_3 and r_3/r_1 are the zeros of the cyclic cubic polynomial x^3 - 3*x^2 - 6*x - 1 of discriminant 3^6. See A214778. Second quotient relations:
(r_1*r_2)/(r_3^2) = 3*r_2 - 6*r_1 - 8, with two other similar relations by cyclically permuting the 3 zeros. The three second quotients are the zeros of the cyclic cubic polynomial x^3 + 24*x^2 + 3*x - 1 of discriminant 3^10.
(r_1^2)/(r_2*r_3) = 1 - 3*(r_2 + r_3), with two other similar relations by cyclically permuting the 3 zeros. (End)
EXAMPLE
rho(9) = 1.87938524157181676810821855464946293987241626852892926618...
MATHEMATICA
RealDigits[2 * Cos[Pi/9], 10, 100][[1]] (* Amiram Eldar, Mar 27 2020 *)
a(n) = 3*a(n-1) + 6*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=26.
+10
7
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
COMMENTS
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.
REFERENCES
R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012. (in review)
FORMULA
G.f.: (2-x-x^2)/(1-3*x-6*x^2-x^3).
EXAMPLE
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
MATHEMATICA
LinearRecurrence[{3, 6, 1}, {2, 5, 26}, 40] (* T. D. Noe, Jul 30 2012 *)
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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6
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
COMMENTS
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.
REFERENCES
R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012. (in review)
FORMULA
G.f.: (2*x+x^2)/(1-3*x-6*x^2-x^3).
MATHEMATICA
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 *)
a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.
+10
5
1, 1, 19, 82, 703, 4096, 29242, 186733, 1266103, 8309143, 55500634, 367187437, 2441886670, 16193659132, 107553444913, 713750040577, 4738726458775, 31453733795086, 208804386436435, 1386041496850144, 9200883498819958, 61076450807299765, 405436597890428431
COMMENTS
The Ramanujan type sequence number 10 for the argument 2Pi/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).
REFERENCES
R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.
FORMULA
G.f.: (1-2*x-8*x^2)/(1-3*x-24*x^2-x^3).
EXAMPLE
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)
MATHEMATICA
LinearRecurrence[{3, 24, 1}, {1, 1, 19}, 30]
a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0) = 2, a(1) = 5, and a(2) = 62.
+10
5
2, 5, 62, 308, 2417, 14705, 102431, 662630, 4460939, 29388368, 195890270, 1297452581, 8623112591, 57204089987, 379864424726, 2521114546457, 16737293922782, 111098495308040, 737511654617345, 4895636145167777, 32498286641627651, 215727639063526946
COMMENTS
The Ramanujan type sequence number 9 for the argument 2Pi/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?
REFERENCES
R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.
FORMULA
G.f.: (2-x-x^2)/(1-3*x-24*x^2-x^3).
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{3, 24, 1}, {2, 5, 62}, 30]
a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.
+10
5
0, 2, 7, 69, 377, 2794, 17499, 119930, 782560, 5243499, 34631867, 230522137, 1527974718, 10151087309, 67355177296, 447219602022, 2968334148479, 19705628071261, 130804123379301, 868315777996646, 5763951923164423, 38262238564792074, 253989877628319020
COMMENTS
The Ramanujan sequence number 11 for the argument 2Pi/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).
REFERENCES
R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.
FORMULA
G.f.: x*(2+x)/(1-3*x-24*x^2-x^3).
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{3, 24, 1}, {0, 2, 7}, 30]
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