login
The number of paths of length 13*n from the origin to the line y = 2*x/11 with unit East and North steps that stay below the line or touch it.
9

%I #43 Oct 06 2021 13:24:11

%S 1,6,593,87143,15149546,2891511017,585739005066,123655688922720,

%T 26908765569970320,5993187329634638043,1359541058523676017369,

%U 313029501692713279534165,72965556751635426636633639,17184586991024424745328563477,4083065013894860643162116395527

%N The number of paths of length 13*n from the origin to the line y = 2*x/11 with unit East and North steps that stay below the line or touch it.

%C Equivalent to nonnegative walks from (0,0) to (13*n,0) with step set [1,2], [1,-11].

%H Alois P. Heinz, <a href="/A300389/b300389.txt">Table of n, a(n) for n = 0..414</a>

%H M. T. L. Bizley, <a href="/A060941/a060941.pdf">Derivation of a new formula for the number of minimal lattice paths from (0, 0) to (km, kn) having just t contacts with the line my = nx and having no points above this line; and a proof of Grossman's formula for the number of paths which may touch but do not rise above this line</a>, Journal of the Institute of Actuaries, Vol. 80, No. 1 (1954): 55-62. [Cached copy]

%H Bryan Ek, <a href="https://arxiv.org/abs/1803.10920">Lattice Walk Enumeration</a>, arXiv:1803.10920 [math.CO], 2018.

%H Bryan Ek, <a href="https://arxiv.org/abs/1804.05933">Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics</a>, arXiv:1804.05933 [math.CO], 2018.

%F G.f. satisfies: f = f^78*t^6 + 5*f^67*t^5 - f^66*t^5 + 6*f^65*t^5 + 10*f^56*t^4 - 4*f^55*t^4 + 20*f^54*t^4 - 5*f^53*t^4 + 15*f^52*t^4 + 10*f^45*t^3 - 6*f^44*t^3 + 24*f^43*t^3 - 12*f^42*t^3 + 30*f^41*t^3 - 10*f^40*t^3 + 20*f^39*t^3 + 5*f^34*t^2 - 4*f^33*t^2 + 12*f^32*t^2 - 9*f^31*t^2 + 18*f^30*t^2 - 12*f^29*t^2 + 20*f^28*t^2 - 10*f^27*t^2 + 15*f^26*t^2 + f^23*t - f^22*t + 2*f^21*t - 2*f^20*t + 3*f^19*t - 3*f^18*t + 4*f^17*t - 4*f^16*t + 5*f^15*t - 5*f^14*t + 6*f^13*t + 1.

%F From _Peter Bala_, Jan 03 2019: (Start)

%F O.g.f.: A(x) = exp( Sum_{n >= 1} (1/13)*binomial(13*n, 2*n)*x^n/n ) - Bizley.

%F Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/13)*binomial(13*n-13*k, 2*n-2*k)*a(k) for n >= 1. (End)

%F The sequence defined by b(n) := [x^n] A(x)^n begins [1, 6, 1222, 282993, 69239846, 17468997381, 4494716943847, 1172353182893367, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 except for p = 11 and p = 13 (checked up to p = 101). - _Peter Bala_, Sep 14 2021

%F a(n) ~ c * 13^(13*n) / (n^(3/2) * 2^(2*n) * 11^(11*n)), where c = 0.0250562444901910770802983936320823301923793538303930752981380507191770309... - _Vaclav Kotesovec_, Sep 16 2021

%e For n=1, the possible walks are EEEEEEEEEEENN, EEEEEEEEEENEN, EEEEEEEEENEEN, EEEEEEEENEEEN, EEEEEEEENEEEEN, EEEEEEENEEEEN.

%t m = 15;

%t Exp[Sum[(1/13) Binomial[13n, 2n] x^n/n, {n, 1, m}]] + O[x]^m // CoefficientList[#, x]& (* _Jean-François Alcover_, Feb 26 2020, after _Peter Bala_ *)

%Y Cf. A001764, A060941, A300386, A300387, A300388.

%K nonn,walk,easy

%O 0,2

%A _Bryan T. Ek_, Mar 04 2018