cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-30 of 38 results. Next

A062148 Second (unsigned) column sequence of triangle A062138 (generalized a=5 Laguerre).

Original entry on oeis.org

1, 14, 168, 2016, 25200, 332640, 4656960, 69189120, 1089728640, 18162144000, 319653734400, 5928123801600, 115598414131200, 2365321396838400, 50685458503680000, 1135354270482432000, 26538906072526848000, 646300418472124416000, 16372943934627151872000
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Examples

			a(3) = (3+1)! * binomial(3+6,6) = 24 * 84 = 2016. - _Indranil Ghosh_, Feb 24 2017
		

Crossrefs

Cf. A001725 (first column of A062138).
Appears in the third left hand column of A167556 multiplied by 120. - Johannes W. Meijer, Nov 12 2009

Programs

  • Magma
    [Factorial(n+1)*Binomial(n+6,6): n in [0..30]]; // G. C. Greubel, Feb 06 2018
  • Mathematica
    Table[Sum[n!/6!, {i, 6, n}], {n, 6, 21}] (* Zerinvary Lajos, Jul 12 2009 *)
  • PARI
    a(n)=(n+1)!*binomial(n+6,6) \\ Indranil Ghosh, Feb 24 2017
    
  • Python
    import math
    f=math.factorial
    def C(n,r):return f(n)/f(r)/f(n-r)
    def A062148(n): return f(n+1)*C(n+6,6) # Indranil Ghosh, Feb 24 2017
    

Formula

E.g.f.: (1+6*x)/(1-x)^8.
a(n) = A062138(n+1, 1) = (n+1)!*binomial(n+6, 6).
If we define f(n,i,x)= Sum_{k=i..n}(Sum_{j=i..k}(binomial(k,j) *Stirling1(n,k)* Stirling2(j,i)*x^(k-j))) then a(n-1) = (-1)^(n-1) * f(n,1,-7), (n>=1). - Milan Janjic, Mar 01 2009
Assuming offset 1: a(n) = n!*binomial(-n,6). - Peter Luschny, Apr 29 2016
From Amiram Eldar, Sep 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 5477/10 - 204*e - 6*gamma + 6*Ei(1), where gamma is Euler's constant (A001620) and Ei(1) is the exponential integral at 1 (A091725).
Sum_{n>=0} (-1)^n/a(n) = 403/10 - 120/e + 6*gamma - 6*Ei(-1), where -Ei(-1) is the negated exponential integral at -1 (A099285). (End)

A143498 Triangle of unsigned 3-Lah numbers.

Original entry on oeis.org

1, 6, 1, 42, 14, 1, 336, 168, 24, 1, 3024, 2016, 432, 36, 1, 30240, 25200, 7200, 900, 50, 1, 332640, 332640, 118800, 19800, 1650, 66, 1, 3991680, 4656960, 1995840, 415800, 46200, 2772, 84, 1, 51891840, 69189120, 34594560, 8648640, 1201200, 96096, 4368
Offset: 3

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062138. This is the case r = 3 of the unsigned r-Lah numbers L(r;n,k). The unsigned 3-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2 and 3 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143499 (r = 4).
The unsigned 3-Lah numbers are related to the 3-Stirling numbers: the lower triangular array of unsigned 3-Lah numbers may be expressed as the matrix product St1(3) * St2(3), where St1(3) = A143492 and St2(3) = A143495 are the arrays of 3-Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^4 * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277.

Examples

			Triangle begins
n\k|......3......4......5......6......7......8
==============================================
3..|......1
4..|......6......1
5..|.....42.....14......1
6..|....336....168.....24......1
7..|...3024...2016....432.....36......1
8..|..30240..25200...7200....900.....50......1
...
T(4,3) = 6. The partitions of {1,2,3,4} into 3 ordered lists, such that the elements 1, 2 and 3 lie in different lists, are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {2}{3}{1,4} and {2}{3}{4,1}.
		

Crossrefs

Cf. A001725 (column 3), A007318, A008275, A008277, A062138, A062148 - A062152 (column 4 to column 8), A062191 (alt. row sums), A062192 (row sums), A105278 (unsigned Lah numbers), A143492, A143495, A143497, A143499.

Programs

  • Magma
    /* As triangle */ [[Factorial(n-3)/Factorial(k-3)*Binomial(n+2, k+2): k in [3..n]]: n in [3.. 15]]; // Vincenzo Librandi, Nov 27 2018
  • Maple
    with combinat: T := (n, k) -> (n-3)!/(k-3)!*binomial(n+2,k+2): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;
  • Mathematica
    T[n_, k_] := (n-3)!/(k-3)!*Binomial[n+2, k+2]; Table[T[n, k], {n, 3, 10}, {k, 3, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

Formula

T(n,k) = (n-3)!/(k-3)!*binomial(n+2,k+2) for n,k >= 3.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 3, with the boundary conditions: T(n,k) = 0 if n < 3 or k < 3; T(3,3) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-3)! = 1/(k-3)!*t^k/(1-t)^(k+3) for k >= 3.
E.g.f: Sum_{n = 3..inf} Sum_{k = 3..n} T(n,k)*x^k*t^n/(n-3)! = (x*t)^3/(1-t)^6*exp(x*t/(1-t)) = (x*t)^3*(1 + (6+x)*t +(42+14*x+x^2)*t^2/2! + ... ).
Generalized Lah identity: (x+5)*(x+6)*...*(x+n+1) = Sum_{k = 3..n} T(n,k)*(x-1)*(x-2)*...*(x-k+3).
The polynomials 1/n!*Sum_{k = 3..n+3} T(n+3,k)*(-x)^(k-3) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,5,x). See A062138.
Array = A143492 * A143495 = abs(A008275) * ( A007318 )^4 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (6,14,24,36, ... ) on the main subdiagonal and zeros everywhere else.

A144699 Triangle of 5-Eulerian numbers.

Original entry on oeis.org

1, 1, 5, 1, 16, 25, 1, 39, 171, 125, 1, 86, 786, 1526, 625, 1, 181, 3046, 11606, 12281, 3125, 1, 372, 10767, 70792, 142647, 92436, 15625, 1, 755, 36021, 380071, 1279571, 1553145, 663991, 78125, 1, 1522, 116368, 1880494, 9818494, 19555438, 15519952, 4608946, 390625
Offset: 5

Views

Author

Peter Bala, Sep 19 2008

Keywords

Comments

This is the case r = 5 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 5-Eulerian numbers count the permutations in Permute(n,n-5) with k excedances. For other cases see A008292 (r = 0 and r = 1), A144696 (r = 2), A144697 (r = 3) and A144698 (r = 4).
An alternative interpretation of the current array due to [Strosser] involves the 5-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-5) to have a 5-excedance at position i (1 <=i <= n-5) if p(i) >= i + 5.
Given a permutation p in Permute(n,n-5), define ~p to be the permutation in Permute(n,n-5) that takes i to n+1 - p(n-i-4). The map ~ is a bijection of Permute(n,n-5) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 5-excedance at position n-i-4. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 5-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-5) with k 5-excedances.
Example: Represent a permutation p:[n-5] -> [n] in Permute(n,n-5) by its image vector (p(1),...,p(n-5)). In Permute(12,7) the permutation p = (1,2,4,12,3,6,5) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 5, 6 and 7). The permutation ~p = (8,7,10,1,9,11,12) has 5-excedances only in the first three positions and the final two positions.

Examples

			Triangle begins
=======================================================
n\k|..0.......1......2......3.........4.......5.......6
=======================================================
5..|..1
6..|..1.......5
7..|..1......16......25
8..|..1......39.....171.....125
9..|..1......86.....786....1526.....625
10.|..1.....181....3046...11606...12281....3125
11.|..1.....372...10767...70792..142647...92436...15625
...
T(7,1) = 16: We represent a permutation p:[n-5] -> [n] in Permute(n,n-5) by its image vector (p(1),...,p(n-5)). The 16 permutations in Permute(7,2) having 1 excedance are (1,3), (1,4), (1,5), (1,6), (1,7), (3,2), (4,2), (5,2), (6,2), (7,2), (2,1), (3,1), (4,1), (5,1), (6,1) and (7,1).
		

References

  • R. Strosser, Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.

Crossrefs

Programs

  • Magma
    m:=5; [(&+[(-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 08 2022
    
  • Maple
    with(combinat):
    T:= (n,k) -> 1/5!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-4)*(j+2)*(j+3)*(j+4)*(j+5),j = 0..k):
    for n from 5 to 13 do
    seq(T(n,k),k = 0..n-5)
    end do;
  • Mathematica
    T[n_, k_] /; 0 < k <= n-5 := 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, 5, 13}, {k, 0, n-5}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)
  • SageMath
    m=5 # A144699
    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 08 2022

Formula

T(n,k) = (1/5!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-4)*(j+2)*(j+3)*(j+4)*(j+5).
T(n,n-k) = (1/5!)*Sum_{j = 5..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-4)*(j-1)*(j-2)*(j-3)*(j-4).
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 >= 5, T(5,k) = 0 for k >= 1.
E.g.f. (with suitable offsets): (1/5)*((1 - x)/(1 - x*exp(t - t*x)))^5 = 1/5 + x*t + (x + 5*x^2)*t^2/2! + (x + 16*x^2 + 25*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_5(x) = 1. It follows that the polynomials R_n(x) for n >= 6 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+4)-th row generating polynomial = (1/5!)*Sum_{k = 1..n} (k+4)!*Stirling2(n,k)*x^(k-1)*(1-x)^(n-k).
For n >= 5,
(1/5)*(x*d/dx)^(n-4) (1/(1-x)^5) = (x/(1-x)^(n+1)) * Sum_{k = 0..n-5} T(n,k)*x^k,
(1/5)*(x*d/dx)^(n-4) (x^5/(1-x)^5) = (1/(1-x)^(n+1)) * Sum_{k = 5..n} T(n,n-k)*x^k,
(1/(1-x)^(n+1)) * Sum_{k = 0..n-5} T(n,k)*x^k = (1/5!) * Sum_{m = 0..infinity} (m+1)^(n-4)*(m+2)*(m+3)*(m+4)*(m+5)*x^m,
(1/(1-x)^(n+1)) * Sum_{k = 5..n} T(n,n-k)*x^k = (1/5!) * Sum_{m = 5..infinity} m^(n-4)*(m-1)*(m-2)*(m-3)*(m-4)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-5} T(n,k)*binomial(x+k,n) = (1/5!)*x^(n-4)*(x-1)*(x-2)*(x-3)*(x-4).
Sum_{k = 5..n} T(n,n-k)* binomial(x+k,n) = (1/5!)*(x+1)^(n-4)*(x+2)*(x+3)*(x+4)*(x+5).
Relation with Stirling numbers (Frobenius-type identities):
T(n+4,k-1) = (1/5!) * Sum_{j = 0..k} (-1)^(k-j)* (j+4)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+4,k-1) = (1/5!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+4)!* binomial(n-j,k)*S(5;n+5,j+5) for n,k >= 0 and
T(n+5,k) = (1/5!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+5)!* binomial(n-j,k)*S(5;n+5,j+5) for n,k >= 0, where S(5;n,k) denotes the 5-Stirling numbers of the second kind, which are given by the formula S(5;n+5,j+5) = 1/j!*Sum_{i = 0..j} (-1)^(j-i)*binomial(j,i)*(i+5)^n for n,j >= 0.

A176732 a(n) = (n+5)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 6, 43, 356, 3333, 34754, 398959, 4996032, 67741129, 988344062, 15434831091, 256840738076, 4536075689293, 84731451264186, 1668866557980343, 34563571477305464, 750867999393119889, 17072113130285524982, 405423357986250112699, 10037458628015142154452, 258639509502117306002581
Offset: 0

Views

Author

Wolfdieter Lang, Jul 14 2010

Keywords

Comments

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=6 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A001725(n+5) = (n+5)!/5!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010).

Examples

			Necklaces and 6 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c6(1), (binomial(4,2)*2)*c6(2), and 1*c6(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c6(n):=A001725(n+5) numbers for the pure 6-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=6: 1/(1-x)^6). This adds up as 9 + 4*2*6 + (6*1)*42 + 3024 = 3333 = a(4).
		

Crossrefs

Cf. A000153, A000261, A001909, A001910 (necklaces and k=5 cords), A176732.

Programs

  • Maple
    a := n -> hypergeom([-n,7],[],1)*(-1)^n:
    seq(simplify(a(n)),n=0..9); # Peter Luschny, Apr 25 2015
  • Mathematica
    Rest[RecurrenceTable[{a[0]==1,a[-1]==0,a[n]==(n+5)a[n-1]+(n-1)a[n-2]},a,{n,20}]] (* Harvey P. Dale, Oct 01 2012 *)

Formula

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^6) = exp(-x)/(1-x)^7, equivalent to the recurrence.
a(n) = A086764(n+6,6).
a(n) = A090010(n), n>0. - R. J. Mathar, Jul 22 2010
a(n) = (-1)^n*hypergeom([-n,7],[],1). - Peter Luschny, Apr 25 2015

A167569 The lower left triangle of the ED2 array A167560.

Original entry on oeis.org

1, 2, 4, 6, 16, 32, 24, 80, 192, 384, 120, 480, 1344, 3072, 6144, 720, 3360, 10752, 27648, 61440, 122880, 5040, 26880, 96768, 276480, 675840, 1474560, 2949120, 40320, 241920, 967680, 3041280, 8110080, 19169280, 41287680, 82575360
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

We discovered that the numbers that appear in the lower left triangle of the ED2 array A167560 (m <= n) behave in a regular way, see the formula below. This rather simple regularity doesn't show up in the upper right triangle of the ED2 array (m > n).

Examples

			The first few triangle rows are:
[1]
[2, 4]
[6, 16, 32]
[24, 80, 192, 384]
[120, 480, 1344, 3072, 6144]
[720, 3360, 10752, 27648, 61440, 122880]
		

Crossrefs

A167560 is the ED2 array.
A047053, 2*A034177 and A167570 are the first three right hand triangle columns.
A000142, 4*A001715, 32*A001725, 384* A049388 and 6144* A049398 are the first five left hand triangle columns.
A167571 equals the row sums.

Programs

  • Maple
    a := proc(n, m): 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Flatten[Table[4^(m - 1)*(m - 1)!*(n + m - 1)!/(2*m - 1)!, {n, 1, 50}, {m, n}]] (* G. C. Greubel, Jun 16 2016 *)

Formula

a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)!.

A145357 Lower triangular array, called S1hat(6), related to partition number array A145356.

Original entry on oeis.org

1, 6, 1, 42, 6, 1, 336, 78, 6, 1, 3024, 588, 78, 6, 1, 30240, 6804, 804, 78, 6, 1, 332640, 62496, 8316, 804, 78, 6, 1, 3991680, 753984, 85176, 9612, 804, 78, 6, 1, 51891840, 8273664, 1021608, 94248, 9612, 804, 78, 6, 1, 726485760, 109118016, 11394432, 1157688, 102024
Offset: 1

Views

Author

Wolfdieter Lang, Oct 17 2008

Keywords

Comments

If in the partition array M31hat(6):=A145356 entries belonging to partitions with the same parts number m are summed one obtains this triangle of numbers S1hat(6). In the same way the signless Stirling1 triangle |A008275| is obtained from the partition array M_2 = A036039.
The first columns are A001725(n+4), A145359, A145360,...

Examples

			Triangle begins:
  [1];
  [6,1];
  [42,6,1];
  [336,78,6,1];
  [3024,588,78,6,1];
  ...
		

Crossrefs

Cf. A145358 (row sums).

Formula

a(n,m) = sum(product(|S1(6;j,1)|^e(n,m,q,j),j=1..n),q=1..p(n,m)) if n>=m>=1, else 0. Here p(n,m)=A008284(n,m), the number of m parts partitions of n and e(n,m,q,j) is the exponent of j in the q-th m part partition of n. |S1(6,n,1)|= A049374(n,1) = A001725(n+4) = (n+4)!/5!.

A161742 Third left hand column of the RSEG2 triangle A161739.

Original entry on oeis.org

1, 4, 13, 30, -14, -504, 736, 44640, -104544, -10644480, 33246720, 5425056000, -20843695872, -5185511654400, 23457840537600, 8506857655296000, -44092609863966720, -22430879475779174400, 130748316971139072000
Offset: 2

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Crossrefs

Equals third left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A129825 and A161743.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.
A001710 (n!/2!), A001715 (n!/3!), A001720 (n!/4!), A001725 (n!/5!), A001730 (n!/6!), A049388 (n!/7!), A049389 (n!/8!), A049398 (n!/9!), A051431 (n!/10!) appear in Maple program.

Programs

  • 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);

Formula

a(n) = sum(((-1)^k/((k+1)!*(k+2)!))*(n!)*A028246(n, k+2)*A008955(k+1, k), k=0..n-2)

A161743 Fourth left hand column of the RSEG2 triangle A161739.

Original entry on oeis.org

1, 10, 73, 425, 1561, -2856, -73520, 380160, 15376416, -117209664, -7506967104, 72162155520, 7045087741056, -80246202992640, -11448278791372800, 149576169325363200, 30017051616972275712, -440857664887810867200
Offset: 3

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Crossrefs

Equals fourth left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A129825 and A161742.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.
A001710 (n!/2!), A001715 (n!/3!), A001720 (n!/4!), A001725 (n!/5!), A001730 (n!/6!), A049388 (n!/7!), A049389 (n!/8!), A049398 (n!/9!), A051431 (n!/10!) appear in Maple program.

Programs

  • 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);

Formula

a(n) = sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n, k+3)*A008955(k+2, k), k = 0..n-3).

A162995 A scaled version of triangle A162990.

Original entry on oeis.org

1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2009

Keywords

Comments

We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
T(n,k) = A173333(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Feb 19 2010
T(n,k) = A094587(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

Examples

			The first few rows of the triangle are:
[1]
[3, 1]
[12, 4, 1]
[60, 20, 5, 1]
		

Crossrefs

Cf. A094587.
A056542(n) equals the row sums for n>=1.
A001710, A001715, A001720, A001725, A001730, A049388, A049389, A049398, A051431 are related to the left hand columns.
A000012, A009056, A002378, A007531, A052762, A052787, A053625 and A159083 are related to the right hand columns.

Programs

  • Haskell
    a162995 n k = a162995_tabl !! (n-1) !! (k-1)
    a162995_row n = a162995_tabl !! (n-1)
    a162995_tabl = map fst $ iterate f ([1], 3)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
  • Maple
    a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)

Formula

a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.

A094646 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -2, 1, 2, -3, 1, 0, 2, -3, 1, 0, 2, -1, -2, 1, 0, 4, 0, -5, 0, 1, 0, 12, 4, -15, -5, 3, 1, 0, 48, 28, -56, -35, 7, 7, 1, 0, 240, 188, -252, -231, 0, 42, 12, 1, 0, 1440, 1368, -1324, -1638, -231, 252, 114, 18, 1, 0, 10080, 11016, -7900, -12790, -3255, 1533, 1050, 240, 25, 1
Offset: 0

Views

Author

Vladeta Jovovic, May 17 2004

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows, given by [ -2, 1, -1, 2, 0, 3, 1, 4, 2, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 23 2006
From Wolfdieter Lang, Jun 23 2011: (Start)
The row polynomials s(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy risefac(x-2,n)=s(n,x), with the rising factorials risefac(x-2,n):=Product_{j=0..n-1} (x-2+j), n >= 1, risefac(x-2,0)=1. Compare this with the formula risefac(x,n)=|S1|(n,x), with the row polynomials |S1|(n,x) of A132393 (unsigned Stirling1).
This is the third triangle of an a-family of Sheffer arrays, call them |S1|(a), with e.g.f. of the row polynomials |S1|(a;x;z) = ((1-z)^a)*exp(-x*log(1-z)). In the notation showing the column e.g.f.s this is Sheffer ((1-z)^a,-log(1-z)). In the umbral notation (see the Roman reference, given under A094645) this is called Sheffer for (exp(a*t),1-exp(-t)). For a=0 this becomes the unsigned Stirling1 triangle |S1|(0) = A132393 with row polynomials |S1|(0;n,x) =: s1(n,x).
E.g.f. column number k (with leading zeros): ((1-x)^a)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is (1-x)^(a-1), and the e.g.f. for the alternating row sums is (1-x)^(a+1).
Row polynomial recurrence:
|S1|(a;n,x)=(x+(n-1-a))*|S1|(a;n-1,m), |S1|(a;0,x)=1.
Meixner identity (see the reference under A060338):
|S1|(a;n,x) - |S1|(a;n,x-1) = n*|S1|(a;n-1,x), n >= 1,
Also (from the corollary 3.7.2 on p. 50 of the Roman reference): |S1|(a;n,x) = (x-a)*|S1|(a;n-1,x+1), n >= 1.
Recurrence: |S1|(a;n,k) = |S1|(a;n-1,k-1) + (n-(a+1))*|S1|(a;n-1,k); |S1|(a;n,k)=0 if n < m, |S1|(a;n,-1)=0, |S1|(a;0,0)=1.
Connection to |Stirling1|=|S1|(0):
|S1|(a;n,k) = Sum_{p=0..a} |S1|(a;a,p)*abs(Stirling1(n-a,k-p)), n >= a.
The exponential convolution identity is
|S1|(a;n,x+y) = Sum_{k=0..n} binomial(n,k)*|S1|(a;k,y)*s1(n-k,x), n >= 0, with symmetry x <-> y.
The Sheffer a- and z-sequences are (see the W. Lang link under A006232): Sha(a;n)=A164555(n)/A027642(n) (independent of a) with e.g.f. x/(1-exp(-x)), and the z-sequence has e.g.f. (exp(a*x)-1)/(exp(-x)-1).
The inverse Sheffer matrix has e.g.f. exp(a*z)*exp(x*(1-exp(-z))), in short notation (exp(a*z),1-exp(-z)),
(or in umbral notation ((1-t)^a,-log(1-t))).
(End)

Examples

			Triangle begins
   1;
  -2,  1;
   2, -3,  1;
   0,  2, -3,  1;
   0,  2, -1, -2,  1;
   0,  4,  0, -5,  0,  1;
   ...
risefac(x-2,3) = (x-2)*(x-1)*x = 2*x-3*x^2+x^3.
-1 = T(4,2) = T(3,1) + 1*T(3,2) =  2 + (-3).
T(4,3) = 2*abs(S1(2,3)) - 3*abs(S1(2,2)) + 1*abs(S1(2,1)) = 2*0 - 3*1 + 1*1 = -2.
		

Crossrefs

Programs

  • Maple
    A094646_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x-n+3, n)), x, k), k=0..n): seq(print(A094646_row(n)), n = 0..6); # Peter Luschny, May 16 2013
  • Mathematica
    Flatten[ Table[ CoefficientList[ Pochhammer[x-2, n], x], {n, 0, 10}]] (* Jean-François Alcover, Sep 26 2011 *)

Formula

E.g.f.: (1-y)^(2-x).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 respectively. - Philippe Deléham, Nov 13 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then |T(n,i)| = |f(n,i,-2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Wolfdieter Lang, Jun 23 2011: (Start)
risefac(x-2,n) = Sum_{k=0..n} T(n,k)*x^k, n >= 0, with the rising factorials (see a comment above).
Recurrence: T(n,k) = T(n-1,k-1) + (n-3)*T(n-1,k); T(n,k)=0 if n < m, T(n,-1)=0, T(0,0)=1.
T(n,k) = 2*abs(S1(n-2,k)) - 3*abs(S1(n-2,k-1)) + abs(S1(n-2,k-2)), n >= 2, with S1(n,k) = Stirling1(n,k) = A048994(n,k).
E.g.f. column number k (with leading zeros):
((1-x)^2)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is 1-x, i.e., [1,-1,0,0,...],
and the e.g.f. for the alternating row sums is (1-x)^3. i.e., [1,-3,3,1,0,0,...]. (End)
Previous Showing 21-30 of 38 results. Next