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 13 results. Next

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.

Crossrefs

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

A305404 Expansion of Sum_{k>=0} (2*k - 1)!!*x^k/Product_{j=1..k} (1 - j*x).

Original entry on oeis.org

1, 1, 4, 25, 217, 2416, 32839, 527185, 9761602, 204800551, 4801461049, 124402647370, 3529848676237, 108859319101261, 3625569585663484, 129689000146431205, 4958830249864725997, 201834650901695603296, 8712774828941647677019, 397596632650906687905565
Offset: 0

Views

Author

Ilya Gutkovskiy, May 31 2018

Keywords

Comments

Stirling transform of A001147.

Crossrefs

Programs

  • Maple
    b:= proc(n, m) option remember;
         `if`(n=0, doublefactorial(2*m-1), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..23);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    nmax = 19; CoefficientList[Series[Sum[(2 k - 1)!! x^k/Product[1 - j x, {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x]
    nmax = 19; CoefficientList[Series[1/Sqrt[3 - 2 Exp[x]], {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS2[n, k] (2 k - 1)!!, {k, 0, n}], {n, 0, 19}]

Formula

E.g.f.: 1/sqrt(3 - 2*exp(x)).
a(n) = Sum_{k=0..n} Stirling2(n,k)*(2*k - 1)!!.
a(n) ~ sqrt(2/3) * n^n / ((log(3/2))^(n + 1/2) * exp(n)). - Vaclav Kotesovec, Jul 01 2018
Conjecture: a(n) = Sum_{k>=0} k^n * binomial(2*k,k) / (2^k * 3^(k + 1/2)). - Diego Rattaggi, Oct 11 2020
O.g.f. conjectural: 1/(1 - x/(1 - 3*x/(1 - 3*x/(1 - 6*x/(1 - 5*x/(1 - 9*x/(1 - 7*x/(1 - ... - (2*n-1)*x/(1 - 3*n*x/(1 - ... )))))))))) - a continued fraction of Stieltjes-type. - Peter Bala, Dec 06 2020
a(0) = 1; a(n) = Sum_{k=1..n} (2 - k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
a(0) = 1; a(n) = a(n-1) - 3*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023

A346985 Expansion of e.g.f. 1 / (7 - 6 * exp(x))^(1/6).

Original entry on oeis.org

1, 1, 8, 113, 2325, 62896, 2109143, 84403033, 3924963750, 207976793991, 12369246804853, 815880360117978, 59107920881218525, 4665585774576259261, 398534278371999103888, 36627974592437584634573, 3603954453161886215458025, 377983931878997401821759456, 42095013846928585982896180123
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Comments

Stirling transform of A008542.
In general, for k >= 1, if e.g.f. = 1 / (k + 1 - k*exp(x))^(1/k), then a(n) ~ n! / (Gamma(1/k) * (k+1)^(1/k) * n^(1 - 1/k) * log(1 + 1/k)^(n + 1/k)). - Vaclav Kotesovec, Aug 14 2021

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (6*n-5)*g(n-1)) end:
    b:= proc(n, m) option remember;
         `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..18);  # Alois P. Heinz, Aug 09 2021
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(7 - 6 Exp[x])^(1/6), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS2[n, k] 6^k Pochhammer[1/6, k], {k, 0, n}], {n, 0, 18}]
  • Maxima
    a[n]:=if n=0 then 1 else (1/n)*sum(binomial(n,k)*(n+5*k)*a[k],k,0,n-1);
    makelist(a[n],n,0,50); /* Tani Akinari, Aug 22 2023 */

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k) * A008542(k).
a(n) ~ n! / (Gamma(1/6) * 7^(1/6) * n^(5/6) * log(7/6)^(n + 1/6)). - Vaclav Kotesovec, Aug 14 2021
For n > 0, a(n) = (1/n)*Sum_{k=0..n-1} binomial(n,k)*(n+5*k)*a(k). - Tani Akinari, Aug 22 2023
O.g.f. (conjectural): 1/(1 - x/(1 - 7*x/(1 - 7*x/(1 - 14*x/(1 - 13*x/(1 - 21*x/(1 - ... - (6*n-5)*x/(1 - 7*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type (S-fraction). - Peter Bala, Aug 25 2023
a(0) = 1; a(n) = a(n-1) - 7*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 17 2023

A346984 Expansion of e.g.f. 1 / (6 - 5 * exp(x))^(1/5).

Original entry on oeis.org

1, 1, 7, 85, 1495, 34477, 983983, 33476437, 1322441575, 59492222077, 3002578396255, 168005805229285, 10321907081030167, 690761732852321677, 50015387402165694607, 3895721046926471861365, 324805103526730206129607, 28861947117644330678207389, 2722944810091827410698112959
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Comments

Stirling transform of A008548.

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (5*n-4)*g(n-1)) end:
    b:= proc(n, m) option remember;
         `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..18);  # Alois P. Heinz, Aug 09 2021
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(6 - 5 Exp[x])^(1/5), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS2[n, k] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k) * A008548(k).
a(n) ~ n! / (Gamma(1/5) * 6^(1/5) * n^(4/5) * log(6/5)^(n + 1/5)). - Vaclav Kotesovec, Aug 14 2021
O.g.f. (conjectural): 1/(1 - x/(1 - 6*x/(1 - 6*x/(1 - 12*x/(1 - 11*x/(1 - 18*x/(1 - ... - (5*n-4)*x/(1 - 6*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type. - Peter Bala, Aug 22 2023
a(0) = 1; a(n) = Sum_{k=1..n} (5 - 4*k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
a(0) = 1; a(n) = a(n-1) - 6*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023

A346983 Expansion of e.g.f. 1 / (5 - 4 * exp(x))^(1/4).

Original entry on oeis.org

1, 1, 6, 61, 891, 16996, 400251, 11217781, 364638336, 13486045291, 559192836771, 25691965808026, 1295521405067181, 71131584836353861, 4224255395774155566, 269791923787785076921, 18439806740525320993551, 1342957106015632474616956, 103824389511747541791086511
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Comments

Stirling transform of A007696.

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (4*n-3)*g(n-1)) end:
    b:= proc(n, m) option remember;
         `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..18);  # Alois P. Heinz, Aug 09 2021
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(5 - 4 Exp[x])^(1/4), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS2[n, k] 4^k Pochhammer[1/4, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k) * A007696(k).
a(n) ~ n! / (Gamma(1/4) * 5^(1/4) * n^(3/4) * log(5/4)^(n + 1/4)). - Vaclav Kotesovec, Aug 14 2021
O.g.f. (conjectural): 1/(1 - x/(1 - 5*x/(1 - 5*x/(1 - 10*x/(1 - 9*x/(1 - 15*x/(1 - ... - (4*n-3)*x/(1 - 5*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type. - Peter Bala, Aug 22 2023
a(0) = 1; a(n) = Sum_{k=1..n} (4 - 3*k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
a(0) = 1; a(n) = a(n-1) - 5*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023

A352117 Expansion of e.g.f. 1/sqrt(2 - exp(2*x)).

Original entry on oeis.org

1, 1, 5, 37, 377, 4921, 78365, 1473277, 31938737, 784384561, 21523937525, 652667322517, 21672312694697, 782133969325801, 30481907097849485, 1275870745561131757, 57083444567425884257, 2718602143583362124641, 137315150097164841942245
Offset: 0

Views

Author

Seiichi Manyama, Mar 05 2022

Keywords

Crossrefs

Programs

  • Mathematica
    m = 18; Range[0, m]! * CoefficientList[Series[(2 - Exp[2*x])^(-1/2), {x, 0, m}], x] (* Amiram Eldar, Mar 05 2022 *)
  • Maxima
    a[n]:=if n=0 then 1 else sum(a[n-k]*(1-k/n/2)*binomial(n,k)*2^k,k,1,n);
    makelist(a[n],n,0,50); /* Tani Akinari, Sep 06 2023 */
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(2-exp(2*x))))
    
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*prod(j=0, k-1, 2*j+1)*stirling(n, k, 2));
    

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * (Product_{j=0..k-1} (2*j+1)) * Stirling2(n,k).
a(n) ~ 2^n * n^n / (log(2)^(n + 1/2) * exp(n)). - Vaclav Kotesovec, Mar 05 2022
Conjectural o.g.f. as a continued fraction of Stieltjes type: 1/(1 - x/(1 - 4*x/(1 - 3*x/(1 - 8*x/(1 - ... - (2*n-1)*x/(1 - 4*n*x/(1 - ... ))))))). Cf. A346982. - Peter Bala, Aug 22 2023
For n > 0, a(n) = Sum_{k=1..n} a(n-k)*(1-k/n/2)*binomial(n,k)*2^k. - Tani Akinari, Sep 06 2023
a(0) = 1; a(n) = a(n-1) - 2*Sum_{k=1..n-1} (-2)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 18 2023

A347015 Expansion of e.g.f. 1 / (1 + 3 * log(1 - x))^(1/3).

Original entry on oeis.org

1, 1, 5, 42, 498, 7644, 144156, 3225648, 83536008, 2457701928, 80970232104, 2953056534768, 118112744060208, 5140622709134496, 241863782829704928, 12232551538417012992, 661818290353375962240, 38140594162828447248000, 2332567001993176540206720, 150880256846462633823648000
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (3*n-2)*g(n-1)) end:
    a:= n-> add(abs(Stirling1(n, k))*g(k), k=0..n):
    seq(a(n), n=0..19);  # Alois P. Heinz, Aug 10 2021
  • Mathematica
    nmax = 19; CoefficientList[Series[1/(1 + 3 Log[1 - x])^(1/3), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Abs[StirlingS1[n, k]] 3^k Pochhammer[1/3, k], {k, 0, n}], {n, 0, 19}]

Formula

a(n) = Sum_{k=0..n} |Stirling1(n,k)| * A007559(k).
a(n) ~ n! * exp(n/3) / (Gamma(1/3) * 3^(1/3) * n^(2/3) * (exp(1/3) - 1)^(n + 1/3)). - Vaclav Kotesovec, Aug 14 2021
a(0) = 1; a(n) = Sum_{k=1..n} (3 - 2*k/n) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023

A365558 Expansion of e.g.f. 1 / (4 - 3 * exp(x))^(2/3).

Original entry on oeis.org

1, 2, 12, 112, 1432, 23272, 458952, 10644552, 283851272, 8555351112, 287585280392, 10666369505992, 432674936431112, 19054822031194952, 905387807689821832, 46166008179076287432, 2514469578906179506952, 145691888630159515550792
Offset: 0

Views

Author

Seiichi Manyama, Sep 09 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Product[3*j + 2, {j, 0, k - 1}] * StirlingS2[n, k], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Sep 11 2023 *)
  • PARI
    a(n) = sum(k=0, n, prod(j=0, k-1, 3*j+2)*stirling(n, k, 2));

Formula

a(n) = Sum_{k=0..n} (Product_{j=0..k-1} (3*j+2)) * Stirling2(n,k).
a(0) = 1; a(n) = Sum_{k=1..n} (3 - k/n) * binomial(n,k) * a(n-k).
a(n) ~ sqrt(3) * Gamma(1/3) * n^(n + 1/6) / (sqrt(Pi) * 2^(11/6) * exp(n) * log(4/3)^(n + 2/3)). - Vaclav Kotesovec, Nov 11 2023
a(0) = 1; a(n) = 2*a(n-1) - 4*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023

A347020 Expansion of e.g.f. 1 / (1 - 3 * log(1 + x))^(1/3).

Original entry on oeis.org

1, 1, 3, 18, 150, 1644, 22116, 353856, 6554376, 138001896, 3254445144, 84979363248, 2433814616592, 75858381808416, 2556180134677152, 92597465283789312, 3588434497019272320, 148134619713440384640, 6489652665043455707520, 300712023388466713739520
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 19; CoefficientList[Series[1/(1 - 3 Log[1 + x])^(1/3), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS1[n, k] 3^k Pochhammer[1/3, k], {k, 0, n}], {n, 0, 19}]

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * A007559(k).
a(n) ~ n! * exp(1/9) / (Gamma(1/3) * 3^(1/3) * n^(2/3) * (exp(1/3) - 1)^(n + 1/3)). - Vaclav Kotesovec, Aug 14 2021
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (3 - 2*k/n) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 11 2023

A375949 Expansion of e.g.f. 1 / (4 - 3 * exp(x))^(4/3).

Original entry on oeis.org

1, 4, 32, 368, 5520, 102064, 2242832, 57095728, 1652211600, 53559908784, 1922581295632, 75700072208688, 3243905700776080, 150289130386531504, 7485459789379535632, 398857142195958963248, 22639650637589839298960, 1363772478150606703714224
Offset: 0

Views

Author

Seiichi Manyama, Sep 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    nmax=17; CoefficientList[Series[1 / (4 - 3 * Exp[x])^(4/3),{x,0,nmax}],x]*Range[0,nmax]! (* Stefano Spezia, Sep 03 2024 *)
  • PARI
    a007559(n) = prod(k=0, n-1, 3*k+1);
    a(n) = sum(k=0, n, a007559(k+1)*stirling(n, k, 2));

Formula

a(n) = Sum_{k=0..n} A007559(k+1) * Stirling2(n,k).
a(n) ~ 3 * sqrt(Pi) * n^(n + 5/6) / (2^(13/6) * Gamma(1/3) * log(4/3)^(n + 4/3) * exp(n)). - Vaclav Kotesovec, Sep 06 2024
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