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.

A038342 G.f.: 1/(1 - 3 x - 3 x^2 + 4 x^3 + x^4 - x^5).

Original entry on oeis.org

1, 3, 12, 41, 146, 511, 1798, 6314, 22187, 77946, 273856, 962142, 3380337, 11876254, 41725295, 146595013, 515037713, 1809501081, 6357387289, 22335644540, 78472648463, 275700866485, 968630080476, 3403123989780
Offset: 0

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Author

Floor van Lamoen

Keywords

Comments

Middle line of 5-wave sequence A038201.
Let M denotes the 5 X 5 matrix = row by row (1,1,1,1,1)(1,1,1,1,0)(1,1,1,0,0)(1,1,0,0,0)(1,0,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n),u(n))=M^n*A where A is the vector (1,1,1,1,1) then a(n)=z(n). - Benoit Cloitre, Apr 02 2002
a(n) appears in the formula for 1/rho(11)^n, with rho(11) := 2*cos(Pi/11) (length ratio (smallest diagonal/side) in the regular 11-gom) when written in the power basis of the degree 5 number field Q(rho(11)): 1/rho(11)^n = a(n)*1 + A230080(n)*rho(11) - A230081(n)*rho(11)^2 - A069006(n-1)* rho(11)^3 + a(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = a(-1). See A230080 with the example for n=4. - Wolfdieter Lang, Nov 04 2013
From Wolfdieter Lang, Nov 20 2013: (Start)
The limit a(n+1)/a(n) for n -> infinity is omega(11) := S(4, x) = 1 - 3*x^2 + x^4 with x = rho(11). omega(11) = 1/(2*cos(Pi*5/11)), approx. 3.51333709. For the Chebyshev S-polynomial see A049310. For rho(11) see the preceding comment. The decimal expansion of omega(11) is given in A231186. omega(11) is an integer in Q(rho(11)) with power basis coefficients [1,0,-3,0,1]. It is known to be the length ratio (longest diagonal)/side in the regular 11-gon.
This limit follows from the a(n)-recurrence and the solutions of X^5 - 3*X^4 - 3*X^3 + 4*X^2 + X - 1 = 0, which are given by the inverse of the known solutions of the minimal polynomial C(11, x) of rho(11) (see A187360). The other four X solutions are 1/rho(11), with coefficients [3,3,-4,-1,1] in the power basis of Q(rho(11)), approx. 0.52110856, 1/(2*cos(Pi*3/11)) with coefficients [-1,-1,1,0,0], approx. 0.763521119, 1/(2*cos(Pi*7/11)) with coefficients [0,-3,3,1,-1], approx. -1.20361562, and 1/(2*cos(Pi*9/11)) with coefficients [0,1,3,0,-1], approx. -0.59435114. These solutions for X are therefore irrelevant for this sequence.
The same limit omega(11) is therefore obtained for the sequences A069006, A230080 and A230081. See the Nov 04 2013 comment.
(End)

References

  • Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002.

Links

Crossrefs

Cf. A006358, A069006, A230080, A230081: same recurrence formula.
Cf. A066170.

Programs

  • Mathematica
    b = {-1, 3, 3, -4, -1, 1}; p[x_] := Sum[x^(n - 1)*b[[7 - n]], {n, 1, 6}] q[x_] := ExpandAll[x^5*p[1/x]] Table[ SeriesCoefficient[ Series[x/q[x], {x, 0, 30}], n], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 19 2006 *)
LinearRecurrence[{3,3,-4,-1,1},{1,3,12,41,146},30] (* Harvey P. Dale, Aug 27 2012 *)

Formula

a(n) = 3a(n-1)+3a(n-2)-4a(n-3)-a(n-4)+a(n-5). Also a(n) = b(4n+2) with b(n) as in 5-wave sequence A038201.
G.f.: 1/(1 - 3 x - 3 x^2 + 4 x^3 + x^4 - x^5) = -1/C(11, x), with C(11, x) the minimal polynomial of 2*cos(Pi/11) (see the name and A187360 for C). - Wolfdieter Lang, Nov 07 2013

Extensions

More terms from Benoit Cloitre, Apr 02 2002
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A240840 Floor(6^n/(1+1/(2*cos(5*Pi/11)))^n).

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 5, 7, 9, 12, 17, 22, 30, 40, 53, 71, 95, 126, 168, 223, 297, 395, 525, 698, 928, 1234, 1640, 2180, 2899, 3854, 5123, 6811, 9055, 12038, 16003, 21275, 28282, 37599, 49984, 66448, 88336, 117433, 156115
Offset: 0

Views

Author

Kival Ngaokrajang, Apr 13 2014

Keywords

Comments

a(n) is the perimeter (rounded down) of a hendecaflake after n iterations, let a(0) = 1. The total number of sides is 11*A000400(n). The total number of holes is A016123(n), n >=1. 1/(2*cos(5*Pi/11)) = A231186.

Links

Crossrefs

Cf. A000400, A016123, A231186, A240523 (pentaflake), A240671 (heptaflake), A240572 (octaflake), A240733 (nonaflake), A240734 (decaflake), A240735 (dodecaflake), A240841 (tridecaflake).

Programs

  • Maple
    A240840:=n->floor(6^n/(1+1/(2*cos(5*Pi/11)))^n); seq(A240840(n), n=0..50); # Wesley Ivan Hurt, Apr 13 2014
  • Mathematica
    Table[Floor[6^n/(1 + 1/(2*Cos[5*Pi/11]))^n], {n, 0, 50}] (* Wesley Ivan Hurt, Apr 13 2014 *)
  • PARI
    {a(n)=floor(6^n/(1+1/(2*cos(5*Pi/11)))^n)}
       for (n=0, 100, print1(a(n), ", "))

A337301 Triangle read by rows in which row n lists the closest integers to diagonal lengths of regular n-gon with unit edge length, n >= 4.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 2, 2, 3, 3, 4, 4, 3, 3, 2, 2, 3, 3, 4, 4, 4, 3, 3, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 2, 2, 3, 4, 4, 4, 4, 4, 4, 4, 3, 2, 2, 3, 4, 4, 5, 5, 5, 5, 4, 4, 3, 2, 2, 3, 4, 4, 5, 5, 5, 5, 5, 4, 4, 3, 2
Offset: 4

Views

Author

Mohammed Yaseen, Aug 22 2020

Keywords

Examples

			Triangle begins:
1;
2, 2;
2, 2, 2;
2, 2, 2, 2;
2, 2, 3, 2, 2;
2, 3, 3, 3, 3, 2;
2, 3, 3, 3, 3, 3, 2;
2, 3, 3, 4, 4, 3, 3, 2;
2, 3, 3, 4, 4, 4, 3, 3, 2;
2, 3, 3, 4, 4, 4, 4, 3, 3, 2;
2, 3, 4, 4, 4, 4, 4, 4, 4, 3, 2;
2, 3, 4, 4, 5, 5, 5, 5, 4, 4, 3, 2;
2, 3, 4, 4, 5, 5, 5, 5, 5, 4, 4, 3, 2;
...
Row n lists the closest integers to the length of the diagonals drawn from a fixed vertex of a regular n-gon with unit edge length, n >= 4.
The lengths of the diagonals drawn from vertex A of a regular 8-gon ABCDEFGH with unit edge length are:
AC = 1.84775...
AD = 2.41421...
AE = 2.61312...
AF = 2.41421...
AG = 1.84775...
So the row for n=8 is 2, 2, 3, 2, 2.

Crossrefs

Cf. A064313.
Decimal expansion of diagonal lengths of regular n-gons with unit edge length:
n=4 A002193.
n=5 A001622.
n=6 A002194, A000038.
n=7 A160389, A231187.
n=8 A179260, A014176, A121601.
n=9 A332437.
n=10 A188593, A104457, A019970, A134945.
n=11 A231186.
n=12 A188887, A090388, A337402, A019973, A214726.

Programs

  • Mathematica
    T[n_,k_]:=Round[Sin[(k+1)*Pi/n]/Sin[Pi/n]]; Flatten[Table[T[n,k],{n,4,16},{k,1,n-3}]] (* Stefano Spezia, Sep 07 2020 *)

Formula

T(n,k) = round(sin((k+1)*Pi/n)/sin(Pi/n)), n >= 4, 1 <= k <= n-3.
Showing 1-3 of 3 results.

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