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Expansion of (5 + 6*x + 3*x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) in powers of x.
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8
5, 16, 45, 130, 377, 1088, 3145, 9090, 26269, 75920, 219413, 634114, 1832625, 5296384, 15306833, 44237570, 127848949, 369490320, 1067846845, 3086134658, 8919094697, 25776662080, 74495936025, 215297250946, 622220603405, 1798250918672, 5197041610021
COMMENTS
Number of tilings of the 2-mod-4 pillow of order n is a perfect square times a(n). [Propp, 1999, p. 272]
REFERENCES
J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 12).
FORMULA
G.f.: (5 + 6*x + 3*x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4).
a(-n) = a(-6 + n). a(-1) = 2, a(-2) = 1, a(-3) = 0. a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4). - Michael Somos, Dec 15 2011
EXAMPLE
G.f. = 5 + 16*x + 45*x^2 + 130*x^3 + 377*x^4 + 1088*x^5 + 3145*x^6 + 9090*x^7 + ...
MAPLE
seq(coeff(series((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
CoefficientList[Series[(5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4), {x, 0, 30}], x] (* Stefano Spezia, Sep 12 2018 *) LinearRecurrence[{2, 2, 2, -1}, {5, 16, 45, 130}, 30] (* Harvey P. Dale, Oct 03 2024 *)
PROG
(PARI) {a(n) = my(m = abs(n+3)); polcoeff( (x - x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + x * O(x^m), m)}; /* Michael Somos, Dec 15 2011 */ (PARI) x='x+O('x^33); Vec((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4)) \\ Altug Alkan, Sep 12 2018 (Python)
from math import log
a0, a1, a2, a3, n = 130, 45, 16, 5, 3
print(0, a3)
print(1, a2)
print(2, a1)
print(3, a0)
while log(a0)/log(10) < 1000:
a0, a1, a2, a3, n = 2*(a0+a1+a2)-a3, a0, a1, a2, n+1
(GAP) a:=[5, 16, 45, 130];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2] +2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018 (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4) )); // G. C. Greubel, Jul 29 2019 (Sage) ((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 29 2019
a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=2, a(3)=5.
+10
5
0, 1, 2, 5, 16, 45, 130, 377, 1088, 3145, 9090, 26269, 75920, 219413, 634114, 1832625, 5296384, 15306833, 44237570, 127848949, 369490320, 1067846845, 3086134658, 8919094697, 25776662080, 74495936025, 215297250946, 622220603405
COMMENTS
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). - T. D. Noe, Dec 23 2008 Case P1 = 2, P2 = -4, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 04 2014
FORMULA
a(n) = round(w^n/2/sqrt(5)) where w = (1+r)/(1-r) = 2.89005363826396... and r = sqrt(sqrt(5)-2) = 0.485868271756...; for n >= 3, a(n) = A071101(n+3). G.f.: -x*(x-1)*(1+x)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4). - R. J. Mathar, Jun 03 2009 Define a Lucas sequence {U(n)} in the ring of Gaussian integers by the recurrence U(n) = (1 + i)*U(n-1) + U(n-2) with U(0) = 0 and U(1) = 1. Then a(n) = |U(n)|^2.
Let a, b denote the zeros of x^2 - (1 + i)*x - 1 and c, d denote the zeros of x^2 - (1 - i)*x - 1.
Then a(n) = (a^n - b^n)*(c^n - d^n)/((a - b)*(c - d)).
a(n) = (alpha(1)^n + beta(1)^n - alpha(2)^n - beta(2)^n)/(2*sqrt(5)), where alpha(1), beta(1) are the roots of x^2 - ( 1 + sqrt(5))*x + 1 = 0, and alpha(2), beta(2) are the roots of x^2 - (1 - sqrt(5))*x + 1 = 0.
The o.g.f. is the Hadamard product of the rational functions x/(1 - (1 + i)x - x^2) and x/(1 - (1 - i)x - x^2). (End)
a(n) = (1/sqrt(5))*(T(n,phi) - T(n,-1/phi)), where phi = 1/2*(1 + sqrt(5)) is the golden ratio and T(n,x) denotes the Chebyshev polynomial of the first kind. Compare with the Fibonacci numbers, A000045, whose terms are given by the Binet formula 1/sqrt(5)*( phi^n - (-1/phi)^n ). a(n) = top right (or bottom left) entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 1; 1, 1]; the off-diagonal elements of M^n give the sequence of Fibonacci numbers. Bottom right entry of the matrix T(n, M) gives A138574. See the remarks in A100047 for the general connection between Chebyshev polynomials and linear divisibility sequences of the fourth order. (End) a(n) = (((phi + sqrt(phi))^n + (phi - sqrt(phi))^n)/2 - (-1)^n * cos(n*arctan(sqrt(phi))))/sqrt(5), where phi=(1+sqrt(5))/2. - Vladimir Reshetnikov, May 11 2016
MAPLE
seq(coeff(series((x*(1-x)*(x+1))/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
Round@Table[(((GoldenRatio + Sqrt[GoldenRatio])^n + (GoldenRatio - Sqrt[GoldenRatio])^n)/2 - (-1)^n Cos[n ArcTan[Sqrt[GoldenRatio]]])/Sqrt[5], {n, 0, 20}] (* or *) LinearRecurrence[{2, 2, 2, -1}, {0, 1, 2, 5}, 20] (* Vladimir Reshetnikov, May 11 2016 *) CoefficientList[Series[-x*(x-1)*(1+x)/(1-2*x-2*x^2-2*x^3+x^4), {x, 0, 20}], x] (* Stefano Spezia, Sep 12 2018 *)
PROG
(PARI) my(x='x+O('x^50)); concat([0], Vec(x*(1-x)*(1+x)/(1 -2*x -2*x^2 -2*x^3 +x^4))) \\ G. C. Greubel, Aug 08 2017 (GAP) a:=[0, 1, 2, 5];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
Inscribe two circles of curvature 2 inside a circle of curvature -1. Sequence gives curvatures of the smallest circles that can be sequentially inscribed in such a diagram.
+10
4
-1, 2, 2, 3, 15, 38, 110, 323, 927, 2682, 7754, 22403, 64751, 187134, 540822, 1563011, 4517183, 13054898, 37729362, 109039875, 315131087, 910745750, 2632104062, 7606921923, 21984412383, 63536130986, 183622826522, 530679817859, 1533693138351, 4432455434478
COMMENTS
The ratio of successive terms approaches the constant phi+sqrt(phi) ~= 2.89005363826396..., where phi is the golden ratio (sqrt(5)+1)/2. The ratio between the curvatures of two successively smaller circles approaches this constant in any apollonian packing as the curvatures increase.
For more comments, references and links, see A189226.
FORMULA
a(n) = 2a(n-1) + 2a(n-2) + 2a(n-3) - a(n-4).
G.f.: -(1-x)*(1 - 3*x - 3*x^2)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4). - Colin Barker, Apr 22 2012
EXAMPLE
After circles of 2, 2, 3, 15 have been inscribed in the diagram, the next smallest circle that can be inscribed has a curvature of 38.
MAPLE
seq(coeff(series((x-1)*(1-3*x-3*x^2)/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
CoefficientList[Series[(-3 z^3 + 4 z - 1)/(z^4 - 2 z^3 - 2 z^2 - 2 z + 1), {z, 0, 100}], z] (* and *) LinearRecurrence[{2, 2, 2, -1}, {-1, 2, 2, 3}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
PROG
(PARI) { for (n=0, 200, if (n>3, a=2*a1 + 2*a2 + 2*a3 - a4; a4=a3; a3=a2; a2=a1; a1=a, if (n==0, a=a4=-1, if (n==1, a=a3=2, if (n==2, a=a2=2, a=a1=3)))); write("b060790.txt", n, " ", a); ) } \\ Harry J. Smith, Jul 12 2009 (GAP) a:=[-1, 2, 2, 3];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
Expansion of (5 + 3*x + x^2 - x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) in powers of x.
+10
4
5, 13, 37, 109, 313, 905, 2617, 7561, 21853, 63157, 182525, 527509, 1524529, 4405969, 12733489, 36800465, 106355317, 307372573, 888323221, 2567301757, 7419639785, 21443156953, 61971873769, 179102039257, 517614500173, 1495933669445, 4323328543981
COMMENTS
Number of tilings of the 0-mod-4 pillow of order n is a perfect square times a(n). [Propp, 1999, p. 271]
REFERENCES
J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 12).
FORMULA
G.f.: (5 + 3*x + x^2 -x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4).
a(-n) = a(-5 + n). a(-1) = a(-2) = 1. a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4). - Michael Somos, Dec 15 2011
EXAMPLE
G.f. = 5 + 13*x + 37*x^2 + 109*x^3 + 313*x^4 + 905*x^5 + 2617*x^6 + 7561*x^7 + ...
MAPLE
seq(coeff(series((5+3*x+x^2-x^3)/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
CoefficientList[Series[(5 + 3*x + x^2 -x^3)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4), {x, 0, 50}], x] (* Stefano Spezia, Sep 12 2018 *) LinearRecurrence[{2, 2, 2, -1}, {5, 13, 37, 109}, 30] (* Harvey P. Dale, Sep 03 2021 *)
PROG
(PARI) {a(n) = my(m = n+2); if( m < 0, m = -1 - m); polcoeff( (1 - x + x^2 - x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + x * O(x^m), m)}; /* Michael Somos, Dec 15 2011 */ (PARI) x='x+O('x^33); Vec((5+3*x+x^2-x^3)/(1-2*x-2*x^2-2*x^3+x^4)) \\ Altug Alkan, Sep 12 2018 (GAP) a:=[5, 13, 37, 109];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=3, a(3)=9, a(4)=25.
+10
2
0, 1, 3, 9, 25, 73, 211, 609, 1761, 5089, 14707, 42505, 122841, 355017, 1026019, 2965249, 8569729, 24766977, 71577891, 206863945, 597847897, 1727812489, 4993470771, 14431398369, 41707515361, 120536956513, 348358269715, 1006774084809, 2909631106713, 8408989965961
FORMULA
a(n) = round(w^n*(1 + 1/sqrt(5))/4) where w = (1+r)/(1-r) = 2.89005363826396... and r = sqrt(sqrt(5)-2) = 0.485868271756... .
G.f.: x*( 1 + x + x^2 - x^3 ) / ( 1 - 2*x - 2*x^2 - 2*x^3 + x^4 ). - R. J. Mathar, Jun 29 2013
MAPLE
seq(coeff(series((x*(1+x+x^2-x^3))/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
LinearRecurrence[{2, 2, 2, -1}, {0, 1, 3, 9, 25}, 50] (* G. C. Greubel, Aug 08 2017 *) CoefficientList[Series[x*( 1+x+x^2-x^3 )/(1-2*x-2*x^2-2*x^3+x^4), {x, 0, 20}], x] (* Stefano Spezia, Sep 12 2018 *)
PROG
(PARI) x='x+O('x^50); concat([0], Vec(x*(1 +x +x^2 -x^3)/(1 -2*x -2*x^2 -2*x^3 +x^4))) \\ G. C. Greubel, Aug 08 2017 (Magma) I:=[0, 1, 3, 9, 25]; [n le 5 select I[n] else 2*Self(n-1)+2*Self(n-2)+2*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Sep 14 2018
Decimal expansion of (1 + sqrt(3) + sqrt(2*sqrt(3)))/2.
+10
2
2, 2, 9, 6, 6, 3, 0, 2, 6, 2, 8, 8, 6, 5, 3, 8, 2, 4, 5, 7, 0, 4, 9, 4, 1, 9, 1, 7, 7, 3, 6, 1, 7, 0, 2, 7, 1, 2, 2, 2, 6, 0, 6, 8, 5, 2, 5, 8, 2, 8, 4, 2, 6, 8, 9, 1, 2, 1, 8, 8, 0, 0, 0, 0, 8, 0, 4, 9, 2, 9, 9, 2, 2, 4, 5, 0, 3, 4, 8, 9, 8, 1
COMMENTS
This constant and its reciprocal are the real solutions of x^4 - 2*x^3 - 2*x + 1 = (x^2 - (sqrt(3)+1)*x + 1)*(x^2 + (sqrt(3)-1)*x + 1) = 0.
Decimal expansion of the largest x satisfying x^2 - (1 + sqrt(3))*x + 1 = 0.
EXAMPLE
2.29663026288653824570494191773617027122260685258284268912188000080492992...
MAPLE
Digits:=100: evalf((1+sqrt(3)+sqrt(2*sqrt(3)))/2); # Muniru A Asiru, Sep 12 2018
MATHEMATICA
RealDigits[(1 + Sqrt[3] + Sqrt[2 Sqrt[3]])/2, 10, 100][[1]] (* Bruno Berselli, Sep 13 2018 *)
PROG
(PARI) (1+sqrt(3)+sqrt(2*sqrt(3)))/2 \\ Altug Alkan, Sep 11 2018
Expansion of (1 - x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4).
+10
1
1, 2, 5, 14, 41, 118, 341, 986, 2849, 8234, 23797, 68774, 198761, 574430, 1660133, 4797874, 13866113, 40073810, 115815461, 334712894, 967338217, 2795659334, 8079605429, 23350493066, 67484177441, 195032892538, 563655520661, 1628994688214, 4707882025385
FORMULA
a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4) for n>3. - Colin Barker, Jul 16 2017
MAPLE
seq(coeff(series((1-x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
PROG
(PARI) Vec((1 - x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + O(x^30)) \\ Colin Barker, Jul 16 2017 (GAP) a:=[1, 2, 5, 14];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
Numerical distance between m-th and (n+m)-th circles in a loxodromic sequence of circles in which each 4 consecutive circles touch.
+10
1
-1, 1, 1, 1, 7, 17, 49, 145, 415, 1201, 3473, 10033, 28999, 83809, 242209, 700001, 2023039, 5846689, 16897249, 48833953, 141132743, 407881201, 1178798545, 3406791025, 9845808799, 28454915537, 82236232177, 237667122001
REFERENCES
Coxeter, H. S. M. "Numerical distances among the circles in a loxodromic sequence." Nieuw Archief voor Wiskunde 16 (1998): 1-10. (Note the word "circles" in the title!)
FORMULA
a(n) = 2(a(n-1)+a(n-2)+a(n-3))-a(n-4).
a(n) = Sum{v=0 to [n/2]} binomial(n, 2v)*F(n-v-2) where F(m) is the m-th Fibonacci number.
G.f.: -(x^3-x^2-3*x+1) / (x^4-2*x^3-2*x^2-2*x+1). - Colin Barker, Sep 23 2013
MAPLE
with(combinat); F:=fibonacci;
f:=n->add(F(n-i)*binomial(n, 2*(i-2)), i=2..n-1);
[seq(f(n), n=3..32)]; # Produces the sequence from a(3) onwards - N. J. A. Sloane, Sep 03 2018
MATHEMATICA
CoefficientList[Series[-(x^3-x^2-3*x+1)/(x^4-2*x^3-2*x^2-2*x+1), {x, 0, 30}], x] (* Stefano Spezia, Sep 12 2018 *)
PROG
(PARI) Vec(-(x^3-x^2-3*x+1)/(x^4-2*x^3-2*x^2-2*x+1) + O(x^100)) \\ Colin Barker, Sep 23 2013 (GAP) a:=[-1, 1, 1, 1];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
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