A370459 Number of unicursal stars with n vertices.
0, 0, 1, 1, 5, 19, 112, 828, 7441, 76579, 871225, 10809051, 144730446, 2079635889, 31912025537, 520913578812, 9013780062785, 164829273635749, 3176388519597555, 64343477504391475, 1366925655386979893, 30390554390984325019, 705740995420852895453
Offset: 3
Keywords
Examples
For n=5, there is only the regular pentagram {5/2}. For n=6, there is only the unicursal hexagram. For n=7, in addition to the two regular heptagrams {7/2} and {7/3}, there are three nontrivial unicursal heptagrams represented by: (0, 2, 4, 1, 6, 3, 5, 0) (0, 2, 5, 1, 3, 6, 4, 0) (0, 2, 5, 1, 4, 6, 3, 0).
Links
- Andrew Howroyd, Table of n, a(n) for n = 3..200
- Gordon FitzGerald, illustration of a unicursal dodecagram.
- Wikipedia, Unicursal hexagram.
Crossrefs
Programs
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PARI
\\ Requires a370068 from A370068. Ro(n)=-(-1)^n + subst(serlaplace(polcoef(((1 - x)^2)/(2*(1 + x)*(1 + (1 - 2*y)*x + 2*y*x^2)) + O(x*x^n), n)), y, 1) Re(n)=subst(serlaplace(polcoef((1 - x - 2*x^2)/(4*(1 + (1 - 2*y)*x + 2*y*x^2)) + O(x*x^n), n)), y, 1) a(n)={if(n<3, 0, (if(n%2, 2*Ro(n\2), Re(n/2)) + a370068(n))/4)} \\ Andrew Howroyd, Mar 01 2024
Formula
Extensions
a(14) onwards from Andrew Howroyd, Feb 26 2024
Comments