A370832
Triangle read by rows: T(n,k) gives the number of parking functions of size n with k lucky cars. 0 <= k <= n.
Original entry on oeis.org
1, 0, 1, 0, 1, 2, 0, 2, 8, 6, 0, 6, 37, 58, 24, 0, 24, 204, 504, 444, 120, 0, 120, 1318, 4553, 6388, 3708, 720, 0, 720, 9792, 44176, 87296, 81136, 33984, 5040, 0, 5040, 82332, 463860, 1203921, 1582236, 1064124, 341136, 40320, 0, 40320, 773280, 5270480, 17164320, 29724000, 28328480, 14602320, 3733920, 362880
Offset: 0
Table begins:
n\k| 0 1 2 3 4 5 6 7 8
---+-------------------------------------------------------------
0 | 1
1 | 0 1
2 | 0 1 2
3 | 0 2 8 6
4 | 0 6 37 58 24
5 | 0 24 204 504 444 120
6 | 0 120 1318 4553 6388 3708 720
7 | 0 720 9792 44176 87296 81136 33984 5040
8 | 0 5040 82332 463860 1203921 1582236 1064124 341136 40320
...
- Alois P. Heinz, Rows n = 0..140, flattened
- Irfan Durmić, Alex Han, Pamela E. Harris, Rodrigo Ribeiro, and Mei Yin, Probabilistic Parking Functions, arXiv:2211.00536 [math.CO], 2022.
- FindStat, St000135: The number of lucky cars of the parking function.
Cf.
A000142 (main diagonal and column k=1 shifted).
-
b:= proc(n) option remember; `if`(n=0, 1,
expand(x*mul((n+1-k)+k*x, k=2..n)))
end:
T:= (n, k)-> coeff(b(n), x, k):
seq(seq(T(n,k), k=0..n), n=0..10); # Alois P. Heinz, Jun 26 2024
-
row[n_] := (x (x - 1)^n Pochhammer[(n + x) / (x - 1), n]) / (n + x);
Table[CoefficientList[Series[row[n], {x, 0, n}], x], {n, 0, 8}] // Flatten
(* Peter Luschny, Jun 27 2024 *)
A121579
Triangle read by rows: T(n,k) is the number of deco polyominoes of height n having along the lower contour exactly k reentrant corners, i.e., a vertical step that is followed by a horizontal step (n>=1, k>=0).
Original entry on oeis.org
1, 2, 5, 1, 16, 8, 65, 52, 3, 326, 344, 50, 1957, 2473, 595, 15, 13700, 19676, 6524, 420, 109601, 173472, 71862, 7840, 105, 986410, 1686912, 823836, 127232, 4410, 9864101, 17981193, 9976686, 1975750, 118125, 945, 108505112, 208769296, 128350992
Offset: 1
T(2,0)=2 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having no reentrant corners along the lower contour.
Triangle starts:
1;
2;
5, 1;
16, 8;
65, 52, 3;
326, 344, 50;
-
Q[1]:=1: for n from 2 to 13 do Q[n]:=sort(expand(subs(x=t,Q[n-1])+(n-1)*x*subs(x=1,Q[n-1]))) od: for n from 1 to 13 do P[n]:=subs(x=1,Q[n]) od: for n from 1 to 13 do seq(coeff(P[n],t,j),j=0..ceil(n/2)-1) od; # yields sequence in triangular form
A321853
a(n) is the sum of the fill times of all 1-dimensional fountains given by the permutations in S_n.
Original entry on oeis.org
0, 1, 10, 86, 756, 7092, 71856, 787824, 9329760, 118956960, 1627067520, 23786386560, 370371536640, 6122231942400, 107109431654400, 1977781262284800, 38445562145894400, 784885857270681600, 16792523049093120000, 375755553108633600000, 8777531590107033600000
Offset: 1
For the permutation 15234, the well takes a total of seven seconds to reach the sink: it takes 4 seconds to fill to 55234, then it takes 1 second to fill to 55334, then it takes 2 seconds to fill to 55444, where it reaches the sink.
For n = 3 the a(3) = 10 sum occurs from summing over the times in the following table:
+-------------+------+-------------+
| permutation | time | final state |
+-------------+------+-------------+
| 123 | 3 | 333 |
| 132 | 2 | 332 |
| 213 | 3 | 333 |
| 231 | 1 | 331 |
| 312 | 1 | 322 |
| 321 | 0 | 321 |
+-------------+------+-------------+
-
List([1..22],n->Factorial(n)*Sum([0..n],k->(n-k)*k/(k+1))); # Muniru A Asiru, Dec 05 2018
-
[Factorial(n)*(&+[(n-k)*k/(k+1): k in [1..n]]): n in [1..25]]; // G. C. Greubel, Dec 04 2018
-
a:=n->factorial(n)*add((n-k)*k/(k+1),k=0..n): seq(a(n),n=1..22); # Muniru A Asiru, Dec 05 2018
-
Table[n!*Sum[(n - k)*k/(k + 1), {k, 1, n - 1}], {n, 1, 21}]
-
vector(25, n, n!*sum(k=0,n, (n-k)*k/(k+1))) \\ G. C. Greubel, Dec 04 2018
-
from sympy.abc import k, a, b
from sympy import factorial
from sympy import Sum
for n in range(1,25): print(int(factorial(n)*Sum((n-k)*k/(k+1), (k, 0, n)).doit().evalf()), end=', ') # Stefano Spezia, Dec 05 2018
-
[factorial(n)*sum((n-k)*k/(k+1) for k in (1..n)) for n in (1..25)] # G. C. Greubel, Dec 04 2018
A367850
Total sum of the block maxima minus the block minima over all partitions of [n].
Original entry on oeis.org
0, 0, 1, 6, 33, 182, 1034, 6122, 37927, 246030, 1669941, 11844324, 87644672, 675494180, 5413500801, 45040155758, 388441330457, 3467619369538, 31998729152474, 304846692965822, 2994781617653439, 30304301968015582, 315536869771786501, 3377398077726963112
Offset: 0
a(3) = 6 = 2 + 1 + 2 + 1 + 0: 123, 12|3, 13|2, 1|23, 1|2|3.
-
b:= proc(n, m, t) option remember; `if`(n=0, [1, 0], (p->
p+[0, p[1]*(n-t)])(b(n-1, m+1, t+1))+m*b(n-1, m, t+1))
end:
a:= n-> b(n, 0, 1)[2]:
seq(a(n), n=0..23);
# second Maple program:
egf:= (z-2)*exp(2*z+exp(z)-1)+(2*z+1)*exp(z+exp(z)-1)+exp(exp(z)-1):
a:= n-> n!*coeff(series(egf, z, n+1), z, n):
seq(a(n), n=0..23);
A368401
Number T(n,k) of permutations of [n] whose sum of cycle maxima minus cycle minima gives k, triangle T(n,k), n>=0, 0<=k<=A002620(n), read by rows.
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 3, 1, 3, 7, 11, 2, 1, 4, 12, 28, 53, 12, 10, 1, 5, 18, 52, 135, 289, 84, 72, 58, 6, 1, 6, 25, 84, 257, 734, 1825, 524, 564, 496, 422, 60, 42, 1, 7, 33, 125, 429, 1407, 4545, 12983, 3520, 3976, 4292, 3950, 3422, 790, 486, 330, 24
Offset: 0
T(3,0) = 1: (1)(2)(3).
T(3,1) = 2: (12)(3), (1)(23).
T(3,2) = 3: (123), (132), (13)(2).
Triangle T(n,k) begins:
1;
1;
1, 1;
1, 2, 3;
1, 3, 7, 11, 2;
1, 4, 12, 28, 53, 12, 10;
1, 5, 18, 52, 135, 289, 84, 72, 58, 6;
1, 6, 25, 84, 257, 734, 1825, 524, 564, 496, 422, 60, 42;
...
-
b:= proc(n, s) option remember; `if`(n=0, 1, (k-> `if`(n>k,
b(n-1, s)+add(b(n-1, subs(h=h+[0, 1], s)), h=s), 0)+
`if`(n>k+1, b(n-1, {s[], [n,1]}), 0)+add(h[2]!*expand(
x^(h[1]-n)*b(n-1, s minus {h})), h=s))(nops(s)))
end:
T:= (n, k)-> coeff(b(n, {}), x, k):
seq(seq(T(n, k), k=0..floor(n^2/4)), n=0..10);
A383875
Number of pairs in the Bruhat order of type A_n.
Original entry on oeis.org
1, 3, 19, 213, 3781, 98407, 3550919
Offset: 0
For n=0, the only element is 1 (identity) so a(0)=1.
For n=1 the elements are 1 (identity) and s1. The order relation consists of pairs (1, 1), (1, s1), and (s1, s1). So a(1) = 3.
For n=2 the line (Hasse) diagram is below.
s1*s2*s1
/ \
s2*s1 s1*s2
| X |
s2 s1
\ /
1
The order relation consists of the six reflexive pairs, the eight pairs shown in the diagram as edges, and the five pairs (1, s2*s1), (1, s1*s2), (1, s1*s2*s1), (s1, s1*s2*s1), and (s2, s1*s2*s1). So a(2) = 6+8+5 = 19.
- A. Bjorner and F. Brenti, Combinatorics of Coxeter Groups, Springer, 2009, 27-64.
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