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 1-7 of 7 results.

A225466 Triangle read by rows, 3^k*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 3, 4, 21, 9, 8, 117, 135, 27, 16, 609, 1431, 702, 81, 32, 3093, 13275, 12015, 3240, 243, 64, 15561, 115479, 171990, 81405, 13851, 729, 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187, 256, 390369, 7998111, 27533142, 29893941, 13121514, 2561706
Offset: 0

Views

Author

Peter Luschny, May 08 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers of order m is in A225468.
From Wolfdieter Lang, Apr 09 2017: (Start)
This is the Sheffer triangle (exp(2*x), exp(3*x) - 1), denoted by S2[3,2]. See also A282629 for S2[3,1]. The stirling2 triangle A048993 is in this notation denoted by S2[1,0].
The a-sequence for this Sheffer triangle has e.g.f. 3*x/log(1+x) and is 3*A006232(n)/A006233(n) (Cauchy numbers of the first kind). For a- and z-sequences for Sheffer triangles see the W. Lang link under A006232, also with references).
The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(2/3)) and gives 2*A284862/A284863.
The first column k sequences divided by 3^k are A000079, A016127, A016297, A025999. For the e.g.f.s and o.g.f.s see below.
The row sums give A284864. The alternating row sums give A284865.
This triangle appears in the o.g.f. G(n, x) of the sequence {(2 + 3*m)^n}{m>=0}, as G(n, x) = Sum{k=0..n} T(n, k)*k!*x^k/(1-x)^(k+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0} (2 + 3*m)^n t^m/m! = exp(t)*Sum_{k=0..n} T(n, k)*t^k.
The corresponding Eulerian number triangle is A225117(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, m)*m!, 0 <= k <= n. (End)

Examples

			[n\k][ 0,     1,      2,       3,       4,      5,     6,    7]
[0]    1,
[1]    2,     3,
[2]    4,    21,      9,
[3]    8,   117,    135,      27,
[4]   16,   609,   1431,     702,      81,
[5]   32,  3093,  13275,   12015,    3240,    243,
[6]   64, 15561, 115479,  171990,   81405,  13851,   729,
[7]  128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see the Maple program): T(4, 2) = 3*T(3, 1) + (3*2+2)*T(3, 2) = 3*117 + 8*135 = 1431.
Boas-Buck recurrence for column k = 2, and n = 4: T(4,2) = (1/2)*(2*(4 + 3*2)*T(3, 2) + 2*6*(-3)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(20*135 + 12*9*(1/6)*9) = 1431. (End)
		

Crossrefs

Programs

  • Maple
    SF_SS := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or  k < 0 then return(0) fi;
    m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:
    seq(print(seq(SF_SS(n, k, 3), k=0..n)), n=0..5);
  • Mathematica
    EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • PARI
    T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial, factorial
    def T(n, k): return sum(binomial(k, j)*(-1)**(j - k)*(2 + 3*j)**n//factorial(k) for j in range(k + 1))
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m) + (m*k+1)*EulerianNumber(n-1,k,m)
    def SF_SS(n, k, m):
        return add(EulerianNumber(n,j,m)*binomial(j,n-k) for j in (0..n))/ factorial(k)
    def A225466(n): return SF_SS(n, k, 3)
    

Formula

T(n, k) = (1/k!)*Sum_{j=0..n} binomial(j, n-k)*A_3(n, j) where A_m(n, j) are the generalized Eulerian numbers A225117.
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A000244.
From Wolfdieter Lang, Apr 09 2017: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*(-1)^(j-k)*(2 + 3*j)^n/k!, 0 <= k <= n.
E.g.f. of triangle: exp(2*z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column k is exp(2*x)*((exp(3*x) - 1)^k)/k! (Sheffer property).
O.g.f. for sequence of column k is 3^k*x^k/Product_{j=0..k} (1 - (2+3*j)*x).
A nontrivial recurrence for the column m=0 entries T(n, 0) = 2^n from the z-sequence given above: T(n,0) = n*Sum_{k=0..n-1} z(k)*T(n-1,k), n >= 1, T(0, 0) = 1.
The corresponding recurrence for columns k >= 1 from the a-sequence is T(n, k) = (n/k)* Sum_{j=0..n-k} binomial(k-1+j, k-1)*a(j)*T(n-1, k-1+j).
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+2) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
(End)
Boas-Buck recurrence for column sequence k: T(n, k) = (1/(n - k))*((n/2)*(4 + 3*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)*(-3)^(n-p)*Bernoulli(n-p)*T(p, k)), for n > k >= 0, with input T(k, k) = 3^k. See a comment and references in A282629, An example is given below. - Wolfdieter Lang, Aug 11 2017

A284859 Row sums of the Sheffer triangle (exp(x), exp(3*x)-1) given in A282629.

Original entry on oeis.org

1, 4, 25, 199, 1876, 20257, 245017, 3266914, 47450923, 743935375, 12497579698, 223619318215, 4240423494685, 84855613320004, 1785410320771933, 39373503608087299, 907548770965519660, 21810536356271794549, 545305573054110017125, 14155835044848094831018
Offset: 0

Views

Author

Wolfdieter Lang, Apr 05 2017

Keywords

Comments

See A282629 for details. These are generalized Bell numbers (A000110) because A282629 is a generalized Stirling2 triangle.

Crossrefs

Programs

  • Mathematica
    T[n_, m_]:= Sum[Binomial[m, k] (-1)^(k - m) (1 + 3k)^n/m!, {k, 0, m}]; Table[Sum[T[n, m], {m, 0, n}], {n, 0, 20}] (* Indranil Ghosh, Apr 10 2017 *)
    Table[Sum[3^k*Binomial[n,k]*BellB[k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 22 2022 *)
  • PARI
    T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
    a(n) = sum(m=0, n, T(n, m)); \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial, factorial
    def T(n, m): return sum([binomial(m, k) * (-1)**(k - m) * (1 + 3*k)**n for k in range(m + 1)])//factorial(m)
    def a(n): return sum([T(n, k) for k in range(n + 1)])
    print([a(n) for n in range(20)]) # Indranil Ghosh, Apr 10 2017

Formula

a(n) = Sum_{m=0..n} A282629(n, m).
E.g.f.: exp(x)*exp(exp(3*x) -1).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(1+3*m)^n, n >= 0. (Dobiński type formula from the A282629(n,m) sum formula, interchanging summations).
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ Bell(n) * (3 + LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 3^n * n^(n + 1/3) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/3)). - Vaclav Kotesovec, Jun 27 2022

A355164 a(n) = exp(-1/3) * Sum_{k>=0} (3*k + 2)^n / (3^k * k!).

Original entry on oeis.org

1, 3, 12, 63, 405, 3024, 25515, 239355, 2465478, 27600669, 333051669, 4303119330, 59202612693, 863285928327, 13288589222508, 215177742933579, 3654114236490393, 64902307993517160, 1202782377224829015, 23207417212751493327, 465302639045308247262, 9677171073270491712513, 208434297638273958963225
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[2 x + (Exp[3 x] - 1)/3], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 3^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
    Table[Sum[Binomial[n, k] 2^(n - k) 3^k BellB[k, 1/3], {k, 0, n}], {n, 0, 22}]

Formula

E.g.f.: exp(2*x + (exp(3*x) - 1) / 3).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^(k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A004212(k).
a(n) ~ 3^(n + 2/3) * n^(n + 2/3) * exp(n/LambertW(3*n) - n - 1/3) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^(n + 2/3)). - Vaclav Kotesovec, Jun 27 2022

A284865 Alternating row sums of the Sheffer triangle S2[3,2] given by A225466.

Original entry on oeis.org

1, -1, -8, -1, 217, 1196, -3725, -115777, -803150, 3402485, 145172737, 1528780238, -1328359499, -320347469485, -5507171702648, -28294413916213, 915647446089037, 28738067698188692, 369693788462739487, -1233559476327263869
Offset: 0

Views

Author

Wolfdieter Lang, Apr 10 2017

Keywords

Comments

This is a generalization of sequence A000587 because S2[3,2] = A225466 is a generalization of the Stirling2 triangle A048993.
For the row sums see A284864.

Crossrefs

Cf. A000587, A225466, A284864, A284860 (case [3,1]).

Formula

a(n) = Sum_{k=0..n} (-1)^k*A225466(n, k).
E.g.f.: exp(2*x)*exp(1 - exp(3*x)) (Sheffer property).
a(n) = (1/e)*Sum_{m>=0} ((-1)^m / m!)*(2+3*m)^n, n >= 0, (Dobiński type formula).

A355162 a(n) = exp(-1) * Sum_{k>=0} (4*k + 2)^n / k!.

Original entry on oeis.org

1, 6, 52, 568, 7312, 107360, 1760576, 31760256, 623137024, 13179872768, 298391335936, 7189153167360, 183428957442048, 4935794590572544, 139571328018628608, 4134634425826115584, 127966201403431518208, 4127825849826169716736, 138477447400991610896384, 4822002684952714247929856
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 19; CoefficientList[Series[Exp[Exp[4 x] + 2 x - 1], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
    Table[Sum[Binomial[n, k] 2^(n + k) BellB[k], {k, 0, n}], {n, 0, 19}]

Formula

E.g.f.: exp(exp(4*x) + 2 x - 1).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n+k) * Bell(k).
a(n) = 2^n * A126390(n). - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 1/2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/2)). - Vaclav Kotesovec, Jun 27 2022

A355163 a(n) = exp(-1) * Sum_{k>=0} (4*k + 3)^n / k!.

Original entry on oeis.org

1, 7, 65, 743, 9921, 150151, 2526593, 46615783, 933072513, 20093861895, 462440842177, 11310514854375, 292627518129985, 7976748158144647, 228308400790500097, 6840702405678586343, 214000748166439723265, 6973447420429351808007, 236204029044752265931585, 8300724166287243795922151
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 19; CoefficientList[Series[Exp[Exp[4 x] + 3 x - 1], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
    Table[Sum[Binomial[n, k] 3^(n - k) 4^k BellB[k], {k, 0, n}], {n, 0, 19}]

Formula

E.g.f.: exp(exp(4*x) + 3 x - 1).
a(0) = 1; a(n) = 3 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 4^k * Bell(k).
a(n) ~ Bell(n) * (4 + 3*LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 3/4) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 3/4)). - Vaclav Kotesovec, Jun 27 2022

A367938 Expansion of e.g.f. exp(exp(3*x) - 1 - 2*x).

Original entry on oeis.org

1, 1, 10, 55, 487, 4654, 51463, 632125, 8536492, 125279785, 1981246555, 33530245984, 603797462677, 11513675558701, 231539488842610, 4893151984630579, 108334206855000739, 2505977899186557502, 60419653270442268643, 1515077412621445514089, 39437350309301393464876, 1063746973172416765272589
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 05 2023

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; CoefficientList[Series[Exp[Exp[3 x] - 1 - 2 x], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = -2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 3^k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 21}]
    Table[Sum[Binomial[n, k] (-2)^(n - k) 3^k BellB[k], {k, 0, n}], {n, 0, 21}]
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(exp(exp(3*x) - 1 - 2*x))) \\ Michel Marcus, Dec 07 2023

Formula

G.f. A(x) satisfies: A(x) = 1 - x * ( 2 * A(x) - 3 * A(x/(1 - 3*x)) / (1 - 3*x) ).
a(n) = exp(-1) * Sum_{k>=0} (3*k-2)^n / k!.
a(0) = 1; a(n) = -2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * (-2)^(n-k) * 3^k * Bell(k).
Showing 1-7 of 7 results.