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.

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A016781 a(n) = (3*n+1)^5.

Original entry on oeis.org

1, 1024, 16807, 100000, 371293, 1048576, 2476099, 5153632, 9765625, 17210368, 28629151, 45435424, 69343957, 102400000, 147008443, 205962976, 282475249, 380204032, 503284375, 656356768, 844596301, 1073741824, 1350125107, 1680700000, 2073071593, 2535525376
Offset: 0

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Author

Keywords

Comments

In general the e.g.f. of {(1 + 3*m)^n}{m>=0} is E(n,x) = exp(x)*Sum{m=0..n} A282629(n, m)*x^m, and the o.g.f. is G(n, x) = (Sum_{m=0..n} A225117(n, n-m)*x^m)/(1-x)^(n+1). - Wolfdieter Lang, Apr 02 2017

Crossrefs

Programs

  • Magma
    [(3*n+1)^5: n in [0..30]]; // Vincenzo Librandi, Sep 21 2011
    
  • Mathematica
    (3Range[0,20]+1)^5 (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,1024,16807,100000,371293,1048576},30] (* Harvey P. Dale, May 13 2012 *)
  • Maxima
    A016781(n):=(3*n+1)^5$
    makelist(A016781(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

Formula

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, May 13 2012
From Wolfdieter Lang, Apr 02 2017: (Start)
O.g.f.: (1+1018*x+10678*x^2+14498*x^3+2933*x^4+32*x^5)/(1-x)^6.
E.g.f: exp(x)*(1+1023*x+7380*x^2+8775*x^3+2835*x^4+243*x^5). (End)
a(n) = A000584(A016777(n)). - Michel Marcus, Apr 06 2017
Sum_{n>=0} 1/a(n) = 2*Pi^5/(3^6*sqrt(3)) + 121*zeta(5)/3^5. - Amiram Eldar, Mar 29 2022

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

A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

Original entry on oeis.org

1, 1, 4, 1, 24, 16, 1, 124, 240, 64, 1, 624, 2656, 1792, 256, 1, 3124, 26400, 33920, 11520, 1024, 1, 15624, 250096, 546560, 331520, 67584, 4096, 1, 78124, 2313360, 8105664, 7822080, 2745344, 372736, 16384, 1, 390624, 21132736, 114627072, 165398016, 88940544, 20299776, 1966080, 65536, 1, 1953124, 191757120, 1574682880, 3270274560, 2529343488, 863256576, 138215424, 10027008, 262144
Offset: 0

Views

Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
For the Sheffer triangle (exp(x), (1/4)*(exp(4*x) - 1)) see A111578, also the P. Bala link where this triangle, T(n, m)/4^m, is named S_{(4,0,1)}. The triangle T(n, m)*m! is given in A285066.
The a-sequence for this Sheffer triangle has e.g.f. 4*x/log(1+x) and is 4*A006232(n)/A006233(n) (Cauchy numbers of the first kind).
The z-sequence has e.g.f. (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)) and is A285062(n)/A285063(n).
The main diagonal gives A000302.
The row sums give A285064. The alternating row sums give A285065.
The first column sequences are A000012, 4*A003463, 4^2*A016234. For the e.g.f.s and o.g.f.s see the formula section.
This triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 4*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 4*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225118 with row reversion.
In general the Sheffer triangle S2[d,a] appears in the reordering of the operator (a*1 + d*E_x) = Sum_{m=0..n} S2(d,a;n,m) x^m*(d_x)^m with the derivative d_x and the Euler operator E_x = x*d_x. For [d,a] = [1,0] this becomes the standard Stirling2 property.

Examples

			The triangle T(n,m) begins:
  n\m  0      1        2         3         4        6        7       8     9
  0:   1
  1:   1      4
  2:   1     24       16
  3:   1    124      240        64
  4:   1    624     2656      1792       256
  5:   1   3124    26400     33920     11520     1024
  6:   1  15624   250096    546560    331520    67584     4096
  7:   1  78124  2313360   8105664   7822080  2745344   372736   16384
  8:   1 390624 21132736 114627072 165398016 88940544 20299776 1966080 65536
  ...
Three term recurrence: T(4, 1) = 4*1 + (1 + 4*1)*124 = 624.
Recurrence for row polynomial R(3, x) (Meixner type): ((1 + 4*x) + 4*x*d_x)*(1 + 24*x + 16*x^2) = 1 + 124*x + 240*x^2 + 64*x^3.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(20*240 + 12*16*(1/6)*16) = 2656. - _Wolfdieter Lang_, Aug 11 2017
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]*4^k*StirlingS2[k, m], {k, 0, n}], {n, 0, 20}, {m, 0, n}] // Flatten (* Indranil Ghosh, May 06 2017 *)
  • PARI
    T(n, m) = sum(k=0, n, binomial(n, k)*4^k*stirling(k, m, 2));
    for(n=0, 20, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, May 06 2017

Formula

Three term recurrence (from the conversion property mentioned above, with [d,a] =[4,1]): T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m. T(n, m) = 4*T(n-1, m-1) + (1 + 4*m)*T(n-1, m) for n >= 1, m = 0..n.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
E.g.f. of the row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)*exp(x*(exp(4*z) - 1)). This is the e.g.f. of the triangle.
E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
O.g.f. for sequence of column m: (4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x) (by Laplace transform of the e.g.f.).
T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in A048993.
A nontrivial recurrence for the column m=0 entries T(n, 0) = 4^n from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)* Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((1 + 4*x) + 4*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 4*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m,m) = 4^m. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A225467 Triangle read by rows, T(n, k) = 4^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, 16, 27, 316, 336, 64, 81, 2320, 4960, 2304, 256, 243, 16564, 63840, 54400, 14080, 1024, 729, 116920, 768496, 1071360, 485120, 79872, 4096, 2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384, 6561, 5758240, 101417920, 322850304
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.
This is the Sheffer triangle (exp(3*x), exp(4*x) - 1). See also the P. Bala link under A225469, the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x) - 1)), which is named there exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017

Examples

			[n\k][  0,      1,       2,        3,        4,       5,      6,     7]
[0]     1,
[1]     3,      4,
[2]     9,     40,      16,
[3]    27,    316,     336,       64,
[4]    81,   2320,    4960,     2304,      256,
[5]   243,  16564,   63840,    54400,    14080,    1024,
[6]   729, 116920,  768496,  1071360,   485120,   79872,   4096,
[7]  2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see the Maple program): T(4, 2) = 4*T(3, 1) + (4*2+3)*T(3, 2) = 4*316 + 11*336 = 4960.
Boas-Buck recurrence for column k = 2, and n = 4: T(4, 2) = (1/2)*(2*(6 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(28*336 + 12*16*(1/6)*16) = 4960. (End)
		

Crossrefs

Cf. A048993 (m=1), A154537 (m=2), A225466 (m=3). A225469 (scaled).
Cf. Columns: A000244, 4*A016138, 16*A018054. A225118.

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, 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]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • PARI
    T(n, k) = sum(m=0, k, binomial(k, m)*(-1)^(m - k)*((3 + 4*m)^n)/k!);
    for(n = 0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 13 2017
    
  • Python
    from sympy import binomial, factorial
    def T(n, k): return sum(binomial(k, m)*(-1)**(m - k)*((3 + 4*m)**n)//factorial(k) for m in range(k + 1))
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 13 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 A225467(n): return SF_SS(n, k, 4)
    

Formula

T(n, k) = (1/k!)*sum_{j=0..n} binomial(j, n-k)*A_4(n, j) where A_m(n, j) are the generalized Eulerian numbers A225118.
For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A190541.
T(n, n) ~ A000302; T(n, n-1) ~ A002700.
From Wolfdieter Lang, Apr 13 2017: (Start)
T(n, k) = Sum_{m=0..k} binomial(k,m)*(-1)^(m-k)*((3+4*m)^n)/k!, 0 <= k <= n.
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^(n-k)*4^k*Stirling2(k, m), 0 <= m <= n.
E.g.f. exp(3*z)*exp(x*(exp(4*z) - 1)) (Sheffer property).
E.g.f. column k: exp(3*x)*((exp(4*x) - 1)^k)/k!, k >= 0.
O.g.f. column k: (4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
Boas-Buck recurrence for column sequence k: T(n, k) = (1/(n - k))*((n/2)*(6 + 4*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, k)), for n > k >= 0, with input T(k, k) = 4^k. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A016223 Expansion of 1/((1-x) * (1-4*x) * (1-7*x)).

Original entry on oeis.org

1, 12, 105, 820, 6081, 43932, 312985, 2212740, 15576561, 109385452, 767096265, 5375266260, 37649233441, 263634112572, 1845796701945, 12922008569380, 90459786608721, 633241412753292, 4432781515242025, 31029837110570100, 217210325789494401, 1520478144588475612
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Maple
    a:=n->sum((7^(n+1-j)-4^(n+1-j))/3, j=0..n+1): seq(a(n), n=0..20); # Zerinvary Lajos, Jan 15 2007
  • PARI
    a(n) = (1-2*4^(n+2)+7^(n+2))/18; \\ Seiichi Manyama, May 03 2025

Formula

a(n) = (1/18) - (16/9)*4^n + (49/18)*7^n. - Antonio Alberto Olivares, Feb 07 2010 [corrected by Seiichi Manyama, May 03 2025]
a(0)=1, a(1)=12, a(n) = 11*a(n-1) - 28*a(n-2) + 1. - Vincenzo Librandi, Feb 10 2011
E.g.f.: exp(x)*(1 - 32*exp(3*x) + 49*exp(6*x))/(2!*3^2). - This is (d^2/dx^2) (exp(x)*(exp(x) - 1)^2 / (2*3^2)). See also the second column of the Sheffer triangle A282629 divided by 3^2. - Wolfdieter Lang, Apr 08 2017
From Seiichi Manyama, May 03 2025: (Start)
a(n) = Sum_{k=0..n} 3^k * binomial(n+2,k+2) * Stirling2(k+2,2).
G.f.: B(x)^3, where B(x) is the g.f. of A383627. (End)

A285068 Denominators of the generalized Bernoulli numbers B[3,1] = 3^n*B(n, 1/3).

Original entry on oeis.org

1, 2, 2, 1, 10, 1, 14, 1, 10, 1, 22, 1, 910, 1, 2, 1, 170, 1, 266, 1, 110, 1, 46, 1, 910, 1, 2, 1, 290, 1, 4774, 1, 170, 1, 2, 1, 639730, 1, 2, 1, 4510, 1, 602, 1, 230, 1, 94, 1, 15470, 1, 22
Offset: 0

Views

Author

Wolfdieter Lang, Apr 28 2017

Keywords

Comments

The numerators are given in A157799.
Because B(n, 2/3) = (-1)^n*B(n, 1/3) (from the e.g.f. z*exp(x*z)/(exp(z)-1) of Bernoulli polynomials {B(n, x)}_{n>=0}) one has for the numbers B[3,2](n) = 3^n*B(n, 2/3) the numerators (-1)^n*A157799(n) and the denominators a(n).
This sequence gives also the denominators of {3^n*B(n)}_{n>=0} with numerators given in A285863.

Examples

			The Bernoulli numbers r(n) = B[3,1](n) begin: 1, -1/2, -1/2, 1, 13/10, -5, -121/14, 49, 1093/10, -809, -49205/22, 20317, 61203943/910, -722813, -5580127/2, 34607305, ...
The Bernoulli numbers B[3,2](n) begin: 1, 1/2, -1/2, -1, 13/10, 5, -121/14, -49, 1093/10, 809, -49205/22, -20317, 61203943/910, 722813, -5580127/2, -34607305, ...
From _Peter Luschny_, Mar 26 2021: (Start)
The generalized Bernoulli numbers as given in the Luschny link are different.
1, -7/2, 23/2, -35, 973/10, -245, 7943/14, -1295, 31813/10, -7721, 288715/22, -13475, 128296423/910, -882557, -4891999/2, 33870025, ...
The numerators of these numbers are in A157811. (End)
		

Crossrefs

Programs

  • Mathematica
    Table[Denominator[3^n*BernoulliB[n, 1/3]], {n, 0, 100}] (* Indranil Ghosh, Jul 18 2017 *)
  • PARI
    a(n) = denominator(3^n * bernfrac(n)); \\ Ruud H.G. van Tol, Jan 31 2024
  • Python
    from sympy import bernoulli, Rational
    def a(n):
        return (3**n * bernoulli(n, Rational(1,3))).as_numer_denom()[1]
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 18 2017
    
  • SageMath
    # uses [gen_bernoulli_number from A157811]
    print([denominator((-1)^n*gen_bernoulli_number(n, 3)) for n in range(23)])
    # Peter Luschny, Mar 26 2021
    

Formula

a(n) = denominator(r(n)) with the rationals (in lowest terms) r(n) = Sum_{k=0..n} ((-1)^k / (k+1))*A282629(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))*A284861(n, k). r(n) = B[3,1](n) = 3^n*B(n, 1/3) with the Bernoulli polynomials A196838/A196839 or A053382/A053383.
a(n) = A157800(n)/3^n, n >= 0.

A021874 Expansion of 1/((1-x) * (1-4*x) * (1-7*x) * (1-10*x)).

Original entry on oeis.org

1, 22, 325, 4070, 46781, 511742, 5430405, 56516790, 580744461, 5916830062, 59935396885, 604729235110, 6084941584541, 61113049957982, 612976296281765, 6142684971387030, 61517309500479021, 615806336417543502, 6162496145690677045, 61655991294017340550, 616777123265962899901
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1 / ((1 - x) (1 - 4 x) (1 - 7 x) (1 - 10 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Jul 11 2013 *)
    LinearRecurrence[{22,-159,418,-280},{1,22,325,4070},30] (* Harvey P. Dale, May 13 2018 *)

Formula

a(n) = (10^(n+3) - 3*7^(n+3) + 3*4^(n+3) - 1)/162. - Yahia Kahloune, Jul 05 2013
E.g.f.: exp(x)*(1000*exp(9*x) - 1029*exp(6*x) + 192*exp(3*x) - 1)/(3!*3^3). This is d^3/dx^3 exp(x)*(exp(3*x - 1))^3/(3!*3^3); see also column m=3 of A282629 divided by 3^3. The o.g.f. is given in the name. - Wolfdieter Lang, Apr 08 2017
a(n) = Sum_{k=0..n} 3^k * binomial(n+3,k+3) * Stirling2(k+3,3). - Seiichi Manyama, May 03 2025

A290312 Third diagonal sequence of the Sheffer triangle A094816 (special Charlier).

Original entry on oeis.org

1, 8, 29, 75, 160, 301, 518, 834, 1275, 1870, 2651, 3653, 4914, 6475, 8380, 10676, 13413, 16644, 20425, 24815, 29876, 35673, 42274, 49750, 58175, 67626, 78183, 89929, 102950, 117335, 133176, 150568, 169609, 190400, 213045, 237651, 264328, 293189, 324350, 357930
Offset: 0

Views

Author

Wolfdieter Lang, Jul 28 2017

Keywords

Comments

See A094816 and A290311.

Crossrefs

Programs

  • PARI
    Vec((1 + 3*x - x^2)/(1 - x)^5 + O(x^60)) \\ Colin Barker, Jul 29 2017

Formula

G.f.: (1 + 3*x - x^2)/(1 - x)^5.
E.g.f.: exp(x)*(1 + 7*x + 14*x^2/2! + 11*x^3/3! + 3*x^4/4!). This is computed from the o.g.f. with eqs. (23)-(25) of the Wolfdieter Lang 2017 link in A282629.
From Colin Barker, Jul 29 2017: (Start)
a(n) = (24 + 70*n + 69*n^2 + 26*n^3 + 3*n^4) / 24.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4.
(End)

A290319 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 4*x)^(-1/4), (-1/4)*log(1 - 4*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 45, 59, 15, 1, 585, 812, 254, 28, 1, 9945, 14389, 5130, 730, 45, 1, 208845, 312114, 122119, 20460, 1675, 66, 1, 5221125, 8011695, 3365089, 633619, 62335, 3325, 91, 1, 151412625, 237560280, 105599276, 21740040, 2441334, 158760, 5964, 120, 1, 4996616625, 7990901865, 3722336388, 823020596, 102304062, 7680414, 355572, 9924, 153, 1, 184874815125, 300659985630, 145717348221, 34174098440, 4608270890, 386479380, 20836578, 722760, 15585, 190, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 08 2017

Keywords

Comments

This generalization of the unsigned Stirling1 triangle A132393 is called here |S1hat[4,1]|.
The signed matrix S1hat[4,1] with elements (-1)^(n-k)*|S1hat[4,1]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[4,1] with elements S2[4,1](n, k)/d^k, where S2[4,1] is Sheffer (exp(x), exp(4*x) - 1), given in A285061. See also the P. Bala link below for the scaled and signed version s_{(4,0,1)}.
For the general |S1hat[d,a]| case see a comment in A286718.

Examples

			The triangle T(n, k) begins:
  n\k         0         1         2        3       4      5    6   7  8 ...
  0:          1
  1:          1         1
  2:          5         6         1
  3:         45        59        15        1
  4:        585       812       254       28       1
  5:       9945     14389      5130      730      45      1
  6:     208845    312114    122119    20460    1675     66    1
  7:    5221125   8011695   3365089   633619   62335   3325   91   1
  8:  151412625 237560280 105599276 21740040 2441334 158760 5964 120  1
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (16 - 3)*T(3, 2) = 59 + 13*15 = 254.
Boas-Buck recurrence for column k=2 and n=4:
T(4, 2) = (4!/2)*(4*(1 + 8*(5/12))*T(2, 2)/2! + 1*(1 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(2*13/3 + 5*15/3!) = 254. (End)
		

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,1], [3,2] and [4,3] is A132393, A028338, A286718, A225470 and A225471, respectively.
Columns k=0..3 give A007696, A024382(n-1), A383700, A383701.
Row sums: A001813. Alternating row sums: A000007.

Programs

  • Mathematica
    FoldList[Join[Table[If[i == 1, 0, #[[i-1]]] + (4*#2 - 3)*#[[i]], {i, Length[#]}], {1}] &, {1}, Range[10]] (* Paolo Xausa, Aug 18 2025 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (4*n - 3)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle): (1 - 4*z)^{-(x + 1)/4}.
E.g.f. of column k is (1 - 4*x)^(-1/4)*((-1/4)*log(1 - 4*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+4), with R(0, x) = 1. Row polynomial R(n, x) = risefac(4,1;x,n) with the rising factorial risefac(d,a;x,n) :=Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0, a_1, ..., a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 4*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*4^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(1 + 4*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A284857 Numerators of the exponential expansion of (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)).

Original entry on oeis.org

1, -1, 11, -49, 1187, -18083, 662407, -3539605, 864309187, -949103125, 289289620393, -4846044126449, 12389144856368069, -69977996793541583, 1191089380720588487, -6783915816877925461, 3295296805315071712171, -169986671194174827887881, 129921413307474873885175559, -149671376459098924087260625
Offset: 0

Views

Author

Wolfdieter Lang, Apr 04 2017

Keywords

Comments

For the denominators see A284858.
This gives the numerators of the z-sequence for the Sheffer triangle (exp(x), exp(3*x) - 1) shown in A282629. For the notion and use of a- and z- sequences for Sheffer triangles see the W. Lang link under A006232. The a-sequence of this Sheffer triangle is given by A006232/A006233.
For the nontrivial decompositions of 1 given by the z-sequence recurrence for the m=0 column repeat(1) of the triangle A282629 see an example there and below.

Examples

			The rationals a(n)/A284858(n) start: 1, -1/6, 11/54, -49/108, 1187/810, -18083/2916, 662407/20412, -3539605/17496, 864309187/590490, -949103125/78732, 289289620393/2598156, -4846044126449/4251528, 12389144856368069/967222620, -69977996793541583/446410440, 1191089380720588487/573956280, -6783915816877925461/229582512, ...
From the z-recurrence for A282629(5, 0) = 1 one finds: 1 = 5*(1*1 + 255*(-1/6) + 945*(11/54) + 594*(-49/108) + 81*(1187/810)).
		

Crossrefs

Cf. A284858 (denominators), A282629, A006232/A006233.

Formula

E.g.f.: (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)) for the rational sequence a(n)/A284858(n), n >= 0.
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