OFFSET
1,1
COMMENTS
We have the formula below because we have 2*n choices for the first step, and (2*n-1)^3 choices for the next three, but have counted exactly 2*n*(2*n-1)*(2*n-2) self-intersecting walks.
REFERENCES
N. Madras and G. Slade, "The Self Avoiding Walk", Birkhäuser, 2013.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = 16*n^4 - 24*n^3 + 8*n^2 + 2*n.
G.f.: 2*x*(1 + 45*x + 123*x^2 + 23*x^3)/(1 - x)^5. - Stefano Spezia, Jul 07 2024
MATHEMATICA
A374397[n_] := 2*n*(4*n*(n - 1)*(2*n - 1) + 1);
Array[A374397, 50] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {2, 100, 726, 2696, 7210}, 50] (* Paolo Xausa, Sep 23 2024 *)
CROSSREFS
KEYWORD
nonn,walk,easy
AUTHOR
Johann Peters, Jul 07 2024
STATUS
approved