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 1-3 of 3 results.

A049444 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -2, 1, 6, -5, 1, -24, 26, -9, 1, 120, -154, 71, -14, 1, -720, 1044, -580, 155, -20, 1, 5040, -8028, 5104, -1665, 295, -27, 1, -40320, 69264, -48860, 18424, -4025, 511, -35, 1, 362880, -663696, 509004, -214676, 54649, -8624, 826, -44, 1, -3628800, 6999840, -5753736
Offset: 0

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Author

Keywords

Comments

T(n, k) = ^2P_n^k in the notation of the given reference with T(0, 0) := 1. The monic row polynomials s(n,x) := Sum_{m=0..n} T(n, k)*x^k which are s(n, x) = Product_{j=0..n-1} (x-(2+j)), n >= 1 and s(0, x)=1 satisfy s(n, x+y) = Sum_{k=0..n} binomial(n, k)*s(k,x)*S1(n-k, y), with the Stirling1 polynomials S1(n, x) = Sum_{m=1..n} (A008275(n, m)*x^m) 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 polynomials for (exp(2*t), exp(t)-1). This translates to the usual exponential Riordan (Sheffer) notation (1/(1+x)^2, log(1+x)).
See A143491 for the unsigned version of this array and A143494 for the inverse. - Peter Bala, Aug 25 2008
Corresponding to the generalized Stirling number triangle of second kind A137650. - Peter Luschny, Sep 18 2011
Unsigned, reversed rows (cf. A145324, A136124) are the dimensions of the cohomology of a complex manifold with a symmetric group (S_n) action. See p. 17 of the Hyde and Lagarias link. See also the Murri link for an interpretation as the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces. - Tom Copeland, Dec 09 2016
The row polynomials s(n, x) = (-1)^n*risingfactorial(2 - x, n) are related to the column sequences of the unsigned Abel triangle A137452(n, k), for k >= 2. See the formula there. - Wolfdieter Lang, Nov 21 2022

Examples

			The Triangle  begins:
n\k       0       1        2       3       4      5      6    7   8 9 ...
0:        1
1:       -2       1
2:        6      -5        1
3:      -24      26       -9       1
4:      120    -154       71     -14       1
5      -720    1044     -580     155     -20      1
6:     5040   -8028     5104   -1665     295    -27      1
7:   -40320   69264   -48860   18424   -4025    511    -35    1
8:   362880 -663696   509004 -214676   54649  -8624    826  -44
9: -3628800 6999840 -5753736 2655764 -761166 140889 -16884 1266 -54 1
...  [reformatted by _Wolfdieter Lang_, Nov 21 2022]
		

References

  • Y. Manin, Frobenius Manifolds, Quantum Cohomology and Moduli Spaces, American Math. Soc. Colloquium Publications Vol. 47, 1999. [From Tom Copeland, Jun 29 2008]
  • S. Roman, The Umbral Calculus, Academic Press, 1984 (also Dover Publications, 2005).

Crossrefs

Unsigned column sequences are A000142(n+1), A001705-A001709. Row sums (signed triangle): n!*(-1)^n, row sums (unsigned triangle): A001710(n-2). Cf. A008275 (Stirling1 triangle).

Programs

  • Haskell
    a049444 n k = a049444_tabl !! n !! k
    a049444_row n = a049444_tabl !! n
    a049444_tabl = map fst $ iterate (\(row, i) ->
       (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 2)
    -- Reinhard Zumkeller, Mar 11 2014
  • Maple
    A049444_row := proc(n) local k,i;
    add(add(Stirling1(n, n-i), i=0..k)*x^(n-k-1),k=0..n-1);
    seq(coeff(%,x,k),k=1..n-1) end:
    seq(print(A049444_row(n)),n=1..7); # Peter Luschny, Sep 18 2011
    A049444:= (n, k)-> add((-1)^(n-j)*(n-j+1)!*binomial(n, j)*Stirling1(j, k), j=0..n):
    seq(print(seq(A049444(n, k), k=0..n)), n=0..11);  # Mélika Tebni, May 02 2022
  • Mathematica
    t[n_, i_] = Sum[(-1)^k*Binomial[n, k]*(k+1)!*StirlingS1[n-k, i], {k, 0, n-i}]; Flatten[Table[t[n, i], {n, 0, 9}, {i, 0, n}]] [[1 ;; 48]]
    (* Jean-François Alcover, Apr 29 2011, after Milan Janjic *)

Formula

T(n, k) = T(n-1, k-1) - (n+1)*T(n-1, k), n >= k >= 0; T(n, k) = 0, n < k; T(n, -1) = 0, T(0, 0) = 1.
E.g.f. for k-th column of signed triangle: ((log(1+x))^k)/(k!*(1+x)^2).
Triangle (signed) = [-2, -1, -3, -2, -4, -3, -5, -4, -6, -5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; triangle (unsigned) = [2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...], where DELTA is Deléham's operator defined in A084938 (unsigned version in A143491).
E.g.f.: (1 + x)^(y-2). - Vladeta Jovovic, May 17 2004 [For row polynomials s(n, y)]
With P(n, t) = Sum_{j=0..n-2} T(n-2,j) * t^j and P(1, t) = -1 and P(0, t) = 1, then G(x, t) = -1 + exp[P(.,t)*x] = [(1+x)^t - 1 - t^2 * x] / [t(t-1)], whose compositional inverse in x about 0 is given in A074060. G(x, 0) = -log(1+x) and G(x, 1) = (1+x) log(1+x) - 2x. G(x, q^2) occurs in formulas on pages 194-196 of the Manin reference. - Tom Copeland, Feb 17 2008
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
T(n, k) = Sum_{j=0..n} (-1)^(n-j)*(n-j+1)!*binomial(n, j)*Stirling1(j, k). - Mélika Tebni, May 02 2022
From Wolfdieter Lang, Nov 24 2022: (Start)
Recurrence for row polynomials {s(n, x)}_{n>=0}: s(0, x) = 1, s(n, x) = (x - 2)*exp(-(d/dx)) s(n-1, x), for n >= 1. This is adapted from the general Sheffer result given by S. Roman, Corollary 3.7.2., p. 50.
Recurrence for column sequence {T(n, k)}{n>=k}: T(n, n) = 1, T(n, k) = (n!/(n-k))*Sum{j=k..n-1} (1/j!)*(a(n-1-j) + k*beta(n-1-j))*T(n-1, k), for k >= 0, where alpha = repeat(-2, 2) and beta(n) = [x^n] (d/dx)log(log(x)/x) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), for n >= 0. This is the adapted Boas-Buck recurrence, also given in Rainville, Theorem 50., p. 141, For the references and a comment see A046521. (End)

Extensions

Second formula corrected by Philippe Deléham, Nov 09 2008

A143491 Unsigned 2-Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764
Offset: 2

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2-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 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned r-Stirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2-Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned r-Stirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r-1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related r-Lah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2-Stirling numbers of the first kind see the Rotbart link. - Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^2,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-2*t),1-exp(-t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444. - Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link). - Tom Copeland, Sep 19 2012

Examples

			Triangle begins
  n\k|.....2.....3.....4.....5.....6.....7
  ========================================
  2..|.....1
  3..|.....2.....1
  4..|.....6.....5.....1
  5..|....24....26.....9.....1
  6..|...120...154....71....14.....1
  7..|...720..1044...580...155....20.....1
  ...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From _Aviv Rotbart_, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
  n\k|.....0.....1.....2.....3.....4.....5.....6.....7
  ====================================================
  2..|.....1
  3..|.....1.....2
  4..|.....1.....5.....6
  5..|.....1.....9....26....24
  6..|.....1....14....71...154...120
  7..|.....1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
   (12)(13)(12)(13) = (312)
   (12)(13)(12)(14) = (41)(23)
   (12)(13)(12)(24) = (432)(1)
   (12)(13)(12)(34) = (342)(1)
   (12)(23)(12)(24) = (13)(24)
   (12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)
		

Crossrefs

Cf. A001705 - A001709 (column 3..7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.
Cf. A094638.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-2)! * add((n-j-1)*abs(stirling1(j,k-2))/j!,j = k-2..n-2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := (n-2)!*Sum[(n-j-1)*Abs[StirlingS1[j, k-2]]/j!, {j, k-2, n-2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* Jean-François Alcover, Apr 16 2013, after Maple *)

Formula

T(n,k) = (n-2)! * Sum_{j = k-2 .. n-2} (n-j-1)*|stirling1(j,k-2)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n-1)!; T(n,3) = (n-1)!*(1/2 + 1/3 + ... + 1/(n-1)).
T(n,k) = Sum_{2 <= i_1 < ... < i_(n-k) < n} (i_1*i_2*...*i_(n-k)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)*...*(x+n-1).
E.g.f. for column (k+2): Sum_{n>=k} T(n+2,k+2)*x^n/n! = (1/k!)*(1/(1-x)^2)*(log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+2) = Sum_{n>=0} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, 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!/2 ( A001710 ). The alternating row sums are (n-2)!.
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-1,i) = |f(n,i,2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row of the triangle = top row of M^(n-2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
2, 1, 0, 0, 0, 0, ...
2, 3, 1, 0, 0, 0, ...
2, 5, 4, 1, 0, 0, ...
2, 7, 9, 5, 1, 0, ...
... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3). - Tom Copeland, Dec 11 2016

A347989 a(n) = [x^n] (2*n)! * Sum_{k=0..2*n} binomial(x+k,k).

Original entry on oeis.org

1, 5, 71, 1665, 54649, 2310945, 119753843, 7353403057, 522289211873, 42137920501677, 3807384320667135, 380929847762489025, 41811136672902061321, 4995760464106519955705, 645541681316043216096315, 89705032647088734873129825, 13340173206548155385625683265, 2114001534402053456524492822485
Offset: 0

Views

Author

Seiichi Manyama, Sep 23 2021

Keywords

Crossrefs

Programs

  • PARI
    a(n) = (2*n)!*polcoef(sum(k=n, 2*n, binomial(x+k, k)), n);
    
  • PARI
    a(n) = (2*n)!*sum(k=n, 2*n, (2*n+1-k)*abs(stirling(k, n, 1))/k!);

Formula

a(n) = (2*n)! * Sum_{k=n..2*n} (2*n+1-k) * |Stirling1(k,n)|/k!.
a(n) = [x^(2*n)] ((2*n)!/n!) * (-log(1 - x))^n/(1 - x)^2.
From Vaclav Kotesovec, Sep 23 2021, updated May 14 2025: (Start)
a(n) = [x^n] Gamma(2*n + x + 2) / Gamma(x + 2).
a(n) ~ c * d^n * (n-1)!, where d = 8*w^2/(2*w-1), w = -LambertW(-1,-exp(-1/2)/2) and c = w^2 * sqrt(2) / (sqrt(w-1)*Pi) = 1.5967712192197964362930380385801737624829174112909160160618... (End)
Showing 1-3 of 3 results.