A103627 - OEIS

%I #37 Sep 08 2022 08:45:17

%S 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,

%T 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,

%U 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

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

%H Robert Israel, <a href="/A103627/b103627.txt">Table of n, a(n) for n = 0..10000</a>

%H J. J. P. Veerman, <a href="https://arxiv.org/abs/math/9701215">Hausdorff Dimension of Boundaries of Self-Affine Tiles in R^n</a>, arXiv:math/9701215 [math.DS], 1997; Bol. Soc. Mex. Mat. 3, Vol. 4, No 2, 1998, 159-182.

%F 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

%F Conjectures from _Colin Barker_, Sep 27 2017: (Start)

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

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

%F (End)

%F a(3*k+1) = 1, a(3*k)=a(3*k+2)=k.  The conjectures easily follow from that. - _Robert Israel_, Jan 11 2018

%F a(n) = binomial(floor(n/3), a(n-1)). - _Jon Maiga_, Nov 24 2018

%p seq(op([k,1,k]),k=0..50); # _Robert Israel_, Jan 11 2018

%t 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]}]]

%t Flatten[Table[{n,1,n},{n,0,30}]] (* _Harvey P. Dale_, Jul 25 2011 *)

%t With[{nn=30},Riffle[Riffle[Range[0,nn],Range[0,nn]],1,{2,-1,3}]] (* _Harvey P. Dale_, Aug 24 2016 *)

%t RecurrenceTable[{a[0] == 0, a[n] == Binomial[Floor[n/3], a[n - 1]]}, a, {n, 50}] (* _Jon Maiga_, Nov 24 2018 *)

%o (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

%o (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

%o (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

%o (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

%K nonn

%O 0,7

%A _Roger L. Bagula_, Mar 25 2005