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 1-10 of 126 results. Next

A028421 Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 6, 22, 18, 4, 24, 100, 105, 40, 5, 120, 548, 675, 340, 75, 6, 720, 3528, 4872, 2940, 875, 126, 7, 5040, 26136, 39396, 27076, 9800, 1932, 196, 8, 40320, 219168, 354372, 269136, 112245, 27216, 3822, 288, 9
Offset: 0

Views

Author

Peter Wiggen (wiggen(AT)math.psu.edu)

Keywords

Comments

Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.
We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that 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 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.
The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).
(End)
For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - Tom Copeland, Feb 28 2014
From Wolfdieter Lang, Nov 25 2018: (Start)
The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).
The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).
The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)
The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
n\k       0        1        2        3        4       5       6      7     8   9 10
------------------------------------------------------------------------------------
0:        1
1:        1        2
2:        2        6        3
3:        6       22       18        4
4:       24      100      105       40        5
5:      120      548      675      340       75       6
6:      720     3528     4872     2940      875     126       7
7:     5040    26136    39396    27076     9800    1932     196      8
8:    40320   219168   354372   269136   112245   27216    3822    288     9
9:   362880  2053152  3518100  2894720  1346625  379638   66150   6960   405  10
10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11
... - _Wolfdieter Lang_, Nov 23 2018
		

Crossrefs

Row sums give A000254(n+1), n >= 0.
Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.
From Johannes W. Meijer, Oct 07 2009: (Start)
A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.
The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).
Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

Programs

  • Maple
    A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:
    seq(seq(A028421(n,k), k=0..n), n=0..8);
    # Johannes W. Meijer, Oct 07 2009, Revised Sep 09 2012
    egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):
    ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):
    seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # Peter Luschny, Jun 12 2022
  • Mathematica
    f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(-ln(1 - x), 10, True) # Peter Luschny, Jan 03 2019

Formula

E.g.f.: d/dt(-log(1-t)/(1-t)^x). - Vladeta Jovovic, Oct 12 2003
The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - Peter Bala, Jul 18 2013
Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - Wolfdieter Lang, Nov 25 2018

Extensions

Edited by Wolfdieter Lang, Nov 23 2018

A269953 Triangle read by rows: T(n, k) = Sum_{j=0..n} binomial(-j-1, -n-1)*S1(j, k) where S1 are the Stirling cycle numbers A132393.

Original entry on oeis.org

1, -1, 1, 1, -1, 1, -1, 2, 0, 1, 1, 0, 5, 2, 1, -1, 9, 15, 15, 5, 1, 1, 35, 94, 85, 40, 9, 1, -1, 230, 595, 609, 315, 91, 14, 1, 1, 1624, 4458, 4844, 2779, 924, 182, 20, 1, -1, 13209, 37590, 43238, 26817, 9975, 2310, 330, 27, 1
Offset: 0

Views

Author

Peter Luschny, Apr 12 2016

Keywords

Comments

Replacing the Stirling cycle numbers in the definition by the Stirling set numbers leads to A105794.
From Wolfdieter Lang, Jun 19 2017: (Start)
The triangle t(n, k) = (-1)^(n-k)*T(n, k) is the matrix product of P = A007318 (Pascal) and s1 = A048994 (signed Stirling1). This is Sheffer (exp(t), log(1+t)).
The present triangle T is therefore the Sheffer triangle (exp(-t), -log(1-t)). Note that P is Sheffer (exp(t), t) (of the Appell type). (End)
The triangle T(n,k) is a representative of the parametric family of triangles T(m,n,k), whose columns are the coefficients of the standard expansion of the function f(x) = (-log(1-x))^(k)*exp(-m*x)/k! for the case m=1. See A381082. - Igor Victorovich Statsenko, Feb 14 2025

Examples

			Triangle starts:
   1;
  -1,  1;
   1, -1,  1;
  -1,  2,  0,  1;
   1,  0,  5,  2,  1;
  -1,  9, 15, 15,  5,  1;
   1, 35, 94, 85, 40,  9,  1.
		

Crossrefs

Columns k=0..4 give A033999, A002741, A381064, A381065, A381066.
Cf. A000166 (row sums), A080956 (diag n,n-1).
KummerU(-n,1-n-x,z): this sequence (z=-1), A094816 (z=1), |A137346| (z=2), A327997 (z=3).

Programs

  • Maple
    A269953 := (n,k) -> add(binomial(-j-1,-n-1)*abs(Stirling1(j,k)), j=0..n):
    seq(print(seq(A269953(n, k), k=0..n)), n=0..9);
    # Alternative:
    egf := exp(-t)*(1-t)^(-x): ser := series(egf, t, 12): p := n -> coeff(ser, t, n):
    seq(n!*seq(coeff(p(n), x, k), k=0..n), n=0..9); # Peter Luschny, Oct 28 2019
  • Mathematica
    Flatten[Table[Sum[Binomial[-j-1,-n-1] Abs[StirlingS1[j,k]], {j,0,n}], {n,0,9},{k,0,n}]]
    (* Or: *)
    p [n_] := HypergeometricU[-n, 1 - n - x, -1];
    Table[CoefficientList[p[n], x], {n, 0, 9}] (* Peter Luschny, Oct 28 2019 *)

Formula

From Wolfdieter Lang, Jun 19 2017: (Start)
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n,k)*x^k: exp(-t)/(1 - t)^x.
E.g.f. of column k sequence: exp(-x)*(-log(1-x))^k/k!, k >= 0. (End)
From Peter Bala, Oct 26 2019: (Start)
Let R(n, x) = (-1)^n*Sum_{k >= 0} binomial(n,k)*k!* binomial(-x,k) the n-th row polynomial of this triangle.
R(n, x) = c_n(-x;-1), where c_n(x;a) denotes the n-th Poisson Charlier polynomial.
The series representation e = Sum_{k >= 0} 1/k! is the case n = 0 of the more general result e = n!*Sum_{k >= 0} 1/(k!*R(n,k)*R(n,k+1)), n = 0,2,3,4,.... (End)
R(n, x) = KummerU(-n, 1-n-x, -1). - Peter Luschny, Oct 28 2019

A269951 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j,-n)*S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 5, 5, 1, 0, 16, 23, 9, 1, 0, 65, 116, 65, 14, 1, 0, 326, 669, 470, 145, 20, 1, 0, 1957, 4429, 3634, 1415, 280, 27, 1, 0, 13700, 33375, 30681, 14084, 3535, 490, 35, 1, 0, 109601, 283072, 284066, 147532, 43939, 7756, 798, 44, 1
Offset: 0

Views

Author

Peter Luschny, Apr 10 2016

Keywords

Examples

			Triangle begins:
  1;
  0,   1;
  0,   2,   1;
  0,   5,   5,   1;
  0,  16,  23,   9,   1;
  0,  65, 116,  65,  14,  1;
  0, 326, 669, 470, 145, 20, 1;
		

Crossrefs

A000522 (col. 1), A073596 (col. 2), A000096 (diag. n-1), A241765 (diag. n-2).
A001339 (row sums), A137597 (unsigned matrix inverse).

Programs

  • Maple
    A269951 := (n,k) -> add((-1)^(n-j)*binomial(-j,-n)*abs(Stirling1(j,k)), j=0..n):
    seq(seq(A269951(n,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[ Table[ Sum[(-1)^(n-j) Binomial[-j,-n] Abs[StirlingS1[j,k]], {j,0,n}], {n,0,9}, {k,0,n}]]

A254882 Triangle read by rows, T(n,k) = Sum_{j=0..k-1} S(n,j+1)*S(n,k-j) where S denotes the Stirling cycle numbers A132393, T(0,0)=1, n>=0, 0<=k<=2n-1.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 1, 0, 4, 12, 13, 6, 1, 0, 36, 132, 193, 144, 58, 12, 1, 0, 576, 2400, 4180, 3980, 2273, 800, 170, 20, 1, 0, 14400, 65760, 129076, 143700, 100805, 46710, 14523, 3000, 395, 30, 1, 0, 518400, 2540160, 5450256, 6787872, 5482456, 3034920, 1184153
Offset: 0

Views

Author

Peter Luschny, Feb 10 2015

Keywords

Examples

			[1]
[0, 1]
[0, 1, 2, 1]
[0, 4, 12, 13, 6, 1]
[0, 36, 132, 193, 144, 58, 12, 1]
[0, 576, 2400, 4180, 3980, 2273, 800, 170, 20, 1]
		

Crossrefs

Programs

  • Maple
    a := n -> (x^n*pochhammer(1+1/x,n))^2:
    c := (n,k) -> coeff(expand(a(n)),x,n-k):
    for n from 0 to 5 do: `if`(n=0,[1],[seq(c(n-1,k),k=-n..n-1)]) od;
    # Second program, a special case of the recurrence given in A246117:
    t := proc(n,k) option remember; if n=0 and k=0 then 1 elif
    k <= 0 or k>n then 0 else iquo(n,2)*t(n-1,k)+t(n-1,k-1) fi end:
    A254882 := (n,k) -> `if`(n=0,1,t(2*n-1,k)):
    seq(print(seq(A254882(n,k), k=0..max(0,2*n-1))), n=0..5);
  • Mathematica
    Flatten[{1,Table[Table[Sum[Abs[StirlingS1[n,j+1]] * Abs[StirlingS1[n,k-j]],{j,0,k-1}],{k,0,2*n-1}],{n,1,10}]}] (* Vaclav Kotesovec, Feb 10 2015 *)
  • Sage
    def A254882(n,k):
        if n == 0: return 1
        return sum(stirling_number1(n,j+1)*stirling_number1(n,k-j) for j in range(k))
    for n in range (5): [A254882(n,k) for k in (0..max(0,2*n-1))]

Formula

T(n+1, n+1) = A129256(n) for n>=0.

A254881 Triangle read by rows, T(n,k) = sum(j=0..k-1, S(n+1,j+1)*S(n,k-j)) where S denotes the Stirling cycle numbers A132393, T(0,0)=1, n>=0, 0<=k<=2n.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 5, 4, 1, 0, 12, 40, 51, 31, 9, 1, 0, 144, 564, 904, 769, 376, 106, 16, 1, 0, 2880, 12576, 23300, 24080, 15345, 6273, 1650, 270, 25, 1, 0, 86400, 408960, 840216, 991276, 748530, 381065, 133848, 32523, 5370, 575, 36, 1, 0, 3628800, 18299520
Offset: 0

Views

Author

Peter Luschny, Feb 10 2015

Keywords

Comments

These are also the coefficients of the polynomials interpolating the sequence k -> n!*((n+k)!/k!)*binomial(n+k-1,k-1) (for fixed n>=0). Divided by n! these polynomials generate the rows of Lah numbers L(n+k, k) = ((n+k)!/k!)* binomial(n+k-1,k-1).

Examples

			[1]
[0, 1, 1]
[0, 2, 5, 4, 1]
[0, 12, 40, 51, 31, 9, 1]
[0, 144, 564, 904, 769, 376, 106, 16, 1]
[0, 2880, 12576, 23300, 24080, 15345, 6273, 1650, 270, 25, 1]
For example in the case n=3 the polynomial (k^6+9*k^5+31*k^4+51*k^3+40*k^2+12*k)/3! generates the Lah numbers 0, 24, 240, 1200, 4200, 11760, 28224, ... (A253285).
		

Crossrefs

The sequences A000012, A002378, A083374, A253285 are the Lah number rows generated by the polynomials divided by n! for n=0, 1, 2, 3 respectivly.

Programs

  • Maple
    # This is a special case of the recurrence given in A246117.
    t := proc(n,k) option remember; if n=0 and k=0 then 1 elif
    k <= 0 or k>n then 0 else iquo(n,2)*t(n-1,k)+t(n-1,k-1) fi end:
    A254881 := (n,k) -> t(2*n,k):
    seq(print(seq(A254881(n,k), k=0..2*n)), n=0..5);
    # Illustrating the comment:
    restart: with(PolynomialTools): with(CurveFitting): for N from 0 to 5 do
    CoefficientList(PolynomialInterpolation([seq([k,N!*((N+k)!/k!)*binomial(N+k-1,k-1)], k=0..2*N)], n), n) od;
  • Mathematica
    Flatten[{1,Table[Table[Sum[Abs[StirlingS1[n+1,j+1]] * Abs[StirlingS1[n,k-j]],{j,0,k-1}],{k,0,2*n}],{n,1,10}]}] (* Vaclav Kotesovec, Feb 10 2015 *)
  • Sage
    def T(n,k):
        if n == 0: return 1
        return sum(stirling_number1(n+1,j+1)*stirling_number1(n,k-j) for j in range(k))
    for n in range (6): [T(n,k) for k in (0..2*n)]

Formula

T(n, n) = A187235(n) for n>=1 (after the explicit formula of Vaclav Kotesovec).

A269954 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(-j,-n)*S1(j,k), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 2, 5, 3, 1, 0, 9, 20, 17, 6, 1, 0, 44, 109, 100, 45, 10, 1, 0, 265, 689, 694, 355, 100, 15, 1, 0, 1854, 5053, 5453, 3094, 1015, 196, 21, 1, 0, 14833, 42048, 48082, 29596, 10899, 2492, 350, 28, 1
Offset: 0

Views

Author

Peter Luschny, Apr 12 2016

Keywords

Examples

			Triangle starts:
  1;
  0,   1;
  0,   0,   1;
  0,   1,   1,   1;
  0,   2,   5,   3,   1;
  0,   9,  20,  17,   6,   1;
  0,  44, 109, 100,  45,  10,  1;
  0, 265, 689, 694, 355, 100, 15, 1;
		

Crossrefs

A000255 (row sums), A000217 (diag. n,n-1), A133252 (diag. n,n-2).
Columns k=0..4 give A000007, A000166(n-1), A300490(n-1), A381067(n-1), A381068(n-1).

Programs

  • Maple
    A269954 := (n, k) -> add(binomial(-j, -n)*abs(Stirling1(j, k)), j=0..n):
    seq(seq(A269954(n, k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Sum[Binomial[-j,-n] Abs[StirlingS1[j,k]],{j,0,n}], {n,0,9},{k,0,n}]]
  • PARI
    T(n, k) = sum(j=0, n, (-1)^(n-j)*binomial(n-1, n-j)*abs(stirling(j, k)));
    for(n=0, 9, for(k=0, n, print1(T(n, k), ", "))); \\ Seiichi Manyama, Feb 13 2025

A308305 a(n) = s(n,n) + s(n,n-1) + s(n,n-2), where s(n,k) are the unsigned Stirling numbers of the first kind (see A132393).

Original entry on oeis.org

1, 2, 6, 18, 46, 101, 197, 351, 583, 916, 1376, 1992, 2796, 3823, 5111, 6701, 8637, 10966, 13738, 17006, 20826, 25257, 30361, 36203, 42851, 50376, 58852, 68356, 78968, 90771, 103851, 118297, 134201, 151658, 170766, 191626, 214342, 239021, 265773, 294711
Offset: 1

Views

Author

Keywords

Comments

Pairwise perpendicular bisectors divide the Euclidean plane into a maximum of a(n) regions. This maximum value a(n) occurs when no three points are collinear and no four points are concyclic in the plane, and with no perpendicular bisectors parallel or coinciding [Zaslavsky, Eq. (1.1)]. This count of regions in the plane is relevant for social science applications to voting preferences based on proximity to candidates on issues.

References

  • T. Zaslavsky, Perpendicular dissections of space. Discrete Comput. Geom. 27 (2002), no. 3, 303-351.

Crossrefs

The unsigned Stirling numbers of the first kind s(n,k) are given in A132393.
The division of space formulation can be generalized to higher dimensions with use of A008275 by Good and Tideman's work.
The maximum number of regions generated by pairwise perpendicular bisectors on a sphere is given by A087645.

Programs

  • Magma
    [(1/24)*(24 - 14*n + 21*n^2 - 10*n^3 + 3*n^4): n in [1..40]]; // Vincenzo Librandi, Jun 30 2019
    
  • Mathematica
    Table[(1/24)(24 - 14 i + 21 i^2 - 10 i^3 + 3 i^4), {i, 40}]
  • PARI
    Vec(x*(1 - 3*x + 6*x^2 - 2*x^3 + x^4) / (1 - x)^5 + O(x^40)) \\ Colin Barker, Jun 30 2019

Formula

a(n) = s(n,n) + s(n,n-1) + s(n,n-2), where s(n,k) are the unsigned Stirling numbers of the first kind.
a(n) = (1/24)*(24 - 14*n + 21*n^2 - 10*n^3 + 3*n^4).
From Colin Barker, Jun 30 2019: (Start)
G.f.: x*(1 - 3*x + 6*x^2 - 2*x^3 + x^4) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
(End)

A354797 Triangle read by rows. T(n, k) = |Stirling1(n, k)| * Stirling2(n + k, n) = A132393(n, k) * A048993(n + k, n).

Original entry on oeis.org

1, 0, 1, 0, 3, 7, 0, 12, 75, 90, 0, 60, 715, 2100, 1701, 0, 360, 7000, 36750, 69510, 42525, 0, 2520, 72884, 595350, 1940295, 2692305, 1323652, 0, 20160, 814968, 9549120, 47030445, 109794300, 120023904, 49329280, 0, 181440, 9801000, 156008160, 1076453763, 3723239520, 6733767040, 6065579520, 2141764053
Offset: 0

Views

Author

Peter Luschny, Jun 06 2022

Keywords

Examples

			Table T(n, k) begins:
[0] 1
[1] 0,     1
[2] 0,     3,      7
[3] 0,    12,     75,      90
[4] 0,    60,    715,    2100,     1701
[5] 0,   360,   7000,   36750,    69510,     42525
[6] 0,  2520,  72884,  595350,  1940295,   2692305,   1323652
[7] 0, 20160, 814968, 9549120, 47030445, 109794300, 120023904, 49329280
		

Crossrefs

Programs

  • Maple
    T := (n, k) -> abs(Stirling1(n, k))*Stirling2(n + k, n):
    for n from 0 to 6 do seq(T(n, k), k = 0..n) od;

Formula

Sum_{k=0..n} (-1)^(n - k)*T(n, k) = n^n. - Werner Schulte, Jun 03 2022 in A000312. [Formerly a conjecture, now proved by Mike Earnest, see link.]
T(n, k) = A132393(n, k) * A354977(n, k) = (1/n!) * Sum_{j=0..n} (-1)^(j + k) * binomial(n, j) * Stirling1(n, k) * j^(n + k).

A271700 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j-1,-n-1)*S1(k,j), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 16, 1, 4, 10, 30, 115, 1, 5, 15, 50, 205, 1021, 1, 6, 21, 77, 336, 1750, 10696, 1, 7, 28, 112, 518, 2814, 17766, 128472, 1, 8, 36, 156, 762, 4308, 28050, 207942, 1734447, 1, 9, 45, 210, 1080, 6342, 42528, 322860, 2746815, 25937683
Offset: 0

Views

Author

Peter Luschny, Apr 14 2016

Keywords

Examples

			Triangle starts:
[1]
[1, 1]
[1, 2, 3]
[1, 3, 6,  16]
[1, 4, 10, 30,  115]
[1, 5, 15, 50,  205, 1021]
[1, 6, 21, 77,  336, 1750, 10696]
[1, 7, 28, 112, 518, 2814, 17766, 128472]
		

Crossrefs

A000027 (col. 1), A000217, A161680 (col. 2), A005581 (col. 3), A211210 (diag. n,n), A211211 (diag. n,n-1).

Programs

  • Maple
    T := (n,k) -> add(abs(Stirling1(k,j))*binomial(-j-1,-n-1)*(-1)^(n-j),j=0..n);
    seq(seq(T(n,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Sum[(-1)^(n-j)Binomial[-j-1,-n-1] Abs[StirlingS1[k,j]],{j,0,n}], {n,0,9},{k,0,n}]]

A368583 Table read by rows: T(n, k) = A124320(n + 1, k) * A132393(n, k).

Original entry on oeis.org

1, 0, 2, 0, 3, 12, 0, 8, 60, 120, 0, 30, 330, 1260, 1680, 0, 144, 2100, 11760, 30240, 30240, 0, 840, 15344, 113400, 428400, 831600, 665280, 0, 5760, 127008, 1169280, 5821200, 16632000, 25945920, 17297280, 0, 45360, 1176120, 13000680, 80415720, 302702400, 696215520, 908107200, 518918400
Offset: 0

Views

Author

Peter Luschny, Jan 10 2024

Keywords

Examples

			Triangle starts:
  [0] [1]
  [1] [0,   2]
  [2] [0,   3,   12]
  [3] [0,   8,   60,     120]
  [4] [0,  30,   330,   1260,   1680]
  [5] [0, 144,  2100,  11760,  30240,  30240]
  [6] [0, 840, 15344, 113400, 428400, 831600, 665280]
		

Crossrefs

Cf. A124320 (rising factorial), A132393 (unsigned Stirling1), A001813 (main diagonal), A052819 (row sums), A227457 (alternating row sums), A368584.

Programs

  • SageMath
    def Trow(n): return [rising_factorial(n+1, k)*stirling_number1(n, k)
                         for k in range(n+1)]
    for n in range(7): print(Trow(n))
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