A208661
Number of paths through the subset array of {1,2,...,n} that have range a subset of {1,2}; see Comments at A208650.
Original entry on oeis.org
1, 4, 16, 144, 3072, 160000, 20736000, 6776582400, 5639493386240, 12041750911647744, 66367674777600000000, 948747002032456704000000, 35319813420678024297185280000, 3435829941215659552226769213849600
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
There are 2 ways to choose 1 or 2 from row 1, then 6 ways to choose 1 or 2 from row 2, then 6 ways to choose 1 or 2 from row 3, then 2 ways to choose 1 or 2 from row 4, so that a(4)=2*6*6*2=144.
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a[1] = 1; z = 15;
a[n_] := (2^n)*Product[Binomial[n - 1, i], {i, 1, n - 1}]
Table[a[n], {n, 1, z}] (* A208661 *)
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 *)
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.
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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 *)
A208651
Number of paths through the subset array whose trace is a permutation of (1,2,...,n); see Comments.
Original entry on oeis.org
1, 2, 12, 216, 11520, 1800000, 816480000, 1067311728000, 3996990937497600, 42672954793151692800, 1293547461212160000000000, 110950032218933108678400000000, 26847804299643702075375747072000000
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,
4 ways to choose a different number from row 2,
1 way to choose remaining number from row 3.
Total: a(3) = 1*4*3 = 12 paths.
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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 *)
Showing 1-4 of 4 results.
Comments