A105927
Let d(n) = A000166(n); then a(n) = ( (n^2+n-1)*d(n) + (-1)^(n-1)*(n-1) )/2.
Original entry on oeis.org
0, 0, 2, 12, 84, 640, 5430, 50988, 526568, 5940576, 72755370, 961839340, 13656650172, 207316760352, 3351430059614, 57487448630220, 1042952206111440, 19954639072648768, 401578933206288978, 8480263630552747596, 187505565234912994340, 4332318322289242716480
Offset: 0
- P. A. MacMahon, Combinatory Analysis, 2 vols., Chelsea, NY, 1960, see p. 108.
-
a:= proc(n) option remember; `if`(n<3, n*(n-1),
n*(n-1)*(a(n-1)+a(n-2))/(n-2))
end:
seq(a(n), n=0..25); # Alois P. Heinz, Jun 03 2014
-
Table[(Subfactorial[n+2]-2Subfactorial[n+1]-Subfactorial[n])/2,{n,0,21}] (* Geoffrey Critzer, Jun 02 2014 *)
-
s(n) = if( n<1, 1, n * s(n-1) + (-1)^n);
a(n) = (s(n + 2) - 2*s(n + 1) - s(n))/2; \\ Indranil Ghosh, Apr 06 2017
A161129
Triangle read by rows: T(n,k) is the number of non-derangements of {1,2,...,n} for which the difference between the largest and smallest fixed points is k (n>=1; 0 <= k <= n-1).
Original entry on oeis.org
1, 0, 1, 3, 0, 1, 8, 3, 2, 2, 45, 8, 9, 8, 6, 264, 45, 44, 42, 36, 24, 1855, 264, 265, 256, 234, 192, 120, 14832, 1855, 1854, 1810, 1704, 1512, 1200, 720, 133497, 14832, 14833, 14568, 13950, 12864, 11160, 8640, 5040, 1334960, 133497, 133496, 131642, 127404
Offset: 1
T(4,1)=3 because we have 1243, 4231, and 2134; T(4,2)=2 because we have 1432 and 3214; T(5,4)=6 because we have 1xyz5 where xyz is any permutation of 234.
Triangle starts:
1;
0, 1;
3, 0, 1;
8, 3, 0, 1;
45, 8, 9, 8, 6;
264, 45, 44, 42, 36, 24;
-
d[0] := 1: for n to 15 do d[n] := n*d[n-1]+(-1)^n end do: T := proc (n, k) if k = 0 then n*d[n-1] elif k < n then (n-k)*(sum(binomial(k-1, j)*d[n-2-j], j = 0 .. k-1)) else 0 end if end proc: for n to 10 do seq(T(n, k), k = 0 .. n-1) end do; # yields sequence in triangular form
-
d = Subfactorial;
T[n_, 0] := n*d[n - 1];
T[n_, k_] := (n - k)*Sum[d[n - j - 2]*Binomial[k - 1, j], {j, 0, k - 1}];
Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 28 2017 *)
A193364
Number of permutations that have a fixed point and contain 123.
Original entry on oeis.org
0, 0, 0, 1, 1, 3, 11, 59, 369, 2665, 21823, 199983, 2028701, 22577141, 273551115, 3585133147, 50540288857, 762641865009, 12265883397719, 209475278413895, 3785852926650453, 72191462591370733, 1448516763956727331, 30507960955933725171, 672958104387944656145
Offset: 0
For n=5 we have 12345, 12354 and 41235, so a(5)=3.
For n=6 we have 123456, 123465, 123546, 123465, 123645, 123654, 412356, 451236, 512346, 541236 and 612354, so a(6)=11.
-
a:= proc(n) option remember;
`if`(n<7, [0$3, 1$2, 3, 11][n+1],
((4*n^3-42*n^2+92*n+39) *a(n-1)
+(32*n^3-2*n^4-163*n^2+223*n+204) *a(n-2)
-(n-4)*(n-7)*(2*n^2-10*n-15) *a(n-3)) / (2*n^2-14*n-3))
end:
seq(a(n), n=0..30); # Alois P. Heinz, Jan 07 2013
-
a[n_] := a[n] = If[n<7, {0, 0, 0, 1, 1, 3, 11}[[n+1]], ((4n^3 - 42n^2 + 92n + 39) a[n-1] + (32n^3 - 2n^4 - 163n^2 + 223n + 204) a[n-2] - (n-4)(n-7) (2n^2 - 10n - 15) a[n-3])/(2n^2 - 14n - 3)];
a /@ Range[0, 30] (* Jean-François Alcover, Mar 15 2021, after Alois P. Heinz *)
A373966
Triangle read by rows: T(n,k) = (-1)^(n+1) * A000166(n) + (-1)^(k) * A000166(k) for n >= 2 and 1 <= k <= n-1.
Original entry on oeis.org
-1, 2, 3, -9, -8, -11, 44, 45, 42, 53, -265, -264, -267, -256, -309, 1854, 1855, 1852, 1863, 1810, 2119, -14833, -14832, -14835, -14824, -14877, -14568, -16687, 133496, 133497, 133494, 133505, 133452, 133761, 131642, 148329, -1334961, -1334960, -1334963, -1334952, -1335005, -1334696, -1336815, -1320128, -1468457
Offset: 2
Triangle begins:
-1;
2, 3;
-9, -8, -11;
44, 45, 42, 53;
-265, -264, -267, -256, -309;
1854, 1855, 1852, 1863, 1810, 2119;
...
-
T[n_,k_]:= (-1)^(n+1)*Subfactorial[n] + (-1)^k*Subfactorial[k]; Table[T[n,k],{n,2,10},{k,n-1}]// Flatten (* Stefano Spezia, Jun 24 2024 *)
Showing 1-4 of 4 results.
Comments