A103627 Let S(n) = {n,1,n}; sequence gives concatenation S(0), S(1), S(2), ...

0, 1, 0, 1, 1, 1, 2, 1, 2, 3, 1, 3, 4, 1, 4, 5, 1, 5, 6, 1, 6, 7, 1, 7, 8, 1, 8, 9, 1, 9, 10, 1, 10, 11, 1, 11, 12, 1, 12, 13, 1, 13, 14, 1, 14, 15, 1, 15, 16, 1, 16, 17, 1, 17, 18, 1, 18, 19, 1, 19, 20, 1, 20, 21, 1, 21, 22, 1, 22, 23, 1, 23, 24, 1, 24, 25, 1, 25, 26, 1, 26, 27, 1, 27, 28, 1, 28, 29

Offset

0,7

Links

Robert Israel, Table of n, a(n) for n = 0..10000

J. J. P. Veerman, Hausdorff Dimension of Boundaries of Self-Affine Tiles in R^n, arXiv:math/9701215 [math.DS], 1997; Bol. Soc. Mex. Mat. 3, Vol. 4, No 2, 1998, 159-182.

Formula

Conjecture: a(n) = (2*n+1 + (2*n-8)*cos((2*n+1)*Pi/3) + sqrt(3)*cos((8*n+1)*Pi/6) + sqrt(3)*sin((2*n+1)*Pi/3))/9. - Wesley Ivan Hurt, Sep 25 2017

Conjectures from Colin Barker, Sep 27 2017: (Start)

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

a(n) = 2*a(n-3) - a(n-6) for n>5.

(End)

a(3*k+1) = 1, a(3*k)=a(3*k+2)=k. The conjectures easily follow from that. - Robert Israel, Jan 11 2018

a(n) = binomial(floor(n/3), a(n-1)). - Jon Maiga, Nov 24 2018

Maple

seq(op([k, 1, k]), k=0..50); # Robert Israel, Jan 11 2018

Mathematica

v[0] = {1, 1, 1} M = {{1, 1, 0}, {0, 1, 0}, {0, 1, 1}} Det[M - x*IdentityMatrix[4] NSolve[Det[M - x*IdentityMatrix[4]] == 0, x] v[n_] := v[n] = M.v[n - 1] a = Flatten[Table[v[n], {n, 0, Floor[200/3]}]]
Flatten[Table[{n, 1, n}, {n, 0, 30}]] (* Harvey P. Dale, Jul 25 2011 *)
With[{nn=30}, Riffle[Riffle[Range[0, nn], Range[0, nn]], 1, {2, -1, 3}]] (* Harvey P. Dale, Aug 24 2016 *)
RecurrenceTable[{a[0] == 0, a[n] == Binomial[Floor[n/3], a[n - 1]]}, a, {n, 50}] (* Jon Maiga, Nov 24 2018 *)

Prog

(PARI) x='x+O('x^90); Vec(x*(1+x^2-x^3+x^4)/((1-x)^2*(1+x+x^2)^2)) \\ G. C. Greubel, Nov 25 2018
(Magma) m:=90; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( x*(1+x^2-x^3+x^4)/((1-x)^2*(1+x+x^2)^2) )); // G. C. Greubel, Nov 25 2018
(Sage) s=(x*(1+x^2-x^3+x^4)/((1-x)^2*(1+x+x^2)^2)).series(x, 90); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 25 2018
(GAP) a:=[0, 1, 0, 1, 1, 1];; for n in [7..90] do a[n]:=2*a[n-3]-a[n-6]; od; Concatenation([0], a); # G. C. Greubel, Nov 25 2018

Crossrefs

Sequence in context: A325537 A072851 A246688 * A344088 A292595 A269596

Adjacent sequences: A103624 A103625 A103626 * A103628 A103629 A103630

Keyword

nonn

Author

Roger L. Bagula, Mar 25 2005

Status

approved