A152664
Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} for which k is the maximal number of initial even entries (0 <= k <= floor(n/2)).
Original entry on oeis.org
1, 1, 1, 4, 2, 12, 8, 4, 72, 36, 12, 360, 216, 108, 36, 2880, 1440, 576, 144, 20160, 11520, 5760, 2304, 576, 201600, 100800, 43200, 14400, 2880, 1814400, 1008000, 504000, 216000, 72000, 14400, 21772800, 10886400, 4838400, 1814400, 518400, 86400
Offset: 1
T(3,0)=4 because we have 123, 132, 312 and 321.
T(4,2)=4 because we have 2413, 2431, 4213 and 4231.
Triangle starts:
1;
1, 1;
4, 2;
12, 8, 4;
72, 36, 12;
360, 216, 108, 36;
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T := proc (n, k) if `mod`(n, 2) = 1 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial(n-k-1, (1/2)*n-1/2) else factorial((1/2)*n)^2*binomial(n-k-1, (1/2)*n-1) end if end proc: for n to 11 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
A152663
Number of leading odd entries in all permutations of {1,2,...,n} (see example).
Original entry on oeis.org
1, 1, 6, 16, 120, 540, 5040, 32256, 362880, 3024000, 39916800, 410572800, 6227020800, 76281004800, 1307674368000, 18598035456000, 355687428096000, 5762136335155200, 121645100408832000, 2211729098342400000, 51090942171709440000, 1030334000462807040000
Offset: 1
a(3) = 6 because in the permutations 123, 132, 213, 231, 312, 321 we have 1+2+0+0+2+1 = 6 leading odd entries.
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ao := proc (n) options operator, arrow; factorial(2*n+1) end proc: ae := proc (n) options operator, arrow: n*factorial(2*n)/(n+1) end proc: a := proc (n) if `mod`(n, 2) = 1 then ao((1/2)*n-1/2) else ae((1/2)*n) end if end proc: seq(a(n), n = 1 .. 20);
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a[n_] := If[OddQ[n], n!, n*n!/(n+2)];
Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Apr 02 2024 *)
A152665
Number of leading even entries in all permutations of {1,2,...,n}.
Original entry on oeis.org
0, 1, 2, 16, 60, 540, 3024, 32256, 241920, 3024000, 28512000, 410572800, 4670265600, 76281004800, 1017080064000, 18598035456000, 284549942476800, 5762136335155200, 99527809425408000, 2211729098342400000, 42575785143091200000, 1030334000462807040000
Offset: 1
The permutation 4,6,2,1,5,3 begins with three even numbers, so would contribute 3 to a(6).
a(3)=2 because in the permutations 123, 132, 213, 231, 312, 321 we have 0+0+1+1+0+0 = 2 leading odd entries.
a(45) = 16: Here are the permutations of 1234, 24 in all:
1(234) total 6, no. of initial even terms = 0
3(124) ditto
21(34) total 2, no. of initial even terms 1*2 = 2
23(14) ditto
24(13) total 2, no. of initial even terms 2 twice = 4
Subtotal from 2*** is 2+2+4 = 8
Subtotal from 4*** is also 2+2+4 = 8
Total a(4) = 8+8 = 16.
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ao := proc (n) options operator, arrow; n*factorial(2*n+1)/(n+2) end proc: ae := proc (n) options operator, arrow; n*factorial(2*n)/(n+1) end proc: a := proc (n) if `mod`(n, 2) = 1 then ao((1/2)*n-1/2) else ae((1/2)*n) end if end proc; seq(a(n), n = 1 .. 20);
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a[n_] := If[OddQ[n], (n-1)*n!/(n+3), n*n!/(n+2)];
Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Apr 29 2023 *)
Showing 1-3 of 3 results.
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