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 1-4 of 4 results.

A226513 Array read by antidiagonals: T(n,k) = number of barred preferential arrangements of k things with n bars (k >=0, n >= 0).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 8, 13, 1, 4, 15, 44, 75, 1, 5, 24, 99, 308, 541, 1, 6, 35, 184, 807, 2612, 4683, 1, 7, 48, 305, 1704, 7803, 25988, 47293, 1, 8, 63, 468, 3155, 18424, 87135, 296564, 545835, 1, 9, 80, 679, 5340, 37625, 227304, 1102419, 3816548, 7087261
Offset: 0

Views

Author

N. J. A. Sloane, Jun 13 2013

Keywords

Comments

The terms of this sequence are also called high-order Fubini numbers (see p. 255 in Komatsu). - Stefano Spezia, Dec 06 2020

Examples

			Array begins:
  1  1   3   13    75    541     4683     47293     545835 ...
  1  2   8   44   308   2612    25988    296564    3816548 ...
  1  3  15   99   807   7803    87135   1102419   15575127 ...
  1  4  24  184  1704  18424   227304   3147064   48278184 ...
  1  5  35  305  3155  37625   507035   7608305  125687555 ...
  1  6  48  468  5340  69516  1014348  16372908  289366860 ...
  ...
Triangle begins:
  1,
  1, 1,
  1, 2, 3,
  1, 3, 8, 13,
  1, 4, 15, 44, 75,
  1, 5, 24, 99, 308, 541,
  1, 6, 35, 184, 807, 2612, 4683,
  1, 7, 48, 305, 1704, 7803, 25988, 47293,
  1, 8, 63, 468, 3155, 18424, 87135, 296564, 545835
  ........
[_Vincenzo Librandi_, Jun 18 2013]

References

  • Z.-R. Li, Computational formulae for generalized mth order Bell numbers and generalized mth order ordered Bell numbers (in Chinese), J. Shandong Univ. Nat. Sci. 42 (2007), 59-63.

Links

Crossrefs

Rows 0-5 give: A000670, A005649, A226515, A226738, A226739, A226740.
Columns 2, 3 = A005563, A226514.
Cf. A053492 (array diagonal), A265609, A346982.

Programs

  • Maple
    T:= (n, k)-> k!*coeff(series(1/(2-exp(x))^(n+1), x, k+1), x, k):
seq(seq(T(d-k, k), k=0..d), d=0..10);  # Alois P. Heinz, Mar 26 2016
  • Mathematica
    T[n_, k_] := Sum[StirlingS2[k, i]*i!*Binomial[n+i, i], {i, 0, k}]; Table[ T[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 26 2016 *)

Formula

T(n,k) = Sum_{i=0..k} S2_k(i)*i!*binomial(n+i,i), where S2_k(i) is the Stirling number of the second kind. - Jean-François Alcover, Mar 26 2016
T(n,k) = k! * [x^k] 1/(2-exp(x))^(n+1). - Alois P. Heinz, Mar 26 2016
Conjectural g.f. for row n as a continued fraction of Stieltjes type: 1/(1 - (n+1)*x/(1 - 2*x/(1 - (n+2)*x/(1 - 4*x/(1 - (n+3)*x/(1 - 6*x/(1 - ... ))))))). Cf. A265609. - Peter Bala, Aug 27 2023
From Seiichi Manyama, Nov 19 2023: (Start)
T(n,0) = 1; T(n,k) = Sum_{j=1..k} (n*j/k + 1) * binomial(k,j) * T(n,k-j).
T(n,0) = 1; T(n,k) = (n+1)*T(n,k-1) - 2*Sum_{j=1..k-1} (-1)^j * binomial(k-1,j) * T(n,k-j). (End)
G.f. for row n: (1/n!) * Sum_{m>=0} (n+m)! * x^m / Product_{j=1..m} (1 - j*x), for n >= 0. - Paul D. Hanna, Feb 01 2024

A226515 Row 2 of array in A226513.

Original entry on oeis.org

1, 3, 15, 99, 807, 7803, 87135, 1102419, 15575127, 242943723, 4145495055, 76797289539, 1534762643847, 32907617073243, 753473367606975, 18347287182129459, 473409784213526967, 12902366605394652363, 370357953441110390895, 11167936445234485414179
Offset: 0

Views

Author

N. J. A. Sloane, Jun 13 2013

Keywords

Links

Crossrefs

Cf. rows 0, 1, 3, 4, 5 of A226513: A000670, A005649, A226738, A226739, A226740.

Programs

  • Magma
    m:=2; [&+[StirlingSecond(n, i)*Factorial(i)*Binomial(m+i, i): i in [0..n]]: n in [0..20]]; // Bruno Berselli, Jun 18 2013
  • Mathematica
    Range[0, 20]! CoefficientList[Series[(2 - Exp@x)^-3, {x, 0, 20}], x] (* Vincenzo Librandi, Jun 18 2013 *)

Formula

E.g.f.: 1/(2 - exp(x))^3 (see the Ahlbach et al. paper, Theorem 4). - Vincenzo Librandi, Jun 18 2013
a(n) = Sum_{i=0..n} S2(n,i)*i!*binomial(2+i,i), where S2 is the Stirling number of the second kind (see the Ahlbach et al. paper, Theorem 3). [Bruno Berselli, Jun 18 2013]
G.f.: 1/Q(0), where Q(k) = 1 - 3*x*(k + 1) - 2*x^2*(k + 1)*(k + 3)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
G.f.: 1/(1 + x)/Q(0,u), where u = x/(1 + x), Q(k,u) = 1 - u*(3*k + 4) - 2*u^2*(k + 1)*(k + 3)/Q(k+1,u); (continued fraction). - Sergei N. Gladkovskii, Oct 03 2013
a(n) ~ n! * n^2 /(16*(log(2))^(n + 3)) * (1 + 3*(1 + log(2))/n). - Vaclav Kotesovec, Oct 08 2013
Conjectural g.f. as a continued fraction of Stieltjes type: 1/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 4*x/(1 - 5*x/(1 - 6*x/(1 - (n+2)*x/(1 - 2*n*x/(1 - ... ))))))))). - Peter Bala, Aug 27 2023
From Seiichi Manyama, Nov 19 2023: (Start)
a(0) = 1; a(n) = Sum_{k=1..n} (2*k/n + 1) * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 3*a(n-1) - 2*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). (End)

Extensions

More terms from Vincenzo Librandi, Jun 18 2013

A226738 Row 3 of array in A226513.

Original entry on oeis.org

1, 4, 24, 184, 1704, 18424, 227304, 3147064, 48278184, 812387704, 14872295784, 294192418744, 6251984167464, 142032703137784, 3434617880825064, 88075274293319224, 2387099326339205544, 68177508876215724664, 2046501717592969431144, 64408432189100396344504
Offset: 0

Views

Author

Vincenzo Librandi, Jun 18 2013

Keywords

Links

Crossrefs

Cf. rows 0, 1, 2, 4, 5 of A226513: A000670, A005649, A226515, A226739, A226740.

Programs

  • Magma
    m:=3; [&+[StirlingSecond(n,i)*Factorial(i)*Binomial(m+i,i): i in [0..n]]: n in [0..20]]; // Bruno Berselli, Jun 18 2013
  • Mathematica
    Range[0, 20]! CoefficientList[Series[(2 - Exp@x)^-4, {x, 0, 20}], x]

Formula

E.g.f.: 1/(2 - exp(x))^4 (see the Ahlbach et al. paper, Theorem 4).
a(n) = sum( S2(n,i)*i!*binomial(3+i,i), i=0..n ), where S2 is the Stirling number of the second kind (see the Ahlbach et al. paper, Theorem 3). [Bruno Berselli, Jun 18 2013]
G.f.: 1/T(0), where T(k) = 1 - x*(k+4)/(1 - 2*x*(k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 28 2013
a(n) ~ n! * n^3 / (96 * log(2)^(n+4)). - Vaclav Kotesovec, Oct 11 2022
Conjectural g.f. as a continued fraction of Stieltjes type: 1/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 4*x/(1 - 6*x/(1 - 6*x/(1 - (n+3)*x/(1 - 2*n*x/(1 - ... ))))))))). - Peter Bala, Aug 27 2023
From Seiichi Manyama, Nov 19 2023: (Start)
a(0) = 1; a(n) = Sum_{k=1..n} (3*k/n + 1) * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 4*a(n-1) - 2*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). (End)

A226740 Row 5 of array in A226513.

Original entry on oeis.org

1, 6, 48, 468, 5340, 69516, 1014348, 16372908, 289366860, 5553635436, 114964523148, 2552305112748, 60474398655180, 1522843616043756, 40605864407444748, 1142786353739186988, 33848016050071188300, 1052381222812017946476, 34266937867683980363148, 1166071764343727862515628
Offset: 0

Views

Author

Vincenzo Librandi, Jun 18 2013

Keywords

Links

Crossrefs

Cf. rows 0, 1, 2, 3, 4 of A226513: A000670, A005649, A226515, A226738, A226739.

Programs

  • Magma
    m:=5; [&+[StirlingSecond(n, i)*Factorial(i)*Binomial(m+i, i): i in [0..n]]: n in [0..20]];
  • Mathematica
    Range[0, 20]! CoefficientList[Series[(2 - Exp@x)^-6, {x, 0, 20}], x]

Formula

E.g.f.: 1/(2 - exp(x))^6 (see the Ahlbach et al. paper, Theorem 4).
a(n) = Sum_{i=0..n} S2(n,i)*i!*binomial(5+i,i), where S2 is the Stirling number of the second kind (see the Ahlbach et al. paper, Theorem 3).
a(n) ~ n! * n^5 / (7680 * log(2)^(n+6)). - Vaclav Kotesovec, Oct 11 2022
Conjectural g.f. as a continued fraction of Stieltjes type: 1/(1 - 6*x/(1 - 2*x/(1 - 7*x/(1 - 4*x/(1 - 8*x/(1 - 6*x/(1 - (n+5)*x/(1 - 2*n*x/(1 - ... ))))))))). - Peter Bala, Aug 27 2023
From Seiichi Manyama, Nov 19 2023: (Start)
a(0) = 1; a(n) = Sum_{k=1..n} (5*k/n + 1) * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 6*a(n-1) - 2*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). (End)
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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