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.

Showing 1-6 of 6 results.

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 10 2007, Oct 15 2008, Oct 17 2008

Keywords

Comments

Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
A094645*A007318 as infinite lower triangular matrices.
Row sums are the factorial numbers. - Roger L. Bagula, Apr 18 2008
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
From Wolfdieter Lang, Jul 20 2017: (Start)
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019

Examples

			Triangle T(n,k) begins:
  1;
  0,    1;
  0,    1,     1;
  0,    2,     3,     1;
  0,    6,    11,     6,    1;
  0,   24,    50,    35,   10,    1;
  0,  120,   274,   225,   85,   15,   1;
  0,  720,  1764,  1624,  735,  175,  21,  1;
  0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
  ...
---------------------------------------------------
Production matrix is
  0, 1
  0, 1, 1
  0, 1, 2,  1
  0, 1, 3,  3,  1
  0, 1, 4,  6,  4,  1
  0, 1, 5, 10, 10,  5,  1
  0, 1, 6, 15, 20, 15,  6, 1
  0, 1, 7, 21, 35, 35, 21, 7, 1
  ...
From _Wolfdieter Lang_, May 09 2017: (Start)
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - _Wolfdieter Lang_, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - _Wolfdieter Lang_, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150

Crossrefs

Essentially a duplicate of A048994. Cf. A008275, A008277, A112007, A130534, A288874, A354795.

Programs

  • Haskell
    a132393 n k = a132393_tabl !! n !! k
    a132393_row n = a132393_tabl !! n
    a132393_tabl = map (map abs) a048994_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x,n)),x,k),k=0..n) end: # Peter Luschny, Nov 28 2010
  • Mathematica
    p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
    Flatten[Table[Abs[StirlingS1[n,i]],{n,0,10},{i,0,n}]] (* Harvey P. Dale, Feb 04 2014 *)
  • Maxima
    create_list(abs(stirling1(n,k)),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    column(n,k) = my(v1, v2); v1 = vector(n-1, i, 0); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, v1[i] = (i+k)*(i+k-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+k)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 \\ generates n first elements of the k-th column starting from the first nonzero element. - Mikhail Kurkov, Mar 05 2025

Formula

T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
From Wolfdieter Lang, Feb 21 2017: (Start)
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
From Wolfdieter Lang, May 28 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1)}_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1.
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link). (End)

A271703 Triangle read by rows: the unsigned Lah numbers T(n, k) = binomial(n-1, k-1)*n!/k! if n > 0 and k > 0, T(n, 0) = 0^n and otherwise 0, for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 6, 1, 0, 24, 36, 12, 1, 0, 120, 240, 120, 20, 1, 0, 720, 1800, 1200, 300, 30, 1, 0, 5040, 15120, 12600, 4200, 630, 42, 1, 0, 40320, 141120, 141120, 58800, 11760, 1176, 56, 1, 0, 362880, 1451520, 1693440, 846720, 211680, 28224, 2016, 72, 1
Offset: 0

Views

Author

Peter Luschny, Apr 14 2016

Keywords

Comments

The Lah numbers can be seen as the case m=1 of the family of triangles T_{m}(n,k) = T_{m}(n-1,k-1)+(k^m+(n-1)^m)*T_{m}(n-1,k) (see the link 'Partition transform').
This is the Sheffer triangle (lower triangular infinite matrix) (1, x/(1-x)), an element of the Jabotinsky subgroup of the Sheffer group. - Wolfdieter Lang, Jun 12 2017

Examples

			As a rectangular array (diagonals of the triangle):
  1,      1,       1,       1,       1,       1,       ... A000012
  0,      2,       6,       12,      20,      30,      ... A002378
  0,      6,       36,      120,     300,     630,     ... A083374
  0,      24,      240,     1200,    4200,    11760,   ... A253285
  0,      120,     1800,    12600,   58800,   211680,  ...
  0,      720,     15120,   141120,  846720,  3810240, ...
A000007, A000142, A001286, A001754, A001755,  A001777.
The triangle T(n,k) begins:
n\k 0       1        2        3        4       5      6     7    8  9 10 ...
0:  1
1:  0       1
2:  0       2        1
3:  0       6        6        1
4:  0      24       36       12        1
5:  0     120      240      120       20       1
6:  0     720     1800     1200      300      30      1
7:  0    5040    15120    12600     4200     630     42     1
8:  0   40320   141120   141120    58800   11760   1176    56    1
9:  0  362880  1451520  1693440   846720  211680  28224  2016   72  1
10: 0 3628800 16329600 21772800 12700800 3810240 635040 60480 3240 90  1
...  - _Wolfdieter Lang_, Jun 12 2017
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., pp. 312, 552.
  • I. Lah, Eine neue Art von Zahlen, ihre Eigenschaften und Anwendung in der mathematischen Statistik, Mitt.-Bl. Math. Statistik, 7:203-213, 1955.
  • T. Mansour, M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, CRC Press, 2016

Crossrefs

Variants: A008297 the main entry for these numbers, A105278, A111596 (signed).
A000262 (row sums). Largest number of the n-th row in A002868.

Programs

  • Maple
    T := (n, k) -> `if`(n=k, 1, binomial(n-1,k-1)*n!/k!):
    seq(seq(T(n, k), k=0..n), n=0..9);
  • Mathematica
    T[n_, k_] := Binomial[n, k]*FactorialPower[n-1, n-k];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2017 *)
  • Sage
    @cached_function
    def T(n,k):
        if k<0 : return 0
        if k==n: return 1
        return T(n-1,k-1) + (k+n-1)*T(n-1,k)
    for n in (0..8): print([T(n,k) for k in (0..n)])

Formula

For a collection of formulas see the 'Lah numbers' link.
T(n, k) = A097805(n, k)*n!/k! = (-1)^k*P_{n, k}(1,1,1,...) where P_{n, k}(s) is the partition transform of s.
T(n, k) = coeff(n! * P(n), x, k) with P(n) = (1/n)*(Sum_{k=0..n-1}(x(n-k)*P(k))), for n >= 1 and P(n=0) = 1, with x(n) = n*x. See A036039. - Johannes W. Meijer, Jul 08 2016
From Wolfdieter Lang, Jun 12 2017: (Start)
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (that is egf of the triangle) is exp(x*t/(1-t)) (a Sheffer triangle of the Jabotinsky type).
E.g.f. column k: (t/(1-t))^k/k!.
Three term recurrence: T(n, k) = T(n-1, k-1) + (n-1+k)*T(n, k-1), n >= 1, k = 0..n, with T(0, 0) =1, T(n, -1) = 0, T(n, k) = 0 if n < k.
T(n, k) = binomial(n, k)*fallfac(x=n-1, n-k), with fallfac(x, n) = Product_{j=0..(n-1)} (x - j), for n >= 1, and 0 for n = 0.
risefac(x, n) = Sum_{k=0..n} T(n, k)*fallfac(k), with risefac(x, n) = Product_{j=0..(n-1)} (x + j), for n >= 1, and 0 for n = 0.
See Graham et al., exercise 31, p. 312, solution p. 552. (End)
Formally, let f_n(x) = Sum_{k>n} (k-1)!*x^k; then f_n(x) = Sum_{k=0..n} T(n, k)* x^(n+k)*f_0^((k))(x), where ^((k)) stands for the k-th derivative. - Luc Rousseau, Dec 27 2020
T(n, k) = Sum_{j=k..n} A354795(n, j)*A360177(j, k). - Mélika Tebni, Feb 02 2023
T(n, k) = binomial(n, k)*(n-1)!/(k-1)! for n, k > 0. - Chai Wah Wu, Nov 30 2023

A008296 Triangle of Lehmer-Comtet numbers of the first kind.

Original entry on oeis.org

1, 1, 1, -1, 3, 1, 2, -1, 6, 1, -6, 0, 5, 10, 1, 24, 4, -15, 25, 15, 1, -120, -28, 49, -35, 70, 21, 1, 720, 188, -196, 49, 0, 154, 28, 1, -5040, -1368, 944, 0, -231, 252, 294, 36, 1, 40320, 11016, -5340, -820, 1365, -987, 1050, 510, 45, 1, -362880, -98208, 34716, 9020, -7645, 3003, -1617, 2970, 825, 55, 1, 3628800
Offset: 1

Views

Author

Keywords

Comments

Triangle arising in the expansion of ((1+x)*log(1+x))^n.
Also the Bell transform of (-1)^(n-1)*(n-1)! if n>1 else 1 adding 1,0,0,0,... as column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 16 2016

Examples

			Triangle begins:
   1;
   1,  1;
  -1,  3,   1;
   2, -1,   6,  1;
  -6,  0,   5, 10,  1;
  24,  4, -15, 25, 15, 1;
  ...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 139.

Crossrefs

Cf. A039621 (second kind), A354795 (variant), A185164, A005727 (row sums), A298511 (central).
Columns: A045406 (column 2), A347276 (column 3), A345651 (column 4).
Diagonals: A000142, A000217, A059302.
Cf. A176118.

Programs

  • Maple
    for n from 1 to 20 do for k from 1 to n do
    printf(`%d,`, add(binomial(l,k)*k^(l-k)*Stirling1(n,l), l=k..n)) od: od:
    # second program:
    A008296 := proc(n, k) option remember; if k=1 and n>1 then (-1)^n*(n-2)! elif n=k then 1 else (n-1)*procname(n-2, k-1) + (k-n+1)*procname(n-1, k) + procname(n-1, k-1) end if end proc:
    seq(print(seq(A008296(n, k), k=1..n)), n=1..7); # Mélika Tebni, Aug 22 2021
  • Mathematica
    a[1, 1] = a[2, 1] = 1; a[n_, 1] = (-1)^n (n-2)!;
    a[n_, n_] = 1; a[n_, k_] := a[n, k] = (n-1) a[n-2, k-1] + a[n-1, k-1] + (k-n+1) a[n-1,k]; Flatten[Table[a[n, k], {n, 1, 12}, {k, 1, n}]][[1 ;; 67]]
    (* Jean-François Alcover, Apr 29 2011 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, n! * polcoeff(((1 + x) * log(1 + x + x * O(x^n)))^k / k!, n))}; /* Michael Somos, Nov 15 2002 */
    
  • Sage
    # uses[bell_matrix from A264428]
    # Adds 1, 0, 0, 0, ... as column 0 at the left side of the triangle.
    bell_matrix(lambda n: (-1)^(n-1)*factorial(n-1) if n>1 else 1, 7) # Peter Luschny, Jan 16 2016

Formula

E.g.f. for a(n, k): (1/k!)[ (1+x)*log(1+x) ]^k. - Len Smiley
Left edge is (-1)*n!, for n >= 2. Right edge is all 1's.
a(n+1, k) = n*a(n-1, k-1) + a(n, k-1) + (k-n)*a(n, k).
a(n, k) = Sum_{m} binomial(m, k)*k^(m-k)*Stirling1(n, m).
From Peter Bala, Mar 14 2012: (Start)
E.g.f.: exp(t*(1 + x)*log(1 + x)) = Sum_{n>=0} R(n,t)*x^n/n! = 1 + t*x + (t+t^2)x^2/2! + (-t+3*t^2+t^3)x^3/3! + .... Cf. A185164. The row polynomials R(n,t) are of binomial type and satisfy the recurrence R(n+1,t) = (t-n)*R(n,t) + t*d/dt(R(n,t)) + n*t*R(n-1,t) with R(0,t) = 1 and R(1,t) = t. Inverse array is A039621.
(End)
Sum_{k=0..n} (-1)^k * a(n,k) = A176118(n). - Alois P. Heinz, Aug 25 2021

Extensions

More terms from James Sellers, Jan 26 2001
Edited by N. J. A. Sloane at the suggestion of Andrew Robbins, Dec 11 2007

A354794 Triangle read by rows. The Bell transform of the sequence {m^m | m >= 0}.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 4, 3, 1, 0, 27, 19, 6, 1, 0, 256, 175, 55, 10, 1, 0, 3125, 2101, 660, 125, 15, 1, 0, 46656, 31031, 9751, 1890, 245, 21, 1, 0, 823543, 543607, 170898, 33621, 4550, 434, 28, 1, 0, 16777216, 11012415, 3463615, 688506, 95781, 9702, 714, 36, 1
Offset: 0

Views

Author

Peter Luschny, Jun 09 2022

Keywords

Comments

For the definition of the Bell transform see A264428. The Bell transform of {(-m)^m | m >= 0} is A039621. The numbers A039621(n, k) are known as the Lehmer-Comtet numbers of 2nd kind. We think it is more natural to use Bell_{n, k}({m^m}) as the basis for the definition (and let the triangle start at (0, 0)).

Examples

			Triangle T(n, k) begins:
[0] 1;
[1] 0,        1;
[2] 0,        1,        1;
[3] 0,        4,        3,       1;
[4] 0,       27,       19,       6,      1;
[5] 0,      256,      175,      55,     10,     1;
[6] 0,     3125,     2101,     660,    125,    15,    1;
[7] 0,    46656,    31031,    9751,   1890,   245,   21,   1;
[8] 0,   823543,   543607,  170898,  33621,  4550,  434,  28,  1;
[9] 0, 16777216, 11012415, 3463615, 688506, 95781, 9702, 714, 36, 1;
		

References

  • Louis Comtet, Advanced Combinatorics. Reidel, Dordrecht, 1974, p. 139-140.

Crossrefs

Cf. A264428, A039621 (signed variant), A195979 (row sums), A000312 (column 1), A045531 (column 2), A281596 (column 3), A281595 (column 4), A000217 (diagonal 1), A215862 (diagonal 2), A354795 (matrix inverse), A137452 (Abel).

Programs

  • Maple
    T := (n, k) -> if n = k then 1 else
    add((-1)^j*(n-j-1)^(n-1)/(j!*(k-1-j)!), j = 0.. k-1) fi:
    seq(seq(T(n, k), k = 0..n), n = 0..9);
    # Alternatively, using the function BellMatrix from A264428:
    BellMatrix(n -> n^n, 9);
    # Or by recursion:
    R := proc(n, k, m) option remember;
       if k < 0 or n < 0 then 0 elif k = 0 then 1 else
       m*R(n, k-1, m) + R(n-1, k, m+1) fi end:
    A039621 := (n, k) -> ifelse(n = 0, 1, R(k-1, n-k, n-k)):
  • Mathematica
    Unprotect[Power]; Power[0, 0] = 1; pow[n_] := n^n;
    R = Range[0, 9]; T[n_, k_] := BellY[n, k, pow[R]];
    Table[T[n, k], {n, R}, {k, 0, n}] // Flatten
  • Python
    from functools import cache
    @cache
    def t(n, k, m):
        if k < 0 or n < 0: return 0
        if k == 0: return n ** k
        return m * t(n, k - 1, m) + t(n - 1, k, m + 1)
    def A354794(n, k): return t(k - 1, n - k, n - k) if n != k else 1
    for n in range(9): print([A354794(n, k) for k in range(n + 1)])

Formula

T(n, k) = Bell_{n, k}(A000312), where Bell_{n, k} is the partial Bell polynomial evaluated over the powers m^m (with 0^0 = 1). See the Mathematica program.
T(n, k) = Sum_{j=0..k-1} (-1)^j*(n-j-1)^(n - 1)/(j! * (k-1-j)!) for 0 <= k < n and T(n, n) = 1.
T(n, k) = r(k-1, n-k, n-k) for n,k >= 1 and T(0, 0) = 1, where r(n, k, m) = m*r(n, k-1, m) + r(n-1, k, m+1) and r(n, 0, m) = 1. (see Vladimir Kruchinin's formula in A039621).
Sum_{k=1..n} binomial(k + x - 1, k-1)*(k-1)!*T(n, k) = (n + x)^(n - 1) for n >= 1.
Sum_{k=1..n} (-1)^(k+j)*Stirling1(k, j)*T(n, k) = n^(n-j)*binomial(n-1, j-1) for n >= 1, which are, up to sign, the coefficients of the Abel polynomials (A137452).
From Werner Schulte, Jun 14 2022 and Jun 19 2022: (Start)
E.g.f. of column k >= 0: (Sum_{i>0} (i-1)^(i-1) * t^i / i!)^k / k!.
Conjecture: T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * A048993(n+i-1, n-1) for 0 < k <= n and T(n, 0) = 0^n for n >= 0; proved by Mike Earnest, see link at A354797. (End)

A360657 Number triangle T associated with 2-Stirling numbers and Lehmer-Comtet numbers (see Comments and Formula section).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 9, 5, 1, 0, 64, 37, 9, 1, 0, 625, 369, 97, 14, 1, 0, 7776, 4651, 1275, 205, 20, 1, 0, 117649, 70993, 19981, 3410, 380, 27, 1, 0, 2097152, 1273609, 365001, 64701, 7770, 644, 35, 1, 0, 43046721, 26269505, 7628545, 1388310, 174951, 15834, 1022, 44, 1
Offset: 0

Views

Author

Werner Schulte, Feb 15 2023

Keywords

Comments

Triangle T is created using 2-Stirling numbers of the first (A049444) and the second (A143494) kind. The unusual construction is as follows:
Define A(n, k) by recurrence A(n, k) = A(n-1, k-1) + (k+1) * A(n-1, k) for 0 < k < n with initial values A(n, n) = 1, n >= 0, and A(n, 0) = 0, n > 0. A without column k = 0 is A143494. Let B = A^(-1) matrix inverse of A. B without column k = 0 is A049444. Now define T(m, k) = Sum_{i=0..m-k} B(m-k, i) * A(m-1+i, m-1) for 0 < k <= m = n/2 and T(m, 0) = 0^m for 0 <= m = n/2; T(i, j) = 0 if i < j or j < 0.
Matrix inverse of T is A360753. - Werner Schulte, Feb 21 2023
Conjecture: the transpose of this array is the upper triangular matrix U in the LU factorization of the array of Stirling numbers of the second kind read as a square array; the corresponding lower triangular array L is the triangle of Stirling numbers of the second kind. See the example section below. - Peter Bala, Oct 10 2023

Examples

			Triangle T(n, k), 0 <= k <= n, starts:
n\k :  0         1         2        3        4       5      6     7   8  9
==========================================================================
  0 :  1
  1 :  0         1
  2 :  0         2         1
  3 :  0         9         5        1
  4 :  0        64        37        9        1
  5 :  0       625       369       97       14       1
  6 :  0      7776      4651     1275      205      20      1
  7 :  0    117649     70993    19981     3410     380     27     1
  8 :  0   2097152   1273609   365001    64701    7770    644    35   1
  9 :  0  43046721  26269505  7628545  1388310  174951  15834  1022  44  1
  etc.
From _Peter Bala_, Oct 10 2023: (Start)
LU factorization of the square array of Stirling numbers of the second kind (apply Xu, Lemma 2.2):
 / 1               \ / 1   1   1   1  ...\    / 1   1   1    1  ... \
 | 1   1           ||      2   5   9  ...|   |  1   3   6   10  ... |
 | 1   3   1       ||          9  37  ...| = |  1   7  25   65  ... |
 | 1   7   6   1   ||             64  ...|   |  1  15  90  350  ... |
 | ...             ||                 ...|   |  ...                 |
(End)
		

Crossrefs

Cf. A000007 (column 0), A000169 (column 1), A055869 (column 2).
Cf. A000012 (main diagonal), A000096 (1st subdiagonal), A360753 (matrix inverse).

Programs

  • PARI
    tabl(m) = {my(n=2*m, A = matid(n), B, T); for( i = 2, n, for( j = 2, i, A[i, j] = A[i-1, j-1] + j * A[i-1, j] ) ); B = A^(-1); T = matrix( m, m, i, j, if( j == 1, 0^(i-1), sum( r = 0, i-j, B[i-j+1, r+1] * A[i-1+r, i-1] ) ) ); }

Formula

For the definition of triangle T see Comments section.
Conjectured formulas:
1. T(n, k) = (Sum_{i=k..n} A354794(n, i) * (i-1)!) / (k-1)! for 0 < k <= n.
2. T(n, k) - k * T(n, k+1) = A354794(n, k) for 0 <= k <= n.
3. T(n, 1) = A000169(n) = n^(n-1) for n > 0.
4. T(n, 2) = A055869(n-1) = n^(n-1) - (n-1)^(n-1) for n > 1.
5. T(n, k) = (Sum_{i=0..k-1} (-1)^i * binomial(k-1, i) * (n-i)^(n-1)) / (k-1)! for 0 < k <= n.
6. Sum_{i=1..n} (-1)^(n-i) * binomial(n-1+k, i-1) * T(n, i) * (i-1)! = (k-1)^(n-1) for n > 0 and k >= 0.
7. Matrix product of A354795 and T without column 0 equals A094587.
8. Matrix product of T and A354795 without column 0 equals A088956.
9. E.g.f. of column k > 0: Sum_{n>=k} T(n, k) * t^(n-1) / (n-1)! = (W(-t)/(-t)) * (Sum_{n>=k} A354794(n, k) * t^(n-1) / (n-1)!) where W is the Lambert_W-function.

A360753 Matrix inverse of A360657.

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 1, -5, 1, 0, 1, 8, -9, 1, 0, 2, 4, 29, -14, 1, 0, 6, 4, -10, 75, -20, 1, 0, 24, 4, -41, -115, 160, -27, 1, 0, 120, -8, -147, -196, -490, 301, -35, 1, 0, 720, -136, -624, -392, -231, -1484, 518, -44, 1
Offset: 0

Views

Author

Werner Schulte, Feb 21 2023

Keywords

Examples

			Triangle T(n, k) for 0 <= k <= n starts:
n\k :  0    1     2     3     4     5      6    7    8  9
=========================================================
  0 :  1
  1 :  0    1
  2 :  0   -2     1
  3 :  0    1    -5     1
  4 :  0    1     8    -9     1
  5 :  0    2     4    29   -14     1
  6 :  0    6     4   -10    75   -20      1
  7 :  0   24     4   -41  -115   160    -27    1
  8 :  0  120    -8  -147  -196  -490    301  -35    1
  9 :  0  720  -136  -624  -392  -231  -1484  518  -44  1
  etc.
		

Crossrefs

Cf. A132013, A215534, A354794, A354795, A360657 (matrix inverse).

Programs

  • PARI
    tabl(m) = {my(n=2*m, A = matid(n), B, C, T); for( i = 2, n, for( j = 2, i, A[i, j] = A[i-1, j-1] + j * A[i-1, j] ) ); B = A^(-1); C = matrix( m, m, i, j, if( j == 1, 0^(i-1), sum( r = 0, i-j, B[i-j+1, r+1] * A[i-1+r, i-1] ) ) ); T = 1/C; }

Formula

Conjectured formulas:
1. Matrix product of A354794 and T without column 0 equals A215534.
2. Matrix product of T and A354794 without column 0 equals A132013.
3. E.g.f. of column k > 0: Sum_{n >= k} T(n, k) * t^(n-1) / (n-1)! = (1 - t) * (Sum_{n >= k} A354795(n, k) * t^(n-1) / (n-1)!).
Showing 1-6 of 6 results.