Displaying 1-8 of 8 results found.
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a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0.
+10
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)
Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p={p_1, p_2, p_3, p_4} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).
+10
7
0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 3, 4, 4, 6, 5, 10, 10, 14, 15, 20, 20, 34, 35, 48, 55, 69, 75, 117, 124, 165, 199, 241, 274, 406, 440, 571, 714, 846, 988, 1417, 1560, 1988, 2548, 2977, 3536, 4965, 5525, 6953, 9061, 10490, 12597, 17443, 19551
COMMENTS
Theory: (Start)
1. Definitions. Let T_(9,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/9,(9-j)*Pi/9) and Area(T_(9,j,0))=sin(j*Pi/9), j in {1,2,3,4}. Associated with T_(9,j,0) are its angle coefficients (j, 9-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(9,j,0) along a line extending between its two corners with even angle coefficient; let H_(9,j,0) denote this half-tile. Similarly, a T_(9,j,r) tile is a linearly scaled version of T_(9,j,0) with sides of length Q^r and Area(T_(9,j,r))=Q^(2*r)*sin(j*Pi/9), r>=0 an integer, where Q is the positive, constant square root Q=sqrt(2*cos(Pi/9)); likewise let H_(9,j,r) denote the corresponding half-tile. Often H_(9,i,r) (i in {1,2,3,4}) can be subdivided into an integral number of each equivalence class H_(9,j,0). But regardless of whether or not H_(9,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3,4, in which the entry m_(i,j) gives the quantity of H_(9,j,0) tiles that should be present in a subdivided H_(9,i,r) tile. The number Q^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_1)^r, where
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that D_r-D_(r-1)-3*D_(r-2)+2*D_(r-3)+D_(r-4)={0,0,0,0}, for r>=4, with initial conditions {D_k}={{0,0,0,1},{0,0,1,1},{0,1,1,2},{1,1,3,3}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_1)^r was defined above. Hence the block D_r corresponds component-wise to the fourth column of M, and a(3*r+p_i)=m_(i,4) gives the quantity of H_(9,4,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)
Combining blocks A_r, B_r, C_r and D_r, from A187495, A187496, A187497 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_1)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r),D_(-r)]=(U_1)^(-r) of (U_1)^r.. Therefore the four sequences need not be causal.
Since U_1 is symmetric, so is M=(U_1)^r, so the block D_r also corresponds to the fourth row of M. Therefore, alternatively, for j=1,2,3,4, a(3r+p_j)=m_(4,j) gives the quantity of H_(9,j,0) tiles that should be present in a H_(9,4,r) tile.
Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of fourth-column entries m_(2,4), m_(3,4) and m_(4,4) (or fourth-row entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_1)^r.
REFERENCES
L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).
FORMULA
Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n >= 12, with initial conditions {a(m)} = {0,0,1,0,1,1,1,1,2,1,3,3}, m=0,1,...,11.
G.f.: -x^2*(1+x)*(x^6+3*x^4+2*x^2+1) / ( (1+x+x^2)*(x^9+3*x^6-1) ).
MAPLE
A052931 := proc(n) if n < 0 then 0; else coeftayl(1/(1-3*x^2-x^3), x=0, n) ; end if; end proc:
A052931a := proc(n) if n mod 3 = 0 then A052931(n/3) ; else 0 ; end if; end proc:
A057078 := proc(n) op(1+(n mod 3), [1, 0, -1]) ; end proc:
A187498 := proc(n) - A057078(n) +A052931a(n) +2*A052931a(n-2) +A052931a(n-3) +3*A052931a(n-4) +2*A052931a(n-5) +A052931a(n-6) +3*A052931a(n-7) -A052931a(n-8) ; %/3 ; end proc:
MATHEMATICA
CoefficientList[Series[-x^2*(1 + x)*(x^6 + 3*x^4 + 2*x^2 + 1)/((1 + x + x^2)*(x^9 + 3*x^6 - 1)), {x, 0, 1000}], x] (* G. C. Greubel, Sep 23 2017 *)
PROG
(PARI) x='x+O('x^50); Vec(-x^2*(1+x)*(x^6+3*x^4+2*x^2+1)/((1+x+x^2)*(x^9+3*x^6-1))) \\ G. C. Greubel, Sep 23 2017
1, 0, 3, 0, 1, 9, 0, 0, 6, 27, 0, 0, 1, 27, 81, 0, 0, 0, 9, 108, 243, 0, 0, 0, 1, 54, 405, 729, 0, 0, 0, 0, 12, 270, 1458, 2187, 0, 0, 0, 0, 1, 90, 1215, 5103, 6561, 0, 0, 0, 0, 0, 15, 540, 5103, 17496, 19683, 0, 0, 0, 0, 0, 1, 135, 2835, 20412, 59049, 59049, 0, 0, 0, 0, 0, 0, 18, 945, 13608, 78732, 196830, 177147
COMMENTS
Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).
Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008
FORMULA
Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).
G.f. of column k: (3*x + x^2)^k.
EXAMPLE
Triangle begins:
1;
0, 3;
0, 1, 9;
0, 0, 6, 27;
0, 0, 1, 27, 81;
0, 0, 0, 9, 108, 243;
...
MATHEMATICA
Table[3^(2*k-n)*Binomial[k, n-k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 19 2021 *)
PROG
(Sage) flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021
CROSSREFS
Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).
Expansion of (1 - x^2)/(1 - 3*x^2 - x^3).
+10
5
1, 0, 2, 1, 6, 5, 19, 21, 62, 82, 207, 308, 703, 1131, 2417, 4096, 8382, 14705, 29242, 52497, 102431, 186733, 359790, 662630, 1266103, 2347680, 4460939, 8309143, 15730497, 29388368, 55500634, 103895601, 195890270, 367187437, 691566411, 1297452581
COMMENTS
Sequence is related to rhombus substitution tilings.
FORMULA
G.f.: (1 - x^2)/(1 - 3*x^2 - x^3).
a(n) = 3*a(n-2)+a(n-3), for n>=3, with a(0)=1, a(1)=0, a(2)=2.
a(n) = a(n-1)+3*a(n-2)-2*a(n-3)-a(n-4), for n>=4, with {a(k)}={1,0,2,1}, k=0,1,2,3.
a(n) = m_(3,3), where (m_(i,j)) = (U_1)^n, i,j=1,2,3,4 and U_1 is the tridiagonal unit-primitive matrix [0, 1, 0, 0; 1, 0, 1, 0; 0, 1, 0, 1; 0, 0, 1, 1].
a(n) = (2^n/3)*(cos^n(Pi/9) + cos^n(5*Pi/9) + cos^n(7*Pi/9)). - Greg Dresden, Sep 24 2022
MAPLE
F:= gfun:-rectoproc({a(n)=3*a(n-2)+a(n-3), a(0)=1, a(1)=0, a(2)=2}, a(n), remember):
MATHEMATICA
CoefficientList[Series[(1-x^2)/(1-3x^2-x^3), {x, 0, 40}], x] (* Harvey P. Dale, Mar 31 2011 *)
LinearRecurrence[{0, 3, 1}, {1, 0, 2}, 50] (* Roman Witula, Aug 20 2012 *)
PROG
(PARI) abs(polsym(1-3*x+x^3, 66)/3) /* Joerg Arndt, Aug 19 2012 */
(Magma) I:=[1, 0, 2, 1]; [n le 4 select I[n] else Self(n-1)+3*Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 22 2015
Expansion of (1+x^2)^2/(1-x^3+x^6).
+10
2
1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0, -2, -1, -1, -2, 0, -1, 0, 1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0, -2, -1, -1, -2, 0, -1, 0, 1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0, -2, -1, -1, -2, 0, -1, 0, 1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0, -2, -1, -1, -2, 0, -1, 0, 1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0, -2, -1, -1, -2, 0, -1, 0, 1, 0, 2, 1, 1, 2, 0, 1, 0, -1, 0
COMMENTS
The denominator is the 18th cyclotomic polynomial. The g.f. is a Chebyshev transform of that of A052931, by the Chebyshev mapping g(x)->(1/(1+x^2))g(x/(1+x^2)). The reciprocal of the 18th cyclotomic polynomial A014027 is given by sum{k=0..n, A099916(n-k)(k/2+1)(-1)^(k/2)(1+(-1)^k)/2}.
FORMULA
a(n)=sum{k=0..floor(n/2), C(n-k, k)(-1)^k*sum{j=0..n-2k, C(j, n-2k-2j)3^(3j-n+2k)}}; a(n)=sum{k=0..n, A014027(n-k)C(2, k/2)(1+(-1)^k)/2}.
Triangle T(n,k) = coefficient [x^n] of x^2/(1-(k+1)*x^2-x^3) for row n, and columns k = 0..n, read by rows.
+10
2
0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 2, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 4, 9, 16, 25, 36, 49, 2, 4, 6, 8, 10, 12, 14, 16, 2, 9, 28, 65, 126, 217, 344, 513, 730, 3, 12, 27, 48, 75, 108, 147, 192, 243, 300, 4, 22, 90, 268, 640, 1314, 2422, 4120, 6588, 10030, 14674
FORMULA
T(n,k) = coefficient [x^n] ( x^2/(1-(k+1)*x^2-x^3) ).
EXAMPLE
The table starts:
0;
0, 0;
1, 1, 1;
0, 0, 0, 0;
1, 2, 3, 4, 5;
1, 1, 1, 1, 1, 1;
1, 4, 9, 16, 25, 36, 49;
2, 4, 6, 8, 10, 12, 14, 16;
2, 9, 28, 65, 126, 217, 344, 513, 730;
3, 12, 27, 48, 75, 108, 147, 192, 243, 300;
MAPLE
t:=taylor(x^2/(1-(k+1)*x^2-x^3), x, 15):
MATHEMATICA
T[n_, k_]:= T[n, k]= Coefficient[Series[x^2/(1-(k+1)*x^2-x^3), {x, 0, n+ 2}], x, n];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten
PROG
(Magma)
m:=12;
R<x>:=PowerSeriesRing(Integers(), m+2);
A117724:= func< n, k | Coefficient(R!( x^2/(1-(k+1)*x^2-x^3) ), n) >;
(SageMath)
P.<x> = PowerSeriesRing(QQ)
return P( x^2/(1-(k+1)*x^2-x^3) ).list()[n]
EXTENSIONS
Sign in definition corrected, offset set to -1 by Assoc. Eds. of the OEIS, Jun 15 2010
Expansion of (1+x^2)^2/(1+x^3+x^6).
+10
1
1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0, 2, -1, 1, -2, 0, -1, 0, 1, 0
COMMENTS
The denominator is the 9th cyclotomic polynomial. The g.f. is a Chebyshev transform of that of (-1)^n* A052931(n) by the Chebyshev mapping g(x)->(1/(1+x^2))g(x/(1+x^2)). The reciprocal of the 9th cyclotomic polynomial A014018 is given by sum{k=0..n, A099917(n-k)(k/2+1)(-1)^(k/2)(1+(-1)^k)/2}.
FORMULA
a(n)=sum{k=0..floor(n/2), C(n-k, k)(-1)^k*sum{j=0..n-2k, C(j, n-2k-2j)3^k(-1/3)^(n-2k)}}; a(n)=sum{k=0..n, A014018(n-k)C(2, k/2)(1+(-1)^k)/2}.
Expansion of x*(1+x) / (1-3*x^2-x^3).
+10
1
0, 1, 1, 3, 4, 10, 15, 34, 55, 117, 199, 406, 714, 1417, 2548, 4965, 9061, 17443, 32148, 61390, 113887, 216318, 403051, 762841, 1425471, 2691574, 5039254, 9500193, 17809336, 33539833, 62928201, 118428835, 222324436, 418214706, 785402143, 1476968554
COMMENTS
Define the 4 X 4 tridiagonal unit-primitive matrix (see [Jeffery]) M=A_{9,1}=[0,1,0,0; 1,0,1,0; 0,1,0,1; 0,0,1,1]; then a(n)=[M^n]_(3,4)=[M^n]_(4,3).
MATHEMATICA
LinearRecurrence[{0, 3, 1}, {0, 1, 1}, 36] (* or *)
CoefficientList[Series[x (1 + x)/(1 - 3 x^2 - x^3), {x, 0, 35}], x] (* Michael De Vlieger, Mar 10 2020 *)
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