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.

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

A225468 Triangle read by rows, 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, 1, 4, 7, 1, 8, 39, 15, 1, 16, 203, 159, 26, 1, 32, 1031, 1475, 445, 40, 1, 64, 5187, 12831, 6370, 1005, 57, 1, 128, 25999, 107835, 82901, 20440, 1974, 77, 1, 256, 130123, 888679, 1019746, 369061, 53998, 3514, 100, 1
Offset: 0

Views

Author

Peter Luschny, May 16 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers. For m = 1 this gives the classical Stirling set numbers A048993. (See the links for details.)
From Peter Bala, Jan 27 2015: (Start)
Exponential Riordan array [ exp(2*z), 1/3*(exp(3*z) - 1)].
Triangle equals P * A111577 = P^(-1) * A075498, where P is Pascal's triangle A007318.
Triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*3^n*binomial((x - 2)/3,n) }n>=0. An example is given below.
This triangle is the particular case a = 3, b = 0, c = 2 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)

Examples

			[n\k][ 0,    1,     2,    3,    4,  5,  6]
[0]    1,
[1]    2,    1,
[2]    4,    7,     1,
[3]    8,   39,    15,    1,
[4]   16,  203,   159,   26,    1,
[5]   32, 1031,  1475,  445,   40,  1,
[6]   64, 5187, 12831, 6370, 1005, 57,  1.
Connection constants: Row 3: [8, 39, 15, 1] so
x^3 = 8 + 39*(x - 2) + 15*(x - 2)*(x - 5) + (x - 2)*(x - 5)*(x - 8). - _Peter Bala_, Jan 27 2015
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).

Programs

  • Maple
    SF_S := 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;
    SF_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
    seq(print(seq(SF_S(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]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • 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_S(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/ (factorial(k)*m^k)
    for n in (0..6): [SF_S(n, k, 3) for k in (0..n)]

Formula

T(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_3(n, j)) / (3^k*k!) with A_3(n,j) = A225117.
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A016127; T(n, 2) ~ A016297; T(n, 3) ~ A025999;
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024212.
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i)*3^(i-k)*binomial(n,i)*Stirling2(i+1,k+1).
E.g.f.: exp(2*z)*exp(x/3*(exp(3*z) - 1)) = 1 + (2 + x)*z + (4 + 7*x + x^2)*z^2/2! + ....
T(n,k) = 1/(3^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(3*j + 2)^n.
O.g.f. for n-th diagonal: exp(-2*x/3)*Sum_{k >= 0} (3*k + 2)^(k + n - 1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k: 1/( (1 - 2*x)*(1 - 5*x)...(1 - (3*k + 2)*x) ). (End)
E.g.f. column k: exp(2*x)*((exp(3*x) - 1)/3)^k, k >= 0. See the Bala link for the S(3,0,2) exponential Riordan aka Sheffer triangle. - Wolfdieter Lang, Apr 10 2017

A225473 Triangle read by rows, k!*S_4(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, 3, 4, 9, 40, 32, 27, 316, 672, 384, 81, 2320, 9920, 13824, 6144, 243, 16564, 127680, 326400, 337920, 122880, 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120, 2187, 821356, 17842272, 114866304, 324065280, 453304320, 309657600, 82575360, 6561
Offset: 0

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).

Examples

			[n\k][0,      1,       2,       3,        4,       5,       6 ]
[0]   1,
[1]   3,      4,
[2]   9,     40,      32,
[3]  27,    316,     672,     384,
[4]  81,   2320,    9920,   13824,     6144,
[5] 243,  16564,  127680,  326400,   337920,  122880,
[6] 729, 116920, 1536992, 6428160, 11642880, 9584640, 2949120.
		

Crossrefs

Cf. A131689 (m=1), A145901 (m=2), A225472 (m=3).

Programs

  • Maple
    SF_SO := 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*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
    seq(print(seq(SF_SO(n, k, 4), 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]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • 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_SO(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
    for n in (0..6): [SF_SO(n, k, 4) for k in (0..n)]

Formula

For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A190541; T(n, n) ~ A047053.
From Wolfdieter Lang, Jul 12 2017: (Start)
T(n, k) = A225467(n, k)*k! = A225469(n, k)*(4^k*k!), 0 <= k <= n.
T(n, k) = Sum_{m=0..n} binomial(k,m)*(-1)^(k-m)*(3 + 4*m)^n.
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, k) = 0 if n < k and T(n, k) =
4*k*T(n-1, k-1) + (3 + 4*k)*T(n-1, k) for n >= 1, k = 0..n (see the Maple program).
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, k)*x^k: exp(3*z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column k: exp(3*x)*(exp(4*x) - 1)^k, k >= 0.
O.g.f. column k: k!*(4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)

A285066 Triangle read by rows: T(n, m) = A285061(n, m)*m!, 0 <= m <= n.

Original entry on oeis.org

1, 1, 4, 1, 24, 32, 1, 124, 480, 384, 1, 624, 5312, 10752, 6144, 1, 3124, 52800, 203520, 276480, 122880, 1, 15624, 500192, 3279360, 7956480, 8110080, 2949120, 1, 78124, 4626720, 48633984, 187729920, 329441280, 268369920, 82575360, 1, 390624, 42265472, 687762432, 3969552384, 10672865280, 14615838720, 9909043200, 2642411520, 1, 1953124, 383514240, 9448097280, 78486589440, 303521218560, 621544734720, 696605736960, 404288962560, 95126814720
Offset: 0

Views

Author

Wolfdieter Lang, Apr 19 2017

Keywords

Comments

This is the Sheffer triangle S2[4,1] = A285061 with column m scaled by m!. This is the fourth member of the triangle family A131689, A145901 and A284861.
This triangle appears in the o.g.f. G(n, x) = Sum_{m=0..n} T(n, m)*x^m/(1-x)^(m+1), n >= 0, of the power sequence {(1+4*m)^n}_{m >= 0}.
The diagonal sequence is A047053. The row sums give A285067. The alternating sum of row n is A141413(n+2), n >= 0.
The first column sequences are: A000012, 4*A003463, 2!*4^2*A016234.

Examples

			The triangle T(n, m) begins:
  n\m 0     1       2        3         4         5         6        7
  0:  1
  1:  1     4
  2:  1    24      32
  3:  1   124     480      384
  4:  1   624    5312    10752      6144
  5:  1  3124   52800   203520    276480    122880
  6:  1 15624  500192  3279360   7956480   8110080   2949120
  7:  1 78124 4626720 48633984 187729920 329441280 268369920 82575360
  ...
row 8: 1 390624 42265472 687762432 3969552384 10672865280 14615838720 9909043200 2642411520
row 9: 1 1953124 383514240 9448097280 78486589440 303521218560 621544734720 696605736960 404288962560 95126814720
...
		

Crossrefs

Programs

  • Mathematica
    T[n_, m_]:=Sum[Binomial[m, k]*(-1)^(k - m)*(1 + 4k)^n, {k, 0, n}]; Table[T[n, m], {n, 0, 10},{m, 0, n}] // Flatten (* Indranil Ghosh, May 02 2017 *)
  • Python
    from sympy import binomial
    def T(n, m):
        return sum([binomial(m, k)*(-1)**(k - m)*(1 + 4*k)**n for k in range(n + 1)])
    for n in range(21):
        print([T(n, m) for m in range(n + 1)])
    # Indranil Ghosh, May 02 2017

Formula

T(n, m) = A285061(n, m)*m! = A111578(n, m)*(4^m*m!), 0 <= m <= n.
T(n, m) = Sum_{k=0..n} binomial(m,k)*(-1)^(k-m)*(1+4*k)^n.
T(n, m) = Sum_{j=0..n} binomial(n-j,m-j)*A225118(n,n-j).
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m and T(n, m) =
4*m*T(n-1, m-1) + (1+4*m)*T(n-1, m) for n >= 1, m=0..n.
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column m: exp(x)*(exp(4*x) - 1)^m, m >= 0.
O.g.f. column m: m!*(4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x), m >= 0.
Showing 1-4 of 4 results.