A336675 Number of paths of length n starting at initial node of the path graph P_10.

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 251, 460, 911, 1690, 3327, 6225, 12190, 22950, 44744, 84626, 164407, 312019, 604487, 1150208, 2223504, 4239225, 8181175, 15621426, 30108147, 57556155, 110820165, 212037241, 407946421, 781074572, 1501844193, 2877011660, 5529362694

Offset: 0

Author

Nachum Dershowitz, Jul 30 2020

Keywords

• nonn
• easy
• walk

Comments

Also the number of paths along a corridor width 10, starting from one side.

In general, a(n,m) = (2^n/(m+1))*Sum_{r=1..m} (1-(-1)^r)*cos(Pi*r/(m+1))^n*(1+cos(Pi*r/(m+1))) gives the number of paths of length n starting at the initial node on the path graph P_m. Here we have m=10. - Herbert Kociemba, Sep 14 2020

Links

• Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015-2016.
• Nachum Dershowitz, Between Broadway and the Hudson: A Bijection of Corridor Paths, arXiv:2006.06516 [math.CO], 2020.
• Index entries for linear recurrences with constant coefficients, signature (1,4,-3,-3,1).

Crossrefs

This is row 10 of A094718. Bisections give A224514 (even part), A216710 (odd part).

Cf. A000004 (row 0), A000007 (row 1), A000012 (row 2), A016116 (row 3), A000045 (row 4), A038754 (row 5), A028495 (row 6), A030436 (row 7), A061551 (row 8), A178381 (row 9), this sequence (row 10), A336678 (row 11), A001405 (limit).

Programs

Maple

X := j -> (-1)^(j/11) - (-1)^(1-j/11):
a := k -> add((2 + X(j))*X(j)^k, j in [1, 3, 5, 7, 9])/11:
seq(simplify(a(n)), n=0..30); # Peter Luschny, Sep 17 2020

Mathematica

a[n_,m_]:=2^(n+1)/(m+1) Module[{x=(Pi r)/(m+1)},Sum[Cos[x]^n (1+Cos[x]),{r,1,m,2}]]
Table[a[n,10],{n,0,40}]//Round (* Herbert Kociemba, Sep 14 2020 *)

PARI

my(x='x+O('x^44)); Vec((1 - 3*x^2 + x^4)/(1 - x - 4*x^2 + 3*x^3 + 3*x^4 - x^5)) \\ Joerg Arndt, Jul 31 2020

Formula

From Stefano Spezia, Jul 30 2020: (Start)

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

a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 3*a(n-4) + a(n-5) for n > 4. (End)

a(n) = (2^n/11)*Sum_{r=1..10} (1-(-1)^r)*cos(Pi*r/11)^n*(1+cos(Pi*r/11)). - Herbert Kociemba, Sep 14 2020