A007174 Erroneous version of A182322.
1, 1, 1, 3, 7, 20, 131, 815, 5142, 36800, 272093, 2077909, 16176607, 127997683, 1025727646, 8310377815, 68217725764, 560527576100, 4556993996246
Offset: 1
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.
A001764 := proc(m) RETURN((3*m)!/(m!*(2*m+1)!)); end; # Gives A001764(m) A047749 := proc(m) local x; if m mod 2 = 1 then x := (m-1)/2; RETURN((3*x+1)!/((x+1)!*(2*x+1)!)); fi; RETURN(A001764(m/2)); end; # Gives A047749(m) A027610 := proc(n) local N; N := 0; N := N + A001764(n)/(12*(n+1)); if n mod 2 = 0 then N := N + 5/24*A001764(n/2); fi; if (n-1) mod 3 = 0 then N := N + 1/3*A001764((n-1)/3); fi; if (n-1) mod 4 = 0 then N := N + 1/4*A001764((n-1)/4); fi; if (n-2) mod 6 = 0 then N := N + 1/6*A001764((n-2)/6); fi; N := N + 3/8*A047749(n); if (2*n-1) mod 3 = 0 then N := N + 1/6*A047749((2*n-1)/3); fi; RETURN(N); end;
Table[Binomial[3 n, 2 n]/(6 (2 n + 1) (2 n + 2)) + If[EvenQ[n], 7 Binomial[3 n/2, n]/(12 (n + 1)), 3 Binomial[3 n/2 - 1/2, n]/(4 (n + 1))] + Switch[Mod[n, 3], 1, Binomial[n - 1, 2 n/3 - 2/3]/(2 n/3 + 1/3), 2, Binomial[n - 1, 2 n/3 - 1/3]/(2 n/3 + 2/3), , 0]/3 + If[1 == Mod[n,4], Binomial[3 n/4 - 3/4, n/2 - 1/2]/(n/2 + 1/2), 0]/4 + If[2 == Mod[n, 6], Binomial[n/2 - 1, n/3 - 2/3]/(n/3 + 1/3), 0]/6, {n, 1, 30}] (* _Robert A. Russell, Apr 11 2012 *)
T(m)={if(m<0||denominator(m)!=1,0,(3*m)!/(m!*(2*m+1)!))}; U(k)={if(k<0||denominator(k)!=1,0,if(k%2,my(m=(k-1)/2);(3*m+1)!/((m+1)!*(2*m+1)!),T(k/2)))}; S(n)=T(n)/(12*(n+1))+5*T(n/2)/24+T((n-1)/3)/3+T((n-1)/4)/4+T((n-2)/6)/6+3*U(n)/8+U((2*n-1)/3)/6; for(k=1,26,print1(S(k),", ")) \\ Hugo Pfoertner, Mar 07 2020
For n=4 the a(4)=3 solutions are the three achiral (there are no chiral) clusters that can be formed from four simplexes in four-space. One has three attached to a fourth, one has four sharing a common triangle, and the last has neither of these properties.
n = 30; e[d_,t_]:=Sum[Binomial[d k,k]/((d-1)k+1)t^k,{k,0,n}] CoefficientList[Series[(10e[4,t^2]e[2,e[4,t^2]t]^3t +30e[4,t^4]t(1+e[4,t^4]t) +20e[1,e[4,t^6] t^2]e[2,e[4,t^6]t^3]t)/60 -(6(e[2,e[4,t^2]t]-1)^2+6e[4,t^4]^2t^2)/24 +(4e[4,t^2]^4t^2+8e[1,e[4,t^6]t^2]e[4,t^6]t^2)/24, {t,0,n}]/t,t] (* Robert A. Russell, Apr 23 2012 *)
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