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 *)
A208653
a(n) = Product_{i=floor((n + 1)/2)..n-1} binomial(n-1, i).
Original entry on oeis.org
1, 1, 1, 3, 4, 50, 90, 5145, 12544, 3429216, 11340000, 15219319500, 68309049600, 457937132487120, 2790771598030416, 94609025993497640625, 783056974947287040000, 135476575389769051389952000, 1523136299736565293430210560, 1354434926051634531310373234715648
Offset: 1
-
m[n_]:=Floor[(n+1)/2]; z=21;
h[n_]:=Product[Binomial[n-1,i],{i,m[n],n-1}]
Table[h[n],{n,1,z}] (* A208653 *)
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