A208650
Number of constant paths through the subset array of {1,2,...,n}; see Comments.
Original entry on oeis.org
1, 2, 6, 36, 480, 15000, 1134000, 211768200, 99131719680, 117595223746560, 356467003200000000, 2779532232516963000000, 56049508602150185041920000, 2935889842347365340037522521600
Offset: 1
Taking n=3:
row 1: {1},{2},{3} ---------> 1,2,3
row 2: {1,2},{1,3},{2,3} ---> 1,1,2,2,3,3
row 3: {1,2,3} -------------> 1,2,3
3 ways to choose a number from row 1,
2 ways to choose same number from row 2,
1 way to choose same number from row 3.
Total: a(3) = 1*2*3 = 6 paths.
-
p[n_]:=Product[Binomial[n-1,k],{k,1,n-1}]
Table[p[n],{n,1,20}] (* A001142(n-1) *)
Table[p[n]*n,{n,1,20}] (* A208650 *)
Table[p[n]*n!,{n,1,20}] (* A208651 *)
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.
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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 1-2 of 2 results.
Comments