A092582 Triangle read by rows: T(n,k) is the number of permutations p of [n] having length of first run equal to k.
1, 1, 1, 3, 2, 1, 12, 8, 3, 1, 60, 40, 15, 4, 1, 360, 240, 90, 24, 5, 1, 2520, 1680, 630, 168, 35, 6, 1, 20160, 13440, 5040, 1344, 280, 48, 7, 1, 181440, 120960, 45360, 12096, 2520, 432, 63, 8, 1, 1814400, 1209600, 453600, 120960, 25200, 4320, 630, 80, 9, 1
Offset: 1
Examples
T(4,3) = 3 because 1243, 1342 and 2341 are the only permutations of [4] having length of first run equal to 3. 1; 1, 1; 3, 2, 1; 12, 8, 3, 1; 60, 40, 15, 4, 1; 360, 240, 90, 24, 5, 1; 2520, 1680, 630, 168, 35, 6, 1; ...
References
- M. Bona, Combinatorics of Permutations, Chapman&Hall/CRC, Boca Raton, Florida, 2004.
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- E. Barcucci, A. Del Lungo and R. Pinzani, "Deco" polyominoes, permutations and random generation, Theoretical Computer Science, 159, 1996, 29-42.
- Olivier Bodini, Antoine Genitrini, and Mehdi Naima, Ranked Schröder Trees, arXiv:1808.08376 [cs.DS], 2018.
- Olivier Bodini, Antoine Genitrini, Cécile Mailler, and Mehdi Naima, Strict monotonic trees arising from evolutionary processes: combinatorial and probabilistic study, hal-02865198 [math.CO] / [math.PR] / [cs.DS] / [cs.DM], 2020.
- Colin Defant and James Propp, Quantifying Noninvertibility in Discrete Dynamical Systems, arXiv:2002.07144 [math.CO], 2020.
- Emeric Deutsch and W. P. Johnson, Create your own permutation statistics, Math. Mag., 77, 130-134, 2004.
Crossrefs
Programs
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GAP
Flat(List([1..11],n->Concatenation([1],List([1..n-1],k->Factorial(n)*k/Factorial(k+1))))); # Muniru A Asiru, Jun 10 2018
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Magma
A092582:= func< n,k | k eq n select 1 else k*Factorial(n)/Factorial(k+1) >; [A092582(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Sep 06 2022
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Mathematica
Drop[Drop[Abs[Map[Select[#, # < 0 &] &, Map[Differences, nn = 10; Range[0, nn]! CoefficientList[Series[(Exp[y x] - 1)/(1 - x), {x, 0, nn}], {x, y}]]]], 1], -1] // Grid (* Geoffrey Critzer, Jun 18 2017 *)
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PARI
{T(n, k) = if( n<1 || k>n, 0, k==n, 1, n! * k /(k+1)!)}; /* Michael Somos, Jun 25 2017 */
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SageMath
def A092582(n,k): return 1 if (k==n) else k*factorial(n)/factorial(k+1) flatten([[A092582(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Sep 06 2022
Formula
T(n, k) = n!*k/(k+1)! for k
Inverse of:
1;
-1, 1;
-1, -2, 1;
-1, -2, -3, 1;
-1, -2, -3, -4, 1;
... where A002260 = (1; 1,2; 1,2,3; ...). - Gary W. Adamson, Feb 22 2012
T(2n,n) = A092956(n-1) for n>0. - Alois P. Heinz, Jun 19 2017
From Alois P. Heinz, Dec 17 2021: (Start)
Sum_{k=1..n} k * T(n,k) = A002627(n).
|Sum_{k=1..n} (-1)^k * T(n,k)| = A055596(n) for n>=1. (End)
From G. C. Greubel, Sep 06 2022: (Start)
T(n, 1) = A001710(n).
T(n, 2) = 2*A001715(n) + [n=2]/3, n >= 2.
T(n, 3) = 3*A001720(n) + [n=3]/4, n >= 3.
T(n, 4) = 4*A001725(n) + [n=4]/5, n >= 4.
T(n, n-1) = A000027(n-1).
T(n, n-2) = A005563(n-1), n >= 3. (End)
Sum_{k=0..n} (k+1) * T(n,k) = A000522(n). - Alois P. Heinz, Apr 28 2023
A157385 A partition product of Stirling_1 type [parameter k = -5] with biggest-part statistic (triangle read by rows).
1, 1, 5, 1, 15, 30, 1, 105, 120, 210, 1, 425, 1800, 1050, 1680, 1, 3075, 18600, 18900, 10080, 15120, 1, 15855, 174300, 338100, 211680, 105840, 151200, 1, 123515, 2227680, 4865700, 4327680, 2540160, 1209600, 1663200, 1, 757755
Offset: 1
Comments
Links
- Peter Luschny, Counting with Partitions.
- Peter Luschny, Generalized Stirling_1 Triangles.
Crossrefs
Formula
T(n,0) = [n = 0] (Iverson notation) and for n > 0 and 1 <= m <= n
T(n,m) = Sum_{a} M(a)|f^a| where a = a_1,..,a_n such that
1*a_1+2*a_2+...+n*a_n = n and max{a_i} = m, M(a) = n!/(a_1!*..*a_n!),
f^a = (f_1/1!)^a_1*..*(f_n/n!)^a_n and f_n = product_{j=0..n-2}(j-n-3).
A049460 Generalized Stirling number triangle of first kind.
1, -5, 1, 30, -11, 1, -210, 107, -18, 1, 1680, -1066, 251, -26, 1, -15120, 11274, -3325, 485, -35, 1, 151200, -127860, 44524, -8175, 835, -45, 1, -1663200, 1557660, -617624, 134449, -17360, 1330, -56, 1, 19958400, -20355120, 8969148, -2231012, 342769, -33320, 2002, -68, 1
Offset: 0
Comments
a(n,m)= ^5P_n^m in the notation of the given reference with a(0,0) := 1.
The monic row polynomials s(n,x) := sum(a(n,m)*x^m,m=0..n) which are s(n,x)= product(x-(5+k),k=0..n-1), n >= 1 and s(0,x)=1 satisfy s(n,x+y) = sum(binomial(n,k)*s(k,x)*S1(n-k,y),k=0..n), with the Stirling1 polynomials S1(n,x)=sum(A008275(n,m)*x^m, m=1..n) and S1(0,x)=1.
In the umbral calculus (see the S. Roman reference given in A048854) the s(n,x) polynomials are called Sheffer for (exp(5*t),exp(t)-1).
Examples
{1}; {-5,1}; {30,-11,1}; {-210,107,-18,1}; ... s(2,x)= 30-11*x+x^2; S1(2,x)= -x+x^2 (Stirling1).
Links
- Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
- D. S. Mitrinovic, M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).
Crossrefs
Programs
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Haskell
a049460 n k = a049460_tabl !! n !! k a049460_row n = a049460_tabl !! n a049460_tabl = map fst $ iterate (\(row, i) -> (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 5) -- Reinhard Zumkeller, Mar 11 2014
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Mathematica
a[n_, m_] := Pochhammer[m+1, n-m] SeriesCoefficient[Log[1+x]^m/(1+x)^5, {x, 0, n}]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 29 2019 *)
Formula
a(n, m)= a(n-1, m-1) - (n+4)*a(n-1, m), n >= m >= 0; a(n, m) := 0, n
Triangle (signed) = [ -5, -1, -6, -2, -7, -3, -8, -4, -9, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, ...]; triangle (unsigned) = [5, 1, 6, 2, 7, 3, 8, 4, 9, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = f(n,i,5), for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008
Extensions
Second formula corrected by Philippe Deléham, Nov 10 2008
A143493 Unsigned 4-Stirling numbers of the first kind.
1, 4, 1, 20, 9, 1, 120, 74, 15, 1, 840, 638, 179, 22, 1, 6720, 5944, 2070, 355, 30, 1, 60480, 60216, 24574, 5265, 625, 39, 1, 604800, 662640, 305956, 77224, 11515, 1015, 49, 1, 6652800, 7893840, 4028156, 1155420, 203889, 22680, 1554, 60, 1, 79833600
Offset: 4
Comments
See A049459 for a signed version of the array. The unsigned 4-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1, 2, 3 and 4 belong to distinct cycles. This is the case r = 4 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143492 (r = 3). See A143496 for the corresponding triangle of 4-Stirling numbers of the second kind.
The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 4-Lah numbers see A143499.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^4,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-4*t),1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049459. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0: triangle T(n,k), read by rows, given by (4,1,5,2,6,3,7,4,8,5,9,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011
Examples
Triangle begins n\k| 4 5 6 7 8 9 ======================================== 4 | 1 5 | 4 1 6 | 20 9 1 7 | 120 74 15 1 8 | 840 638 179 22 1 9 | 6720 5944 2070 355 30 1 ... T(6,5) = 9. The 9 permutations of {1,2,3,4,5,6} with 5 cycles such that 1, 2, 3 and 4 belong to different cycles are: (1,5)(2)(3)(4)(6), (1,6)(2)(3)(4)(5), (2,5)(1)(3)(4)(6), (2,6)(1)(3)(4)(5), (3,5)(1)(2)(4)(6), (3,6)(1)(2)(4)(5), (4,5)(1)(2)(3)(6), (4,6)(1)(2)(3)(5) and (5,6)(1)(2)(3)(4).
Links
- Broder Andrei Z., The r-Stirling numbers, Discrete Math. 49, 241-259 (1984)
- A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
- Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
- Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).
- Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.
Crossrefs
Programs
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Maple
with combinat: T := (n, k) -> (n-4)! * add(binomial(n-j-1,3)*abs(stirling1(j,k-4))/j!,j = k-4..n-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
Formula
T(n,k) = (n-4)! * Sum_{j = k-4 .. n-4} C(n-j-1,3)*|stirling1(j,k-4)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 4, with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4.
Special cases:
T(n,4) = (n-1)!/3!.
T(n,5) = (n-1)!/3!*(1/4 + ... + 1/(n-1)).
T(n,k) = sum {4 <= i_1 < ...< i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(7,5) = Sum_{4 <= i < j < 7} (i*j) = 4*5 + 4*6 + 5*6 = 74.
Row g.f.: Sum_{k = 4..n} T(n,k)*x^k = x^4*(x+4)*(x+5)* ... *(x+n-1).
E.g.f. for column (k+4): Sum_{n = k..inf} T(n+4,k+4)*x^n/n! = 1/k!*1/(1-x)^4 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+4) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+4,k+4)*x^k*t^n/n! = 1 + (4+x)*t/1! + (20+9*x+x^2)*t^2/2! + .... This array is the matrix product St1 * P^3, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/4! ( A001720 ). The alternating row sums are (n-2)!/2!.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n+4,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008
A144698 Triangle of 4-Eulerian numbers.
1, 1, 4, 1, 13, 16, 1, 32, 113, 64, 1, 71, 531, 821, 256, 1, 150, 2090, 6470, 5385, 1024, 1, 309, 7470, 40510, 65745, 33069, 4096, 1, 628, 25191, 221800, 612295, 592884, 194017, 16384, 1, 1267, 81853, 1113919, 4835875, 7843369, 4915423, 1101157, 65536
Offset: 4
Comments
This is the case r = 4 of the r-Eulerian numbers, denoted by A(r;n,k), defined as follows. Let [n] denote the ordered set {1,2,...,n} and let r be a nonnegative integer. Let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that the permutation p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number A(r;n,k) is defined as the number of permutations in Permute(n,n-r) with k excedances. Thus the 4-Eulerian numbers are the number of permutations in Permute(n,n-4) with k excedances. For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144697 (r = 3) and A144699 (r = 5).
An alternative interpretation of the current array due to [Strosser] involves the 4-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapter 4, Section 3]). We define a permutation p in Permute(n,n-4) to have a 4-excedance at position i (1 <= i <= n-4) if p(i) >= i + 4.
Given a permutation p in Permute(n,n-4), define ~p to be the permutation in Permute(n,n-4) that takes i to n+1 - p(n-i-3). The map ~ is a bijection of Permute(n,n-4) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 4-excedance at position n-i-3. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 4-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries are the number of permutations in Permute(n,n-4) with k 4-excedances.
Example: Represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). In Permute(10,6) the permutation p = (1,2,4,10,3,6) does not have an excedance in the first two positions (i = 1 and 2) or in the final two positions (i = 5 and 6). The permutation ~p = (5,8,1,7,9,10) has 4-excedances only in the first two positions and the final two positions.
Examples
Triangle begins ===+============================================= n\k| 0 1 2 3 4 5 6 ===+============================================= 4 | 1 5 | 1 4 6 | 1 13 16 7 | 1 32 113 64 8 | 1 71 531 821 256 9 | 1 150 2090 6470 5385 1024 10 | 1 309 7470 40510 65745 33069 4096 ... T(6,1) = 13: We represent a permutation p:[n-4] -> [n] in Permute(n,n-4) by its image vector (p(1),...,p(n-4)). The 13 permutations in Permute(6,2) having 1 excedance are (1,3), (1,4), (1,5), (1,6), (3,2), (4,2), (5,2), (6,2), (2,1), (3,1), (4,1), (5,1) and (6,1).
References
- R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.
Links
- G. C. Greubel, Rows n = 4..54 of the triangle, flattened
- J. F. Barbero G., J. Salas, and E. J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013-2015.
- Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 9.
- Sergi Elizalde, Descents on quasi-Stirling permutations, arXiv:2002.00985 [math.CO], 2020.
- D. Foata and M. Schutzenberger, Théorie Géometrique des Polynômes Eulériens, arXiv:math/0508232 [math.CO], 2005; Lecture Notes in Math., no. 138, Springer Verlag, 1970.
- L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
- Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO], 2012. - From _N. J. A. Sloane_, Aug 21 2012
Crossrefs
Programs
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Magma
m:=4; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 04 2022
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Maple
with(combinat): T:= (n,k) -> 1/4!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4),j = 0..k): for n from 4 to 12 do seq(T(n,k),k = 0..n-4) end do;
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Mathematica
T[n_, k_] /; 0 < k <= n-4 := T[n, k] = (k+1) T[n-1, k] + (n-k) T[n-1, k-1]; T[, 0] = 1; T[, _] = 0; Table[T[n, k], {n, 4, 12}, {k, 0, n-4}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)
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SageMath
m=4 # A144698 def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) ) flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 04 2022
Formula
T(n,k) = (1/4!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-3)*(j+2)*(j+3)*(j+4).
T(n,n-k) = (1/4!)*Sum_{j = 4..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-3)*(j-1)*(j-2)*(j-3).
Recurrence relation:
T(n,k) = (k + 1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 4, T(4,k) = 0 for k >= 1. Special cases: T(n,n-4) = 4^(n-4); T(n,n-5) = 5^(n-3) - 4^(n-3) - (n-3)*4^(n-4).
E.g.f. (with suitable offsets): 1/4*[(1 - x)/(1 - x*exp(t - t*x))]^4 = 1/4 + x*t + (x + 4*x^2)*t^2/2! + (x + 13*x^2 + 16*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x + 1)*R_n(x) + x*(1 - x)*d/dx(R_n(x)) with R_4(x) = 1. It follows that the polynomials R_n(x) for n >= 5 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+3)-th row generating polynomial = (1/4!)*Sum_{k = 1..n} (k+3)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 4,
1/4*(x*d/dx)^(n-3) (1/(1-x)^4) = x/(1-x)^(n+1) * Sum_{k = 0..n-4} T(n,k)*x^k,
1/4*(x*d/dx)^(n-3) (x^4/(1-x)^4) = 1/(1-x)^(n+1) * Sum_{k = 4..n} T(n,n-k)*x^k,
1/(1-x)^(n+1) * Sum {k = 0..n-4} T(n,k)*x^k = (1/4!) * Sum_{m = 0..inf} (m+1)^(n-3)*(m+2)*(m+3)*(m+4)*x^m,
1/(1-x)^(n+1) * Sum {k = 4..n} T(n,n-k)*x^k = (1/4!) * Sum_{m = 4..inf} m^(n-3)*(m-1)*(m-2)*(m-3)*x^m,
Worpitzky-type identities:
Sum_{k = 0..n-4} T(n,k)*binomial(x+k,n) = (1/4!)*x^(n-3)*(x-1)*(x-2)*(x-3).
Sum_{k = 4..n} T(n,n-k)* binomial(x+k,n) = (1/4!)*(x+1)^(n-3)*(x+2)*(x+3)*(x+4).
Relation with Stirling numbers (Frobenius-type identities):
T(n+3,k-1) = (1/4!) * Sum_{j = 0..k} (-1)^(k-j)* (j+3)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+3,k-1) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+3)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 1 and
T(n+4,k) = 1/4! * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+4)!* binomial(n-j,k)*S(4;n+4,j+4) for n,k >= 0, where S(4;n,k) denotes the 4-Stirling numbers of the second kind A143496(n,k).
For n >=4, the shifted row polynomial t*R(n,t) = (1/4)*D^(n-3)(f(x,t)) evaluated at x = 0, where D is the operator (1-t)*(1+x)*d/dx and f(x,t) = (1+x*t/(t-1))^(-4). - Peter Bala, Apr 22 2012
A303697 Number T(n,k) of permutations p of [n] whose difference between sum of up-jumps and sum of down-jumps equals k; triangle T(n,k), n>=0, min(0,1-n)<=k<=max(0,n-1), read by rows.
1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 1, 4, 5, 4, 5, 4, 1, 1, 11, 19, 19, 20, 19, 19, 11, 1, 1, 26, 82, 100, 101, 100, 101, 100, 82, 26, 1, 1, 57, 334, 580, 619, 619, 620, 619, 619, 580, 334, 57, 1, 1, 120, 1255, 3394, 4339, 4420, 4421, 4420, 4421, 4420, 4339, 3394, 1255, 120, 1
Offset: 0
Comments
An up-jump j occurs at position i in p if p_{i} > p_{i-1} and j is the index of p_i in the increasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are larger than p_{i-1}. A down-jump j occurs at position i in p if p_{i} < p_{i-1} and j is the index of p_i in the decreasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are smaller than p_{i-1}. First index in the lists is 1 here.
Examples
Triangle T(n,k) begins: : 1 ; : 1 ; : 1, 0, 1 ; : 1, 1, 2, 1, 1 ; : 1, 4, 5, 4, 5, 4, 1 ; : 1, 11, 19, 19, 20, 19, 19, 11, 1 ; : 1, 26, 82, 100, 101, 100, 101, 100, 82, 26, 1 ; : 1, 57, 334, 580, 619, 619, 620, 619, 619, 580, 334, 57, 1 ;
Links
- Alois P. Heinz, Rows n = 0..125, flattened
Crossrefs
Programs
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Maple
b:= proc(u, o) option remember; expand(`if`(u+o=0, 1, add(b(u-j, o+j-1)*x^(-j), j=1..u)+ add(b(u+j-1, o-j)*x^( j), j=1..o))) end: T:= n-> (p-> seq(coeff(p, x, i), i=ldegree(p)..degree(p)))( `if`(n=0, 1, add(b(j-1, n-j), j=1..n))): seq(T(n), n=0..12);
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Mathematica
b[u_, o_] := b[u, o] = Expand[If[u+o == 0, 1, Sum[b[u-j, o+j-1] x^-j, {j, 1, u}] + Sum[b[u+j-1, o-j] x^j, {j, 1, o}]]]; T[0] = {1}; T[n_] := x^n Sum[b[j-1, n-j], {j, 1, n}] // CoefficientList[#, x]& // Rest; T /@ Range[0, 12] // Flatten (* Jean-François Alcover, Feb 20 2021, after Alois P. Heinz *)
A062260 Third (unsigned) column sequence of triangle A062140 (generalized a=4 Laguerre).
1, 21, 336, 5040, 75600, 1164240, 18627840, 311351040, 5448643200, 99891792000, 1917922406400, 38532804710400, 809188898918400, 17739910476288000, 405483668029440000, 9650511299100672000
Offset: 0
Links
Programs
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Magma
[Factorial(n+2)*Binomial(n+6, 6)/2: n in [0..30]]; // G. C. Greubel, Feb 06 2018
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Maple
a:=n->sum((n-j)*n!/6!, j=5..n): seq(a(n), n=6..21); # Zerinvary Lajos, Apr 29 2007
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Mathematica
Table[(n + 2)! Binomial[n + 6, 6]/2, {n, 0, 20}] (* Wesley Ivan Hurt, Jan 23 2017 *)
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PARI
{ f=1; for (n=0, 100, f*=n + 2; write("b062260.txt", n, " ", f*binomial(n + 6, 6)/2) ) } \\ Harry J. Smith, Aug 03 2009
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Sage
[binomial(n,6)*factorial (n-4)/2 for n in range(6, 22)] # Zerinvary Lajos, Jul 07 2009
Formula
E.g.f.: (1+12*x+15*x^2)/(1-x)^9.
a(n) = A062140(n+2, 2) = (n+2)!*binomial(n+6, 6)/2!.
If we define f(n,i,x) = Sum_{k=1..n} Sum_{j=i..k} binomial(k,j) * Stirling1(n,k) * Stirling2(j,i) * x^(k-j) then a(n-2) = (-1)^n * f(n,2,-7), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = binomial(n,6)*(n-4)!/2, n >= 6. - Zerinvary Lajos, Jul 07 2009
A144891 Lower triangular array called S1hat(5) related to partition number array A144890.
1, 5, 1, 30, 5, 1, 210, 55, 5, 1, 1680, 360, 55, 5, 1, 15120, 3630, 485, 55, 5, 1, 151200, 29820, 4380, 485, 55, 5, 1, 1663200, 321300, 39570, 5005, 485, 55, 5, 1, 19958400, 3225600, 421800, 43320, 5005, 485, 55, 5, 1, 259459200, 38808000, 4265100, 470550, 46445, 5005
Offset: 1
Comments
Examples
[1];[5,1];[30,5,1];[210,55,5,1];[1680,360,55,5,1];...
Links
- W. Lang, First 10 rows of the array and more.
- W. Lang, Combinatorial Interpretation of Generalized Stirling Numbers, J. Int. Seqs. Vol. 12 (2009) 09.3.3.
Crossrefs
A144892 (row sums).
A161742 Third left hand column of the RSEG2 triangle A161739.
1, 4, 13, 30, -14, -504, 736, 44640, -104544, -10644480, 33246720, 5425056000, -20843695872, -5185511654400, 23457840537600, 8506857655296000, -44092609863966720, -22430879475779174400, 130748316971139072000
Offset: 2
Crossrefs
Programs
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Maple
nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 2 to nmax do a(n):=sum(((-1)^k/((k+1)!*(k+2)!)) *(n!)*A028246(n,k+2)* A008955(k+1,k),k=0..n-2) od: seq(a(n),n=2..nmax);
A161743 Fourth left hand column of the RSEG2 triangle A161739.
1, 10, 73, 425, 1561, -2856, -73520, 380160, 15376416, -117209664, -7506967104, 72162155520, 7045087741056, -80246202992640, -11448278791372800, 149576169325363200, 30017051616972275712, -440857664887810867200
Offset: 3
Crossrefs
Programs
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Maple
nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 3 to nmax do a(n) := sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n,k+3)* A008955(k+2,k), k=0..n-3) od: seq(a(n),n=3..nmax);
Comments
Emeric Deutsch, Feb 23 2008