A208654
Number of palindromic paths through the subset array of {1,2,...,n}; see Comments.
Original entry on oeis.org
1, 2, 18, 144, 12000, 540000, 388962000, 108425318400, 650403212820480, 1175952237465600000, 57409367332363200000000, 691636564481660937216000000, 270540272566435932512004833280000
Offset: 1
For n=4, write
row 1: 1,2,3,4
row 2: 1,2; 1,3; 1,4; 2,3; 2,4; 3;4
row 3: 1,2,3; 1,2,4; 1,3,4; 2,3,4
row 4: 1,2,3,4
To form a palindromic path of length 4, there are 4 ways to choose 1st term from row 1, then 12 ways to choose 2nd term from row 2, then 3 ways to choose 3rd term, then 1 way to finish, so that a(4)=4*12*3*1=144.
-
m[n_] := Floor[(n + 1)/2]; z = 21;
g[n_] := Product[i*Binomial[n, i], {i, 1, m[n]}]
h[n_] := Product[Binomial[n - 1, i], {i, m[n], n - 1}]
Table[g[n], {n, 1, z}] (* A208652 *)
Table[h[n], {n, 1, z}] (* A208653 *)
Table[g[n] h[n], {n, 1, 2 z/3}] (* A208654 *)
Table[g[n] h[n]/n, {n, 1, 2 z/3}] (* A208655 *)
A208652
Product{i*C(n,i) : 1<=i<=floor[(n+1)/2]}.
Original entry on oeis.org
1, 2, 18, 48, 3000, 10800, 4321800, 21073920, 51849745920, 342921600000, 5062554438480000, 45444644517888000, 3960533401513405516800, 48272757747179554068480, 24547182197052103038450000000
Offset: 1
-
m[n_]:=Floor[(n+1)/2];z=21;
g[n_]:=Product[i*Binomial[n,i],{i,1,m[n]}]
Table[g[n],{n,1,z}] (* A208652 *)
Table[Gamma[1 + n]^(n/2)/BarnesG[1 + n]*((1 + (-1)^n)/2 + (1 - (-1)^n) * Sqrt[Gamma[1 + n]]/(2*Gamma[(1 + n)/2])), {n, 1, 15}] (* Vaclav Kotesovec, Apr 10 2024 *)
A208655
Number of palindromic paths starting with 1 through the subset array of {1,2,...,n}; see Comments.
Original entry on oeis.org
1, 1, 6, 36, 2400, 90000, 55566000, 13553164800, 72267023646720, 117595223746560000, 5219033393851200000000, 57636380373471744768000000, 20810790197418148654769602560000, 1578992018570629416640340512656998400
Offset: 1
For n=4, write
row 1: 1,2,3,4
row 2: 1,2; 1,3; 1,4; 2,3; 2,4; 3;4
row 3: 1,2,3; 1,2,4; 1,3,4; 2,3,4
row 4: 1,2,3,4
To form a palindromic path of length 4 and starting with 1, there is 1 way to choose 1st term from row 1, then 12 ways to choose 2nd term from row 2, then 3 ways to choose 3rd term, then 1 way to finish. Thus, a(4)=1*12*3*1=36.
-
m[n_] := Floor[(n + 1)/2]; z = 21;
g[n_] := Product[i*Binomial[n, i], {i, 1, m[n]}]
h[n_] := Product[Binomial[n - 1, i], {i, m[n], n - 1}]
Table[g[n], {n, 1, z}] (* A208652 *)
Table[h[n], {n, 1, z}] (* A208653 *)
Table[g[n] h[n], {n, 1, 2 z/3}] (* A208654 *)
Table[g[n] h[n]/n, {n, 1, 2 z/3}] (* A208655 *)
Showing 1-3 of 3 results.
Comments