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 1-8 of 8 results.

A286718 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 18 2017

Keywords

Comments

This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018

Examples

			The triangle T(n, k) begins:
n\k        0        1        2       3      4     5    6  7 8 ...
O:         1
1:         1        1
2:         4        5        1
3:        28       39       12       1
4:       280      418      159      22      1
5:      3640     5714     2485     445     35     1
6:     58240    95064    45474    9605   1005    51    1
7:   1106560  1864456   959070  227969  28700  1974   70  1
8:  24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

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.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • Mathematica
    T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[, -1] = 0; T[0, 0] = 1; T[n, k_] /; nJean-François Alcover, Jun 20 2018 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*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) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;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 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017

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

A225118 Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 4}(x) in descending order.

Original entry on oeis.org

1, 3, 1, 9, 22, 1, 27, 235, 121, 1, 81, 1996, 3446, 620, 1, 243, 15349, 63854, 40314, 3119, 1, 729, 112546, 963327, 1434812, 422087, 15618, 1, 2187, 806047, 12960063, 37898739, 26672209, 4157997, 78117, 1, 6561, 5705752, 162711868, 840642408, 1151050534
Offset: 0

Views

Author

Peter Luschny, May 02 2013

Keywords

Comments

The row sums equal the quadruple factorial numbers A047053 and the alternating row sums, i.e., sum((-1)^k*T(n,k),k=0..n), are up to a sign A079858. - Johannes W. Meijer, May 04 2013

Examples

			[0]  1
[1]  3*x   +    1
[2]  9*x^2 +   22*x   +    1
[3] 27*x^3 +  235*x^2 +  121*x   + 1
[4] 81*x^4 + 1996*x^3 + 3446*x^2 + 620*x + 1
...
The triangle T(n, k) begins:
n\k
0:    1
1:    3      1
2:    9     22        1
3:   27    235      121        1
4:   81   1996     3446      620        1
5:  243  15349    63854    40314     3119       1
6:  729 112546   963327  1434812   422087   15618     1
7: 2187 806047 12960063 37898739 26672209 4157997 78117 1
...
row n=8: 6561 5705752 162711868 840642408 1151050534 442372648 39531132 390616 1,
row n=9: 19683 40156777 1955297356 16677432820 39523450714 29742429982 6818184988 367889284 1953115 1.
... - _Wolfdieter Lang_, Apr 12 2017
		

Crossrefs

Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,3}(x) = A225117. A123125, A225467, A225469, A225473.

Programs

  • Maple
    gf := proc(n, k) local f; f := (x,t) -> x*exp(t*x/k)/(1-x*exp(t*x));
    series(f(x,t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
    collect(simplify(%), x) end:
    seq(print(seq(coeff(gf(n, 4), x, n-k), k=0..n)), n=0..6);
    # Recurrence:
    P := proc(n,x) option remember; if n = 0 then 1 else
      (n*x+(1/4)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
      expand(%) fi end:
    A225117 := (n,k) -> 4^n*coeff(P(n,x),x,n-k):
    seq(print(seq(A225117(n,k), k=0..n)), n=0..5);  # Peter Luschny, Mar 08 2014
  • Mathematica
    gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 4], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • Sage
    @CachedFunction
    def EB(n, k, x):  # Modified cardinal B-splines
        if n == 1: return 0 if (x < 0) or (x >= 1) else 1
        return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
    def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
        R. = ZZ[]
        if x == 0: return 1
        return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
    [EulerianPolynomial(n, 4).coefficients()[::-1] for n in (0..5)]

Formula

G.f. of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 4; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
From Wolfdieter Lang, Apr 12 2017 : (Start)
E.g.f. of row polynomials (rising powers of x): (1-x)*exp(3*(1-x)*z)/(1-y*exp(4*(1-x)*z)), i.e. e.g.f. of the triangle.
E.g.f. for the row polynomials with falling powers of x (A_{n, 4}(x) of the name): (1-x)*exp((1-x)*z)/(1 - x*exp(4*(1-x)*z)).
T(n, k) = Sum_{j=0..k} (-1)^(k-j) * binomial(n+1,k-j) * (3+4*j)^n, 0 <= k <= n.
Recurrence: T(n, k) = (4*(n-k) + 1)*T(n-1, k-1) + (3 + 4*k)*T(n-1, k), n >= 1, with T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k. (End)
In terms of Euler's triangle = A123125: T(n, k) = Sum_{m=0..n} (binomial(n, m)*3^(n-m)*4^m*Sum_{p=0..k} (-1)^(k-p)*binomial(n-m, k-p)*A123125(m, p)), 0 <= k <= n. - Wolfdieter Lang, Apr 13 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)

A157817 Numerator of Bernoulli(n, 1/4).

Original entry on oeis.org

1, -1, -1, 3, 7, -25, -31, 427, 127, -12465, -2555, 555731, 1414477, -35135945, -57337, 2990414715, 118518239, -329655706465, -5749691557, 45692713833379, 91546277357, -7777794952988025, -1792042792463, 1595024111042171723, 1982765468311237, -387863354088927172625
Offset: 0

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Author

N. J. A. Sloane, Nov 08 2009

Keywords

Comments

From Wolfdieter Lang, Apr 28 2017: (Start)
The rationals r(n) = Sum_{k=0..n} ((-1)^k / (k+1))*A285061(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))*A225473(n, k) define generalized Bernoulli numbers, named B[4,1](n), in terms of the generalized Stirling2 numbers S2[4,1]. The numerators of r(n) are a(n) and the denominators A141459(n). r(n) = B[4,1](n) = 4^n*B(n, 1/4) with the Bernoulli polynomials B(n, x) = Bernoulli(n, x) from A196838/A196839 or A053382/A053383.
The generalized Bernoulli numbers B[4,3](n) = Sum_{k=0..n} ((-1)^k/(k+1))* A225467(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))*A225473(n, k) satisfy
B[4,3](n) = 4^n*B(n, 3/4) = (-1)^n*B[4,1](n). They have numerators (-1)^n*a(n) and also denominators A141459(n). (End)

Crossrefs

For denominators see A157818 and A141459.

Programs

  • Mathematica
    Table[Numerator[BernoulliB[n, 1/4]], {n, 0, 50}] (* Vincenzo Librandi, Mar 16 2014 *)

Formula

From Wolfdieter Lang, Apr 28 2017: (Start)
a(n) = numerator(Bernoulli(n, 1/4)) with denominator A157818(n) (see the name).
a(n) = numerator(4^n*Bernoulli(n, 1/4)) with denominator A141459(n) = A157818(n)/4^n.
a(n)*(-1)^n = numerator(4^n*Bernoulli(n, 3/4)) with denominator A141459(n).
(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

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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

A285059 Numerators of the exponential expansion of (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)).

Original entry on oeis.org

1, -3, 9, -363, 6411, -46569, 3615627, -108267435, 2044658079, -27994845375, 5887932942123, -90460390681593, 475997756735954241, -3681053425472669991, 14270353890553782297, -2661381204559253577387, 880641541680797362210263
Offset: 0

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Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

For the denominators see A285060.
This gives one third of the numerators of the z-sequence for the Sheffer triangle (exp(3*x), exp(4*x) - 1) shown in A225467. For the notion and use of a- and z- sequences for Sheffer triangles see the W. Lang link under A006232, also for references. The a-sequence of this Sheffer triangle is given by 4*A006232/A006233.
For the nontrivial recurrence of {3^n} given by the z-sequence for the m = 0 column of the triangle A225467 see the example for n = 3 below.

Examples

			The rationals a(n)/A285060(n) start: 1, -3/8, 9/16, -363/256, 6411/1280, -46569/2048, 3615627/28672, -108267435/131072, 2044658079/327680, -27994845375/524288, ...
From the z-recurrence for A225467(3, 0) = 3^3 = 27 one finds: 3^3 = 3*3*(1*9 + 40*(-3/8) + 16*(9/16)).
		

Crossrefs

Cf. A006232/A006233 (a-sequence), A284857/A284858 (case [3,1]), A225467.

Formula

E.g.f.: (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)) for the rational sequence a(n)/A285060(n), n >= 0.

A285060 Denominators of the exponential expansion of (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)).

Original entry on oeis.org

1, 8, 16, 256, 1280, 2048, 28672, 131072, 327680, 524288, 11534336, 16777216, 7633633280, 4697620480, 1342177280, 17179869184, 365072220160, 171798691840, 45698452029440, 7696581394432
Offset: 0

Views

Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

For the numerators see A285059, also for the rationals A285059(n)/a(n).
The z-sequence for the Sheffer triangle S2[4,3] = A225467 is 3*A285059(n)/a(n).

Examples

			See A285059.
		

Crossrefs

Formula

E.g.f.: (4/(3*log(1+x)))*(1 - 1/(1+x)^(3/4)) for the rational sequence A285059(n)/a(n), n >= 0.
Showing 1-8 of 8 results.