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.

A225469 Triangle read by rows, 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, 1, 9, 10, 1, 27, 79, 21, 1, 81, 580, 310, 36, 1, 243, 4141, 3990, 850, 55, 1, 729, 29230, 48031, 16740, 1895, 78, 1, 2187, 205339, 557571, 299131, 52745, 3689, 105, 1, 6561, 1439560, 6338620, 5044536, 1301286, 137592, 6524, 136, 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 (see the links for details).
This is the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x -1))). See the P. Bala link where this is called exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017

Examples

			[n\k][ 0,     1,     2,     3,    4,   5,  6]
[0]    1,
[1]    3,     1,
[2]    9,    10,     1,
[3]   27,    79,    21,     1,
[4]   81,   580,   310,    36,    1,
[5]  243,  4141,  3990,   850,   55,  1,
[6]  729, 29230, 48031, 16740, 1895, 78,  1.
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225468 (m=3).
Cf. Columns: A000244, A016138, A018054.

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, 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]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[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_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, 4) for k in (0..n)]

Formula

T(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_4(n, j)) / (4^k*k!) where A_4(n,j) = A225118.
For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A016138; T(n, 2) ~ A018054.
T(n, n) ~ A000012; T(n, n-1) ~ A014105.
From Wolfdieter Lang, Apr 13 2017: (Start)
E.g.f.: exp(3*z)*exp((x/4)*(exp(4*z -1))). Sheffer triangle (see a comment above).
E.g.f. column k: exp(3*x)*(exp(4*x) -1)^k/(4^k*k!), k >= 0 (Sheffer property).
O.g.f. column k: x^m/Product_{j=0..k} (1 - (3+4*j)*x), k >= 0.
(End)

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

A225117 Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 3}(x) in descending order.

Original entry on oeis.org

1, 2, 1, 4, 13, 1, 8, 93, 60, 1, 16, 545, 1131, 251, 1, 32, 2933, 14498, 10678, 1018, 1, 64, 15177, 154113, 262438, 88998, 4089, 1, 128, 77101, 1475736, 4890287, 3870352, 692499, 16376, 1, 256, 388321, 13270807, 77404933, 117758659, 50476003, 5175013, 65527, 1
Offset: 0

Views

Author

Peter Luschny, May 02 2013

Keywords

Comments

The row sums equal the triple factorial numbers A032031 and the alternating row sums, i.e., Sum_{k=0..n}(-1)^k*T(n,k), are up to a sign A000810. - Johannes W. Meijer, May 04 2013

Examples

			[0]  1
[1]  2*x   +   1
[2]  4*x^2 +  13*x   +    1
[3]  8*x^3 +  93*x^2 +   60*x   +   1
[4] 16*x^4 + 545*x^3 + 1131*x^2 + 251*x + 1
...
The triangle T(n, k) begins:
n \ k 0      1        2        3       4      5     6 7 ...
0:    1
1:    2      1
2:    4     13        1
3:    8     93       60        1
5:   16    545     1131      251       1
6:   32   2933    14498    10678    1018      1
7:   64  15177   154113   262438   88998   4089     1
8:  128  77101  1475736  4890287 3870352 692499 16376 1
...  - _Wolfdieter Lang_, Apr 08 2017
Three term recurrence: T(2,1) = (3*(2-1)+1)*2 + (3*1+2)*1 = 13. - _Wolfdieter Lang_, Apr 10 2017
		

Crossrefs

Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,4}(x) = A225118.

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, 3), x, n-k), k=0..n)), n=0..6);
    # Recurrence
    P := proc(n,x) option remember; if n = 0 then 1 else
      (n*x+(1/3)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
      expand(%) fi end:
    A225117 := (n,k) -> 3^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, 3], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • PARI
    T(n, k) = sum(j=0, n - k, (-1)^(n - k - j)*binomial(n + 1, n - k - j)*(1 + 3*j)^n);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial
    def T(n,k): return sum((-1)**(n - k - j)* binomial(n + 1, n - k - j)*(1 + 3*j)**n for j in range(n - 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 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, 3).coefficients()[::-1] for n in (0..5)]
    

Formula

Generating function 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 = 3; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
From Wolfdieter Lang, Apr 10 2017: (Start)
T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+1, n-k-j)*(1+3*j)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..n-k} (-1)^(n-k-m)*binomial(n-m, k)*A284861(n, m), 0 <= k <= n.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are R(n, x) = (x-1)^n*Sum_{m=0} A284861(n, m)*(1/(x-1))^m, n >= 0, i.e. the row polynomials of A284861 in the variable 1/(x-1) multiplied by (x-1)^n.
The row polynomials with falling powers are P(n, x) = (1-x)^n*Sum_{m=0..n} A284861(n, m)*(x/(1-x))^m, n >= 0.
The e.g.f. of the row polynomials in falling powers of x (A_{n, 3}(x) of the name) is exp((1-x)*z)/(1 - (x/(1 - x)) * (exp(3*(1-x)*z) - 1)) = (1-x)*exp((1-x)*z)/(1 - x*exp(3*(1-x)*z)).
The e.g.f. of the row polynomials R(n, x) (rising powers of x) is then (1-x)*exp(2*(1-x)*z)/(1 - x*exp(3*(1-x)*z)).
Three term recurrence: T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = (3*(n-k)+1)*T(n-1, k-1) + (3*k+2)*T(n-1, k) for n >= 1, k=0..n. (End)

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

A000810 Expansion of e.g.f. (sin x + cos x)/cos 3x.

Original entry on oeis.org

1, 1, 8, 26, 352, 1936, 38528, 297296, 7869952, 78098176, 2583554048, 31336418816, 1243925143552, 17831101321216, 825787662368768, 13658417358350336, 722906928498737152, 13551022195053101056
Offset: 0

Views

Author

Keywords

Crossrefs

(-1)^(n*(n-1)/2)*a(n) gives the alternating row sums of A225118. - Wolfdieter Lang, Jul 12 2017

Programs

  • Mathematica
    CoefficientList[Series[(Sin[x]+Cos[x])/Cos[3*x], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 25 2013 *)
    Table[Abs[EulerE[n, 1/3]] 6^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 21 2015 *)
  • PARI
    x='x+O('x^66); v=Vec(serlaplace( (sin(x)+cos(x)) / cos(3*x) ) ) \\ Joerg Arndt, Apr 27 2013
  • Sage
    from mpmath import mp, lerchphi
    mp.dps = 32; mp.pretty = True
    def A000810(n): return abs(3^n*2^(n+1)*lerchphi(-1,-n,1/3))
    [int(A000810(n)) for n in (0..17)]  # Peter Luschny, Apr 27 2013
    

Formula

a(2n) = A000436(n).
(-1)^n*a(2n+1)=1-sum_{i=0,1,...,n-1} (-1)^i*binomial(2n+1,2i+1)*3^(2n-2i)*a(2i+1). - R. J. Mathar, Nov 19 2006
a(n) = | 3^n*2^(n+1)*lerchphi(-1,-n,1/3) |. - Peter Luschny, Apr 27 2013
a(n) ~ n!*2^(n+1)*3^(n+1/2)/Pi^(n+1) if n is even and a(n) ~ n!*2^(n+1)*3^n/Pi^(n+1) if n is odd. - Vaclav Kotesovec, Jun 25 2013
a(n) = (-1)^floor(n/2)*3^n*skp(n, 1/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014

A079858 E.g.f. 1/(cos(2*x) - sin(2*x)).

Original entry on oeis.org

1, 2, 12, 88, 912, 11552, 176832, 3150208, 64188672, 1470996992, 37459479552, 1049279715328, 32063706796032, 1061443378552832, 37841217707753472, 1445427909919080448, 58892032991566036992, 2549444593020567683072
Offset: 0

Views

Author

Michael Somos, Jan 20 2003

Keywords

Crossrefs

Cf. A001586.
(-1)^(n*(n-1)/2)*a(n) gives the alternating row sums of A225118. - Wolfdieter Lang, Jul 12 2017

Programs

  • Mathematica
    CoefficientList[Series[1/(Cos[2*x]-Sin[2*x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 07 2013 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(1/(cos(2*x+x*O(x^n))-sin(2*x+x*O(x^n))),n))
    
  • PARI
    x='x+O('x^66); v=Vec(serlaplace( 1/(cos(2*x)-sin(2*x)) ) ) \\ Joerg Arndt, Apr 27 2013
  • Sage
    from mpmath import mp, lerchphi
    mp.dps = 32; mp.pretty = True
    def A079858(n): return abs(2*8^n*lerchphi(-1,-n,1/4))
    [int(A079858(n)) for n in (0..17)]  # Peter Luschny, Apr 27 2013
    

Formula

E.g.f.: 1/(cos(2*x) - sin(2*x)).
a(n) = 2^n * A001586(n).
a(n) = | 2*8^n*lerchphi(-1,-n,1/4) |. - Peter Luschny, Apr 27 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - 8*x^2*(k+1)^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
a(n) ~ 4 * n^(n+1/2) * (8/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
E.g.f.: 1/(1-2*x)*(1 + 2*x^2/((1-2*x)*W(0) - x )), where W(k) = x + (k+1)/( 1 - 2*x/( 2*k+3 - x*(2*k+3)/W(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Dec 27 2013

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.

A337997 Triangle read by rows, generalized Eulerian polynomials evaluated at x = 1.

Original entry on oeis.org

1, 0, 1, 0, 2, 8, 0, 6, 48, 162, 0, 24, 384, 1944, 6144, 0, 120, 3840, 29160, 122880, 375000, 0, 720, 46080, 524880, 2949120, 11250000, 33592320, 0, 5040, 645120, 11022480, 82575360, 393750000, 1410877440, 4150656720
Offset: 0

Views

Author

Peter Luschny, Oct 07 2020

Keywords

Examples

			Polynomial triangle starts:
[0] 1
[1] 0, 1
[2] 0, 1+x, x^2+6*x+1
[3] 0, x^2+4*x+1, x^3+23*x^2+23*x+1, 8*x^3+93*x^2+60*x+1
[4] 0, x^3+11*x^2+11*x+1, x^4+76*x^3+230*x^2+76*x+1, 16*x^4+545*x^3+1131*x^2+251*x+
1, 81*x^4+1996*x^3+3446*x^2+620*x+1
Integer triangle starts:
[0] 1
[1] 0,    1
[2] 0,    2,      8
[3] 0,    6,     48,      162
[4] 0,   24,    384,     1944,     6144
[5] 0,  120,   3840,    29160,   122880,    375000
[6] 0,  720,  46080,   524880,  2949120,  11250000,   33592320
[7] 0, 5040, 645120, 11022480, 82575360, 393750000, 1410877440, 4150656720
		

Crossrefs

Programs

  • Maple
    # Two alternative implementations are given in the link.
    GeneralizedEulerianPolynomial := proc(n, k, x) local S;
       if n = 0 then  return 1 fi;
       S := m -> add((-1)^j*binomial(n+1,j)*(k*(m-j)+1)^n*signum(k*(m-j)+1),j=0..n+1);
       add(S(m)*x^m, m=0..n)/2 end:
    T := (n, k) -> subs(x=1, GeneralizedEulerianPolynomial(n, k, x)):
    for n from 0 to 6 do seq(T(n, k), k=0..n) od;

Formula

The polynomials are defined P(0,0,x)=1 and P(n,k,x) = (1/2)*Sum_{m=0..n} S(m)*x^m where S(m) = Sum_{j=0..n+1}(-1)^j*binomial(n+1,j)*(k*(m-j)+1)^n*signum(k*(m-j)+1).
T(n, k) = P(n, k, 1).
T(n, k) = n!*k^n. - Hugo Pfoertner, Oct 07 2020
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