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.

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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

A051562 Second unsigned column of triangle A051380.

Original entry on oeis.org

0, 1, 19, 299, 4578, 71394, 1153956, 19471500, 343976400, 6366517200, 123418922400, 2503748556000, 53091962697600, 1175271048201600, 27123099523027200, 651708291206649600, 16282170039031142400
Offset: 0

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Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=9) ~ exp(-x)/x^2*(1 - 19/x + 299/x^2 - 4578/x^3 + 71394/x^4 - 1153956/x^5 + 19471500/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051380.

Crossrefs

Cf. A049389 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 8; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051380(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^9.
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-9,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[8]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A051564 Second unsigned column of triangle A051523.

Original entry on oeis.org

0, 1, 21, 362, 6026, 101524, 1763100, 31813200, 598482000, 11752855200, 240947474400, 5154170774400, 114942011990400, 2669517204076800, 64496340380102400, 1619153396908185600, 42188624389562112000
Offset: 0

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Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=10) ~ exp(-x)/x^2*(1 - 21/x + 362/x^2 - 6026/x^3 + 101524/x^4 - 1763100/x^5 + 31813200/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051523.

Crossrefs

Cf. A049398 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[n_] := n!*Sum[(-1)^k*Binomial[-10, k]/(n - k), {k, 0, n - 1}]; Array[f, 17, 0]
    Range[0, 16]! CoefficientList[ Series[-Log[(1 - x)]/(1 - x)^10, {x, 0, 16}], x]
    (* Or, using elementary symmetric functions: *)
    f[k_] := k + 9; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051523(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^10.
a(n) = n!*Sum_{k=0..n-1}((-1)^k*binomial(-10,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[9]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

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

A136394 Triangle read by rows: T(n,k) is the number of permutations of an n-set having k cycles of size > 1 (0<=k<=floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 5, 1, 20, 3, 1, 84, 35, 1, 409, 295, 15, 1, 2365, 2359, 315, 1, 16064, 19670, 4480, 105, 1, 125664, 177078, 56672, 3465, 1, 1112073, 1738326, 703430, 74025, 945, 1, 10976173, 18607446, 8941790, 1346345, 45045, 1, 119481284, 216400569, 118685336
Offset: 0

Views

Author

Vladeta Jovovic, May 03 2008

Keywords

Examples

			Triangle (n,k) begins:
  1;
  1;
  1,    1;
  1,    5;
  1,   20,    3;
  1,   84,   35;
  1,  409,  295,  15;
  1, 2365, 2359, 315;
  ...
		

Crossrefs

Programs

  • Maple
    egf:= proc(k::nonnegint) option remember; x-> exp(x)* ((-x-ln(1-x))^k)/k! end; T:= (n,k)-> coeff(series(egf(k)(x), x=0, n+1), x, n) *n!; seq(seq(T(n,k), k=0..n/2), n=0..30); # Alois P. Heinz, Aug 14 2008
    # second Maple program:
    b:= proc(n) option remember; expand(`if`(n=0, 1, add(b(n-i)*
          `if`(i>1, x, 1)*binomial(n-1, i-1)*(i-1)!, i=1..n)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
    seq(T(n), n=0..15);  # Alois P. Heinz, Sep 25 2016
    # third Maple program:
    T:= proc(n, k) option remember; `if`(k<0 or k>2*n, 0,
          `if`(n=0, 1, add(T(n-i, k-`if`(i>1, 1, 0))*
           mul(n-j, j=1..i-1), i=1..n)))
        end:
    seq(seq(T(n,k), k=0..n/2), n=0..15);  # Alois P. Heinz, Jul 16 2017
  • Mathematica
    max = 12; egf = Exp[x*(1-y)]/(1-x)^y; s = Series[egf, {x, 0, max}, {y, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]*n!; t[0, 0] = t[1, 0] = 1; Table[t[n, k], {n, 0, max}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Jan 28 2014 *)

Formula

E.g.f.: exp(x*(1-y))/(1-x)^y. Binomial transform of triangle A008306. exp(x)*((-x-log(1-x))^k)/k! is e.g.f. of k-th column.
From Alois P. Heinz, Jul 13 2017: (Start)
T(2n,n) = A001147(n).
T(2n+1,n) = A051577(n) = (2*n+3)!!/3 = A001147(n+2)/3. (End)
From Alois P. Heinz, Aug 17 2023: (Start)
Sum_{k=0..floor(n/2)} k * T(n,k) = A001705(n-1) for n>=1.
Sum_{k=0..floor(n/2)} (-1)^k * T(n,k) = A159964(n-1) for n>=1. (End)

A067318 Sum of the reflection lengths of all permutations of n letters.

Original entry on oeis.org

0, 1, 7, 46, 326, 2556, 22212, 212976, 2239344, 25659360, 318540960, 4261576320, 61148511360, 937030429440, 15275952518400, 264030355814400, 4823280687052800, 92865738644582400, 1879691760950169600, 39905092126771200000, 886664974825728000000
Offset: 1

Views

Author

H. Nick Hann (nickhann(AT)aol.com), Jan 15 2002

Keywords

Comments

The reflection length of a permutation is the minimum number of transpositions needed to express the permutation.
May also be called the "weight" of the symmetric group S_n.
a(n) is the number of n+1-permutations that have at least 3 cycles. a(n) = Sum_{k=3..n+1} A132393(n+1,k). Cf. A001563 (n-permutations with at least 2 cycles). - Geoffrey Critzer, Dec 01 2013
The number of covering relations in the middle order on S_n. - Bridget Tenner, Jul 11 2025

Examples

			a(3)=7 since the permutations are 1, (12), (13), (23), (12)(13) and (13)(12). The sum of reflection lengths of all elements in S_3 is 0+1+1+1+2+2=7.
The terms satisfy the series: x/(1-x) = x/((1+x)*(1+2*x)*(1+3*x)) + 7*x^2/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + 46*x^3/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)*(1+5*x)) + 326*x^4/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)*(1+5*x)*(1+6*x)) + ... - _Paul D. Hanna_, Aug 28 2012
		

References

  • N. Hann, Average Weight of a Random Permutation, preprint, 2002. [Apparently unpublished]

Crossrefs

Programs

  • Maple
    ZL :=[S, {S = Set(Cycle(Z),3 <= card)}, labelled]: seq(combstruct[count](ZL, size=n), n=2..22); # Zerinvary Lajos, Mar 25 2008
  • Mathematica
    a[n_] := n!*(n - HarmonicNumber[n]); Table[a[n], {n, 1, 21}](* Jean-François Alcover, Feb 10 2012 *)
    nn=22;Drop[Range[0,nn]!CoefficientList[Series[1/(1-x)-1-Log[1/(1-x)]-Log[1/(1-x)]^2/2!,{x,0,nn}],x],2] (* Geoffrey Critzer, Dec 01 2013 *)
  • Maxima
    A067318(n):=n*n! - abs(stirling1(n+1, 2))$
    makelist(A067318(n),n,1,30); /* Martin Ettl, Nov 03 2012 */
  • PARI
    {a(n)=if(n==0,0,if(n==1, 1, 1-polcoeff(sum(k=1, n-1, a(k)*x^k/prod(j=1, k+2, (1+j*x+x*O(x^n)) ) ), n)))} /* Paul D. Hanna, Aug 28 2012 */
    

Formula

a(n) = n!*(0/1+1/2+...+(n-1)/n) = n!*(n - H_n), where H_n = Sum_{k=1..n} 1/k; a(1) = 0, a(2) = 1, a(n) = n*a(n-1) + (n-1)*(n-1)!.
a(n) = n*n! - abs(stirling1(n+1, 2)) (cf. A000254). E.g.f.: (x+(1-x)*log(1-x))/(1-x)^2. - Vladeta Jovovic, Feb 01 2003
a(n) = T(n, n-1) for the triangle read by rows: [0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 30 2003
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n/Product_{k=1..n+2} (1+k*x). - Paul D. Hanna, Aug 28 2012
a(n) = A062119(n) - A001705(n-1). - Anton Zakharov, Sep 22 2016

Extensions

Definition and example edited by Bridget Tenner, Jul 11 2025

A112486 Coefficient triangle for polynomials used for e.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 1, 1, 2, 5, 3, 6, 26, 35, 15, 24, 154, 340, 315, 105, 120, 1044, 3304, 4900, 3465, 945, 720, 8028, 33740, 70532, 78750, 45045, 10395, 5040, 69264, 367884, 1008980, 1571570, 1406790, 675675, 135135, 40320, 663696, 4302216, 14777620, 29957620
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

The k-th diagonal of |A008275| appears as the k-th column in |A008276| with k-1 leading zeros.
The recurrence, given below, is derived from (d/dx)g1(k,x) - g1(k,x)= x*(d/dx)g1(k-1,x) + g1(k-1,x), k >= 1, with input g(-1,x):=0 and initial condition g1(k,0)=1, k >= 0. This differential recurrence for the e.g.f. g1(k,x) follows from the one for unsigned Stirling1 numbers.
The column sequences start with A000142 (factorials), A001705, A112487- A112491, for m=0,...,5.
The main diagonal gives (2*k-1)!! = A001147(k), k >= 1.
This computation was inspired by the Bender article (see links), where the Stirling polynomials are discussed.
The e.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| with k-1 leading zeros is g1(k-1,x) = exp(x)*Sum_{m=0..k-1} a(k,m)*(x^(k-1+m))/(k-1+m)!.
a(k,n) = number of lists with entries from [n] such that (i) each element of [n] occurs at least once and at most twice, (ii) for each i that occurs twice, all entries between the two occurrences of i are > i, and (iii) exactly k elements of [n] occur twice. Example: a(1,2)=5 counts 112, 121, 122, 211, 221, and a(2,2)=3 counts 1122,1221,2211. - David Callan, Nov 21 2011

Examples

			Triangle begins:
    1;
    1,    1;
    2,    5,     3;
    6,   26,    35,    15;
   24,  154,   340,   315,   105;
  120, 1044,  3304,  4900,  3465,   945;
  720, 8028, 33740, 70532, 78750, 45045, 10395;
k=3 column of |A008276| is [0,0,2,11,35,85,175,...] (see A000914), its e.g.f. exp(x)*(2*x^2/2! + 5* x^3/3! + 3*x^4/4!).
		

Crossrefs

Cf. A112007 (triangle for o.g.f.s for unsigned Stirling1 diagonals). A112487 (row sums).

Programs

  • Maple
    A112486 := proc(n,k)
        if n < 0 or k<0 or  k> n then
            0 ;
        elif n = 0 then
            1 ;
        else
            (n+k)*procname(n-1,k)+(n+k-1)*procname(n-1,k-1) ;
        end if;
    end proc: # R. J. Mathar, Dec 19 2013
  • Mathematica
    A112486 [n_, k_] := A112486[n, k] = Which[n<0 || k<0 || k>n, 0, n == 0, 1, True, (n+k)*A112486[n-1, k]+(n+k-1)*A112486[n-1, k-1]]; Table[A112486[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after R. J. Mathar *)

Formula

a(k, m) = (k+m)*a(k-1, m) + (k+m-1)*a(k-1, m-1) for k >= m >= 0, a(0, 0)=1, a(k, -1):=0, a(k, m)=0 if k < m.
From Tom Copeland, Oct 05 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = -(1 + t)
P(3,t) = 2 + 5 t + 3 t^2
P(4,t) = -( 6 + 26 t + 35 t^2 + 15 t^3)
P(5,t) = 24 + 154 t +340 t^2 + 315 t^3 + 105 t^4
Apparently, P(n,t) = (-1)^(n+1) PW[n,-(1+t)] where PW are the Ward polynomials A134991. If so, an e.g.f. for the polynomials is
A(x,t) = -(x+t+1)/t - LW{-((t+1)/t) exp[-(x+t+1)/t]}, where LW(x) is a suitable branch of the Lambert W Fct. (e.g., see A135338). The comp. inverse in x (about x = 0) is B(x) = x + (t+1) [exp(x) - x - 1]. See A112487 for special case t = 1. These results are a special case of A134685 with u(x) = B(x), i.e., u_1=1 and (u_n)=(1+t) for n>0.
Let h(x,t) = 1/(dB(x)/dx) = 1/[1+(1+t)*(exp(x)-1)], an e.g.f. in x for row polynomials in t of signed A028246 , then P(n,t), is given by
(h(x,t)*d/dx)^n x, evaluated at x=0, i.e., A(x,t)=exp(x*h(u,t)*d/du) u, evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t).
The e.g.f. A(x,t) = -v * Sum_{j>=1} D(j-1,u) (-z)^j / j! where u=-(x+t+1)/t, v=1+u, z=(1+t*v)/(t*v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = -1/[t*(v-A)].(End)
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=t+1, and (a_n)=t*P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 =0. - Tom Copeland, Oct 08 2011
The row polynomials R(n,x) may be calculated using R(n,x) = 1/x^(n+1)*D^n(x), where D is the operator (x^2+x^3)*d/dx. - Peter Bala, Jul 23 2012
For n>0, Sum_{k=0..n} a(n,k)*(-1/(1+W(t)))^(n+k+1) = (t d/dt)^(n+1) W(t), where W(t) is Lambert W function. For t=-x, this gives Sum_{k>=1} k^(k+n)*x^k/k! = - Sum_{k=0..n} a(n,k)*(-1/(1+W(-x)))^(n+k+1). - Max Alekseyev, Nov 21 2019
Conjecture: row polynomials are R(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling2(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(n+j+k)/((n+j-k)!*(i-j)!*k!). - Mikhail Kurkov, Apr 21 2025

A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 50, 26, 7, 1, 274, 154, 47, 9, 1, 1764, 1044, 342, 74, 11, 1, 13068, 8028, 2754, 638, 107, 13, 1, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1
Offset: 1

Views

Author

Johannes W. Meijer, Oct 05 2009

Keywords

Comments

The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2+6*n+3*n^2)/x^2 - (6+22*n+18*n^2+ 4*n^3)/x^3 + ... ) is discussed in A028421. The formula for the asymptotic expansion leads for n = 1, 2, 3, .., to the left hand columns of the triangle given above.
The recurrence relations of the right hand columns of this triangle lead to Pascal's triangle A007318, their a(n) formulas lead to Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; cf. A080663, A165676, A165677, A165678 and A165679.
The row sums of this triangle lead to A093344. Surprisingly the e.g.f. of the row sums Egf(x) = (exp(1)*Ei(1,1-x) - exp(1)*Ei(1,1))/(1-x) leads to the exponential integrals in view of the fact that E(x,m=1,n=1) = Ei(n=1,x). We point out that exp(1)*Ei(1,1) = A073003.
The Maple programs generate the coefficients of the triangle given above. The first one makes use of a relation between the triangle coefficients, see the formulas, and the second one makes use of the asymptotic expansions of the E(x,m=2,n).
Amarnath Murthy discovered triangle A093905 which is the reversal of our triangle.
A165675 is an extended version of this triangle. Its reversal is A105954.
Triangle A094587 is generated by the asymptotic expansions of E(x,m=1,n).

Crossrefs

A093905 is the reversal of this triangle.
A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564 are the first ten left hand columns.
A080663, n>=2, is the third right hand column.
A165676, A165677, A165678 and A165679 are the next right hand columns, A093344 gives the row sums.
A073003 is Gompertz's constant.
A094587 is generated by the asymptotic expansions of E(x, m=1, n).
Cf. A165675, A105954 (Quet) and A067176 (Bottomley).
Cf. A007318 (Pascal), A028421 (Wiggen), A126671 (Wood).

Programs

  • Maple
    nmax:=9; for n from 1 to nmax do a(n, n) := 1 od: for n from 2 to nmax do a(n, 1) := n*a(n-1, 1) + (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m = 1..n), n = 1..nmax);
    # End program 1
    nmax := nmax+1: m:=2; with(combinat): EA := proc(x, m, n) local E, i; E:=0: for i from m-1 to nmax+2 do E := E + sum((-1)^(m+k1+1) * binomial(k1, m-1) * n^(k1-m+1) * stirling1(i, k1), k1=m-1..i) / x^(i-m+1) od: E:= exp(-x)/x^(m) * E: return(E); end: for n1 from 1 to nmax do f(n1-1) := simplify(exp(x) * x^(nmax+3) * EA(x, m, n1)); for m1 from 0 to nmax+2 do b(n1-1, m1) := coeff(f(n1-1), x, nmax+2-m1) od: od: for n1 from 0 to nmax-1 do for m1 from 0 to n1-m+1 do a(n1-m+2, m1+1) := abs(b(m1, n1-m1)) od: od: seq(seq(a(n, m), m = 1..n),n = 1..nmax-1);
    # End program 2
    # Maple programs revised by Johannes W. Meijer, Sep 22 2012

Formula

a(n,m) = (n-m+1)*a(n-1,m) + a(n-1,m-1), for 2 <= m <= n-1, with a(n,n) = 1 and a(n,1) = n*a(n-1,1) + (n-1)!.
a(n,m) = product(i, i= m..n)*sum(1/i, i = m..n).

A165675 Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 5, 1, 24, 50, 26, 7, 1, 120, 274, 154, 47, 9, 1, 720, 1764, 1044, 342, 74, 11, 1, 5040, 13068, 8028, 2754, 638, 107, 13, 1, 40320, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 362880, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1
Offset: 0

Views

Author

Johannes W. Meijer, Oct 05 2009

Keywords

Comments

Previous name: Extended triangle related to the asymptotic expansions of the E(x, m = 2, n).
For the definition of the hyperharmonic numbers see the formula section.
This triangle is the same as triangle A165674 except for the extra left-hand column T(n, 0) = n!. The T(n) formulas for the right-hand columns generate the coefficients of this extra left-hand column.
Leroy Quet discovered triangle A105954 which is the reversal of our triangle.
In square format, row k gives the (n-1)-st elementary symmetric function of {k, k+1, k+2,..., k+n}, as in the Mathematica section. - Clark Kimberling, Dec 29 2011

Examples

			Triangle T(n, k) begins:
  [0]    1;
  [1]    1,     1;
  [2]    2,     3,    1;
  [3]    6,    11,    5,    1;
  [4]   24,    50,   26,    7,   1;
  [5]  120,   274,  154,   47,   9,   1;
  [6]  720,  1764, 1044,  342,  74,  11,  1;
  [7] 5040, 13068, 8028, 2754, 638, 107, 13, 1;
Seen as an array (the triangle arises when read by descending antidiagonals):
  [0] 1,  1,   2,    6,    24,    120,     720,     5040, ...
  [1] 1,  3,  11,   50,   274,   1764,   13068,   109584, ...
  [2] 1,  5,  26,  154,  1044,   8028,   69264,   663696, ...
  [3] 1,  7,  47,  342,  2754,  24552,  241128,  2592720, ...
  [4] 1,  9,  74,  638,  5944,  60216,  662640,  7893840, ...
  [5] 1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, ...
  [6] 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, ...
  [7] 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, ...
		

Crossrefs

A105954 is the reversal of this triangle.
A165674, A138771 and A165680 are related triangles.
A080663 equals the third right hand column.
A000142 equals the first left hand column.
A093345 are the row sums.
Columns include A165676, A165677, A165678 and A165679.

Programs

  • Maple
    nmax := 8; for n from 0 to nmax do a(n, 0) := n! od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax);
    # Johannes W. Meijer, revised Nov 27 2012
    # Shows the array format, using hyperharmonic numbers.
    H := proc(n, k) option remember; if n = 0 then 1/(k+1)
    else add(H(n - 1, j), j = 0..k) fi end:
    seq(lprint(seq((k + 1)!*H(n, k), k = 0..7)), n = 0..7);
    # Shows the array format, using the hypergeometric formula.
    A := (n, k) -> (k+1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1):
    seq(lprint(seq(simplify(A(n, k)), k = 0..7)), n = 0..7);
    # Peter Luschny, Jul 03 2022
  • Mathematica
    a[n_] := SymmetricPolynomial[n - 1, t[n]]; z = 10;
    t[n_] := Table[k - 1, {k, 1, n}]; t1 = Table[a[n], {n, 1, z}]  (* A000142 *)
    t[n_] := Table[k,     {k, 1, n}]; t2 = Table[a[n], {n, 1, z}]  (* A000254 *)
    t[n_] := Table[k + 1, {k, 1, n}]; t3 = Table[a[n], {n, 1, z}]  (* A001705 *)
    t[n_] := Table[k + 2, {k, 1, n}]; t4 = Table[a[n], {n, 1, z}]  (* A001711 *)
    t[n_] := Table[k + 3, {k, 1, n}]; t5 = Table[a[n], {n, 1, z}]  (* A001716 *)
    t[n_] := Table[k + 4, {k, 1, n}]; t6 = Table[a[n], {n, 1, z}]  (* A001721 *)
    t[n_] := Table[k + 5, {k, 1, n}]; t7 = Table[a[n], {n, 1, z}]  (* A051524 *)
    t[n_] := Table[k + 6, {k, 1, n}]; t8 = Table[a[n], {n, 1, z}]  (* A051545 *)
    t[n_] := Table[k + 7, {k, 1, n}]; t9 = Table[a[n], {n, 1, z}]  (* A051560 *)
    t[n_] := Table[k + 8, {k, 1, n}]; t10 = Table[a[n], {n, 1, z}] (* A051562 *)
    t[n_] := Table[k + 9, {k, 1, n}]; t11 = Table[a[n], {n, 1, z}] (* A051564 *)
    t[n_] := Table[k + 10, {k, 1, n}];t12 = Table[a[n], {n, 1, z}] (* A203147 *)
    t = {t1, t2, t3, t4, t5, t6, t7, t8, t9, t10};
    TableForm[t]  (* A165675 in square format *)
    m[i_, j_] := t[[i]][[j]];
    (* A165675 as a sequence *)
    Flatten[Table[m[i, n + 1 - i], {n, 1, 10}, {i, 1, n}]]
    (* Clark Kimberling, Dec 29 2011 *)
    A[n_, k_] := (k + 1)*((n + k)! / n!)*HypergeometricPFQ[{-k, 1, 1}, {2, n + 1}, 1];
    Table[A[n, k], {n, 0, 7}, {k, 0, 7}] // TableForm (* Peter Luschny, Jul 03 2022 *)
  • Python
    from functools import cache
    @cache
    def Trow(n: int) -> list[int]:
        if n == 0:
            return [1]
        row = Trow(n - 1) + [1]
        for m in range(n - 1, 0, -1):
            row[m] = (n - m + 1) * row[m] + row[m - 1]
        row[0] *= n
        return row
    for n in range(9): print(Trow(n))  # Peter Luschny, Feb 27 2025

Formula

The hyperharmonic numbers are H(n, k) = Sum_{j=0..k} H(n - 1, j), with base condition H(0, k) = 1/(k + 1).
T(n, k) = (n - k + 1)*T(n - 1, k) + T(n - 1, k - 1), 1 <= k <= n-1, with T(n, 0) = n! and T(n, n) = 1.
From Peter Luschny, Jul 03 2022: (Start)
The rectangular array is given by:
A(n, k) = (k + 1)!*H(n, k).
A(n, k) = (k + 1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1). (End)
From Werner Schulte, Feb 26 2025: (Start)
T(n, k) = n * T(n-1, k) + (n-1)! / (k-1)! for 0 < k < n.
T(n, k) = (Sum_{i=k..n} 1/i) * n! / (k-1)! for 0 < k <= n.
Matrix inverse M = T^(-1) is given by: M(n, n) = 1, M(n, n-1) = 1 - 2 * n for n > 0, M(n, n-2) = (n-1)^2 for n > 1, and M(i, j) = 0 otherwise. (End)

Extensions

New name from Peter Luschny, Jul 03 2022

A105954 Array read by descending antidiagonals: A(n, k) = (n + 1)! * H(k, n + 1), where H(n, k) is a higher-order harmonic number, H(0, k) = 1/k and H(n, k) = Sum_{j=1..k} H(n-1, j), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 5, 11, 6, 1, 7, 26, 50, 24, 1, 9, 47, 154, 274, 120, 1, 11, 74, 342, 1044, 1764, 720, 1, 13, 107, 638, 2754, 8028, 13068, 5040, 1, 15, 146, 1066, 5944, 24552, 69264, 109584, 40320, 1, 17, 191, 1650, 11274, 60216, 241128, 663696, 1026576, 362880
Offset: 0

Views

Author

Leroy Quet, Jun 26 2005

Keywords

Comments

Antidiagonal sums are A093345 (n! * (1 + Sum_{i=1..n}((1/i)*Sum_{j=0..i-1} 1/j!))). - Gerald McGarvey, Aug 27 2005
A recasting of A093905 and A067176. - R. J. Mathar, Mar 01 2009
The triangular array of this sequence is the reversal of A165675 which is related to the asymptotic expansion of the higher order exponential integral E(x,m=2,n); see also A165674. - Johannes W. Meijer, Oct 16 2009

Examples

			A(2, 2) = (1 + (1 + 1/2) + (1 + 1/2 + 1/3))*6 = 26.
Array A(n, k) begins:
  [n\k]  0       1       2        3        4        5          6
  -------------------------------------------------------------------
  [0]    1,      1,      1,       1,       1,       1,         1, ...
  [1]    1,      3,      5,       7,       9,       11,       13, ...
  [2]    2,     11,     26,      47,      74,      107,      146, ...
  [3]    6,     50,    154,     342,     638,     1066,     1650, ...
  [4]   24,    274,   1044,    2754,    5944,    11274,    19524, ...
  [5]  120,   1764,   8028,   24552,   60216,   127860,   245004, ...
  [6]  720,  13068,  69264,  241128,  662640,  1557660,  3272688, ...
  [7] 5040, 109584, 663696, 2592720, 7893840, 20355120, 46536624, ...
		

Crossrefs

Column 0 = A000142 (factorial numbers).
Column 1 = A000254 (Stirling numbers of first kind s(n, 2)) starting at n=1.
Column 2 = A001705 (Generalized Stirling numbers: a(n) = n!*Sum_{k=0..n-1}(k+1)/(n-k)), starting at n=1.
Column 3 = A001711 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*(k+1)*3^k*stirling1(n+1, k+1)).
Column 4 = A001716 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*(k+1)*4^k*stirling1(n+1, k+1)).
Column 5 = A001721 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*binomial(k+1, 1)*5^k*stirling1(n+1, k+1)).
Column 6 = A051524 (2nd unsigned column of A051338) starting at n=1.
Column 7 = A051545 (2nd unsigned column of A051339) starting at n=1.
Column 8 = A051560 (2nd unsigned column of A051379) starting at n=1.
Column 9 = A051562 (2nd unsigned column of A051380) starting at n=1.
Column 10= A051564 (2nd unsigned column of A051523) starting at n=1.
2nd row is A005408 (2n - 1, starting at n=1).
3rd row is A080663 (3n^2 - 1, starting at n=1).
Main diagonal gives A384024.

Programs

  • Maple
    H := proc(n, k) option remember; if n = 0 then 1/k else add(H(n - 1, j), j = 1..k) fi end: A := (n, k) -> (n + 1)!*H(k, n + 1):
    # Alternative with standard harmonic number:
    A := (n, k) -> if k = 0 then n! else (harmonic(n + k) - harmonic(k - 1))*(n + k)! / (k - 1)! fi:
    for n from 0 to 7 do seq(A(n, k), k = 0..6) od;
    # Alternative with hypergeometric formula:
    A := (n, k) -> (n+1)*((n + k)! / k!)*hypergeom([-n, 1, 1], [2, k+1], 1):
    seq(print(seq(simplify(A(n, k)), k = 0..6)), n=0..7); # Peter Luschny, Jul 01 2022
  • Mathematica
    H[0, m_] := 1/m; H[n_, m_] := Sum[H[n - 1, k], {k, m}]; a[n_, m_] := m!H[n, m]; Flatten[ Table[ a[i, n - i], {n, 10}, {i, n - 1, 0, -1}]]
    Table[ a[n, m], {m, 8}, {n, 0, m + 1}] // TableForm (* to view the table *)
    (* Robert G. Wilson v, Jun 27 2005 *)
  • PARI
    a(n, k) = polcoef(prod(j=0, n, 1+(j+k)*x), n); \\ Seiichi Manyama, May 19 2025

Formula

A(n, k) = (Harmonic(n + k) - Harmonic(k - 1))*(n + k)!/(k - 1)! if k > 0, otherwise n!.
From Gerald McGarvey, Aug 27 2005, edited by Peter Luschny, Jul 02 2022: (Start)
E.g.f. for column k: -log(1 - x)/(x*(1 - x)^k).
Row 3 is r(n) = 4*n^3 + 18*n^2 + 22*n + 6.
Row 4 is r(n) = 5*n^4 + 40*n^3 + 105*n^2 + 100*n + 24.
Row 5 is r(n) = 6*n^5 + 75*n^4 + 340*n^3 + 675*n^2 + 548*n + 120.
Row 6 is r(n) = 7*n^6 + 126*n^5 + 875*n^4 + 2940*n^3 + 4872*n^2 + 3528*n + 720.
Row 7 is r(n) = 8*n^7 + 196*n^6 + 1932*n^5 + 9800*n^4 + 27076*n^3 + 39396*n^2 + 26136*n + 5040.
The sum of the polynomial coefficients for the n-th row is |S1(n, 2)|, which are the unsigned Stirling1 numbers which appear in column 1.
A(m, n) = Sum_{k=1..m} n*A094645(m, n)*(n+1)^(k-1). (A094645 is Generalized Stirling number triangle of first kind, e.g.f.: (1-y)^(1-x).) (End)
In Gerard McGarvey's formulas for the row coefficients we find Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; see A165674. - Johannes W. Meijer, Oct 16 2009
A(n, k) = (n + 1)*((n + k)! / k!)*hypergeom([-n, 1, 1], [2, k + 1], 1). - Peter Luschny, Jul 01 2022
A(n,k) = [x^n] Product_{j=0..n} (1 + (j+k)*x). - Seiichi Manyama, May 19 2025

Extensions

More terms from Robert G. Wilson v, Jun 27 2005
Edited by Peter Luschny, Jul 02 2022
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