A176806 Consider asymmetric 1-D random walk with set of possible jumps {-1,+1,+2}. Sequence gives number of paths of length n ending at origin.

1, 0, 2, 3, 6, 20, 35, 105, 238, 588, 1512, 3630, 9339, 23166, 58487, 148148, 373230, 949416, 2406248, 6122142, 15591856, 39729000, 101432982, 259049230, 662421643, 1695149220, 4341026900, 11125755615, 28530984915, 73213888650, 187980163110, 482906682675

Offset: 0

Author

Sergey Perepechko, Apr 26 2010

Keywords

• nonn
• easy

Comments

It appears that a(n) is the coefficient of x^n in the expansion (1+x^2+x^3)^n. - Joerg Arndt, Jul 01 2011 [For the proof see the formula section. - Wolfdieter Lang, Nov 05 2018]

Examples

			a(3) = 3: (+2-1-1) or (-1+2-1) or (-1-1+2).
From _Wolfdieter Lang_, Nov 05 2018: (Start)
a(8) = (1/8!)*(d/dt)^8 (1 + t^2 + t^3)^8 becomes for t = 0: 238. (See the comment with the conjecture by _Joerg Arndt_, now proved.)
a(8) = 168 + 70 = 238, the row sum of row n = 8 of A321203, arising from the two [e2, e3] pairs [1, 2] and [4, 0], given in row n = 8 of A321201.
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Crossrefs

Cf. A001005, A321201, A321203.

Programs

Maple

a:=n->add(binomial(n,k)*binomial(k,3*k-n),k=floor((n+2)/3)..floor(n/2));

Mathematica

Table[Sum[Binomial[n, k]*Binomial[k, 3*k-n], {k, Floor[(n+2)/3], Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 01 2016 *)

Formula

a(n) = Sum_{k=floor((n+2)/3)..floor(n/2)} binomial(n,k)*binomial(k,3*k-n).

G.f. g(x) satisfies (31*x^3 + 18*x^2 - x - 4)*g(x)^3 + (x+3)*g(x) + 1 = 0.

Recurrence: 2*n*(2*n-1)*(52*n-79)*a(n) + (n-1)*(52*n^2-79*n+36)*a(n-1) - 6*(n-1)*(156*n^2-315*n+106)*a(n-2) - 31*(n-1)*(n-2)*(52*n-27)*a(n-3) = 0.

a(n) ~ c * d^n / sqrt(Pi*n), where d = 2.61071861327603934981864900838405862... is the root of the equation -31 - 18*d + d^2 + 4*d^3 = 0 and c = 0.57803237802255683003114674597591616... is the root of the equation -31 + 324*c^2 - 1248*c^4 + 1664*c^6 = 0. - Vaclav Kotesovec, Mar 01 2016

From Wolfdieter Lang, Nov 05 2018: (Start)

G.f. G(x) = x*(d/dx)log(F^{[-1]}) = f(x)/(1 - (x*f(x))^2 - 2*(x*f(x))^3) = f(x)/(3 - 2*f(x) + (x*f(x))^2), where f(x) = F^{[-1]}(x)/x, and F^{[-1]})(x) is the compositional inverse of F(y) = x/(1 + x^2 + x^3); that is, F(F^{[-1]}(x)) = x, identically. This G can be proved to solve the equation given above for the g.f. g, if one applies the identity for f (as done above in the last formula for G): (x*f(x))^3 + (x*f(x))*2 - f(x) + 1 = 0 (following from the equation for F^{[-1]}). The expansion of f is given in A001005.

The g.f. G(x) can be computed from the general Lagrange series for the function h(t) with derivative h(t)" = 1/phi(t), where phi(t) = (1 + t^2 + t^3), and the inversion of x = y/phi(y) = F(y). Then one finds G(x) = (d/dx)h(F^{[-1]}(x)) = (1/phi(F^{[-1]}(x)))*(d/dx)F^{[-1]}(x), which becomes with the above mentioned identity for f(x) = F^{[-1]}(x)/x the result G(x) = f(x)/(3 - 2*f(x) + (x*f(x))^2).

From this special Lagrange series derivation and the proof that the g.f. g from above coincides with G, the conjecture, given by Joerg Arndt as a comment above, has been proved. This uses [t^n]phi(t)^n = (1/n!)*(d/dt)^n phi(t)^n, evaluated at t = 0, which appears in the considered Lagrange series.

a(n) = Sum_{2*e + 3*e3 = n} n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 1 and a(1) = 0. This is the row sum of the irregular table A321203 of these multinomial numbers for the solutions for the pairs (e2, e3). The pairs of solutions are given in A321201.

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