cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A005386 Area of n-th triple of squares around a triangle.

Original entry on oeis.org

1, 3, 16, 75, 361, 1728, 8281, 39675, 190096, 910803, 4363921, 20908800, 100180081, 479991603, 2299777936, 11018898075, 52794712441, 252954664128, 1211978608201, 5806938376875, 27822713276176, 133306628004003, 638710426743841, 3060245505715200
Offset: 1

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Author

Jean Meeus

Keywords

Comments

a(n)*(-1)^(n+1) is the r=-3 member of the r-family of sequences S_r(n), n>=1, defined in A092184 where more information can be found.
The sequence is the case P1 = 3, P2 = -10, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Essentially the same as A003769.
First differences of A099025.
Cf. A100047.

Programs

  • Magma
    I:=[1, 3, 16]; [n le 3 select I[n] else 4*Self(n-1) +4*Self(n-2) -Self(n-3): n in [1..41]]; // G. C. Greubel, Nov 16 2022
    
  • Maple
    A005386:=-(-1+z)/(z+1)/(z**2-5*z+1); [Conjectured by Simon Plouffe in his 1992 dissertation.]
    a:= n-> (Matrix([[0,1,3]]). Matrix(3, (i,j)-> if (i=j-1) then 1 elif j=1 then [4,4,-1][i] else 0 fi)^(n))[1,1]: seq(a(n), n=1..25); # Alois P. Heinz, Aug 05 2008
  • Mathematica
    a[n_]:= Module[{n1=1, n2=0}, Do[{n1, n2}={Sqrt[3]*n1+n2, n1}, {n-1}];n1^2];
    Table[a[n], {n,30}]
    a[n_]:= Round[((5+Sqrt[21])/2)^n/7]; Table[a[n], {n, 30}]
    Rest@(CoefficientList[Series[x/(1-x*(Sqrt[3]+x)), {x, 0, 30}], x])^2
    Abs[ChebyshevU[Range[1,40]-1, I*Sqrt[3]/2]]^2 (* G. C. Greubel, Nov 16 2022 *)
  • SageMath
    def A005386(n): return abs(chebyshev_U(n-1, i*sqrt(3)/2))^2
    [A005386(n) for n in range(1,40)] # G. C. Greubel, Nov 16 2022

Formula

G.f.: x*(1-x)/((1+x)*(1-5*x+x^2)).
a(n) = 4*a(n-1) + 4*a(n-2) - a(n-3), a(1)=1, a(2)=3, a(3)=16.
a(n) = (2/7)*(T(n, 5/2) - (-1)^n) with twice Chebyshev's polynomials of the first kind evaluated at x=5/2: 2*T(n, 5/2) = A003501(n) = ((5+sqrt(21))^n + (5-sqrt(21))^n)/2^n. - Wolfdieter Lang, Oct 18 2004
a(2*n) = A003690(n). a(2*n+1) = A004253(n)^2. - Alexander Evnin, Mar 11 2012
From Peter Bala, Apr 03 2014: (Start)
a(n) = |U(n-1, sqrt(3)*i/2)|^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 5/2; 1, 3/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)

Extensions

Edited by Peter J. C. Moses, Apr 23 2004
More terms from Pab Ter (pabrlos(AT)yahoo.com), May 09 2004

A136211 Denominators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].

Original entry on oeis.org

1, 4, 5, 19, 24, 91, 115, 436, 551, 2089, 2640, 10009, 12649, 47956, 60605, 229771, 290376, 1100899, 1391275, 5274724, 6665999, 25272721, 31938720, 121088881, 153027601, 580171684, 733199285, 2779769539, 3512968824
Offset: 1

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Author

Gary W. Adamson, Dec 21 2007

Keywords

Comments

A136210(n)/A136211(n) tends to 0.791287847... = [0; 1, 3, 1, 3, 1, 3, ...] = (sqrt(21) - 3)/2 = the inradius of a right triangle with hypotenuse 3, legs 2 and sqrt(21).
The number 0.791287847... = (sqrt(21) - 3)/2 arises in finding a number which is 5 less than its square; the result is: 2.791287847... because (2.791287847...)^2 = 7.791287847... In general the quadratic equation for finding such numbers is x^2 - x = N, so x = (1 + sqrt(1 + 4N))/2. - Alexander R. Povolotsky, Dec 23 2007
Prepending a 1 to the sequence gives [1, 1, 4, 5, 19, 24, ...]. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) with the parameters R = 3 and Q = -1. It is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all natural numbers n and m. - Peter Bala, May 14 2014

Examples

			a(4) = 19 = 3*a(3) + a(2) = 3*5 + 4.
a(5) = 24 = a(4) + a(3) = 19 + 5.
T^3 = [19, 72; 24, 91], where the bottom row [24, 91] = [a(5), a(6)].
		

Crossrefs

Cf. A136210.

Programs

  • Mathematica
    Denominator[NestList[(3/(3+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
    a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Denominator; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 15 2014 *)
  • PARI
    x='x + O('x^25); Vec(x*(1+4*x-x^3)/(1-5*x^2+x^4)) \\ G. C. Greubel, Feb 18 2017

Formula

a(1) = 1, a(2) = 4, then for n>2, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Let T = the 2 X 2 matrix [1, 3; 1, 4]. Then T^n = [A136210(2n-1), A136210(2n); a(2n-1), a(2n)].
From R. J. Mathar, May 18 2008: (Start)
O.g.f.: x*(1+4*x-x^3)/(1-5*x^2+x^4).
a(2*n) = A004253(n+1).
a(2*n+1) = A004254(n). (End)
a(n)*a(n+1) = A099025(n). - R. K. Guy, May 18 2008
{-a(n) + 5 a(n + 2) - a(n + 4), a(0) = 1, a(1) = 4, a(2) = 5, a(3) = 19}. - Robert Israel, May 14 2008

A003769 Number of perfect matchings (or domino tilings) in K_4 X P_n.

Original entry on oeis.org

3, 16, 75, 361, 1728, 8281, 39675, 190096, 910803, 4363921, 20908800, 100180081, 479991603, 2299777936, 11018898075, 52794712441, 252954664128, 1211978608201, 5806938376875, 27822713276176, 133306628004003, 638710426743841, 3060245505715200
Offset: 1

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Author

Keywords

References

  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.

Crossrefs

Essentially the same as A005386. First differences of A099025.

Programs

  • PARI
    Vec(x*(3 + 4*x - x^2) / ((1 + x)*(1 - 5*x + x^2)) + O(x^40)) \\ Colin Barker, Dec 16 2017

Formula

a(n) = 4a(n-1) + 4a(n-2) - a(n-3), n>3.
a(n) = (1/7)*(6*A030221(n) - A054477(n) + 2(-1)^n).
G.f.: x*(3+4*x-x^2)/((1+x)*(1-5*x+x^2)). - R. J. Mathar, Dec 16 2008
a(n) = 2^(-1-n)*((-1)^n*2^(2+n) + (5-sqrt(21))^(1+n) + (5+sqrt(21))^(1+n)) / 7. - Colin Barker, Dec 16 2017
Showing 1-3 of 3 results.