This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A333759 #17 Apr 07 2020 10:38:14 %S A333759 1,1,11,191,11346,2002405,1112939654,1878223479450 %N A333759 Number of self-avoiding closed paths in the n X n grid graph which pass through all vertices on four (left, right, upper, lower) sides of the graph. %C A333759 a(11) = 152567999801505122456. %e A333759 a(2) = 1; %e A333759 +--+ %e A333759 | | %e A333759 +--+ %e A333759 a(3) = 1; %e A333759 +--+--+ %e A333759 | | %e A333759 + + %e A333759 | | %e A333759 +--+--+ %e A333759 a(4) = 11; %e A333759 +--+--+--+ +--+--+--+ +--+--+--+ %e A333759 | | | | | | %e A333759 +--*--* + +--* *--+ +--* + %e A333759 | | | | | | %e A333759 +--*--* + +--* *--+ +--* + %e A333759 | | | | | | %e A333759 +--+--+--+ +--+--+--+ +--+--+--+ %e A333759 +--+--+--+ +--+--+--+ +--+--+--+ %e A333759 | | | | | | %e A333759 + *--*--+ + *--* + + *--+ %e A333759 | | | | | | | | %e A333759 + *--*--+ + * * + + *--+ %e A333759 | | | | | | | | %e A333759 +--+--+--+ +--+ +--+ +--+--+--+ %e A333759 +--+--+--+ +--+--+--+ +--+ +--+ %e A333759 | | | | | | | | %e A333759 + + + + + *--* + %e A333759 | | | | | | %e A333759 + *--* + + + + *--* + %e A333759 | | | | | | | | | | %e A333759 +--+ +--+ +--+--+--+ +--+ +--+ %e A333759 +--+ +--+ +--+ +--+ %e A333759 | | | | | | | | %e A333759 + *--* + + * * + %e A333759 | | | | | | %e A333759 + + + *--* + %e A333759 | | | | %e A333759 +--+--+--+ +--+--+--+ %o A333759 (Python) %o A333759 # Using graphillion %o A333759 from graphillion import GraphSet %o A333759 import graphillion.tutorial as tl %o A333759 def A333759(n): %o A333759 universe = tl.grid(n - 1, n - 1) %o A333759 GraphSet.set_universe(universe) %o A333759 cycles = GraphSet.cycles() %o A333759 points = [i for i in range(1, n * n + 1) if i % n < 2 or ((i - 1) // n + 1) % n < 2] %o A333759 for i in points: %o A333759 cycles = cycles.including(i) %o A333759 return cycles.len() %o A333759 print([A333759(n) for n in range(2, 10)]) %Y A333759 Main diagonal of A333758. %Y A333759 Cf. A333466. %K A333759 nonn,more %O A333759 2,3 %A A333759 _Seiichi Manyama_, Apr 04 2020