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 1-9 of 9 results.

A285064 Row sums of Sheffer triangle S2[4,1] = A285061.

Original entry on oeis.org

1, 5, 41, 429, 5329, 75989, 1215481, 21453693, 412820385, 8579772325, 191166679497, 4538638641997, 114238219541617, 3035305413035125, 84819458105387417, 2484842038066995485, 76101249873390595905, 2430497813260105226053, 80769536433102942870377, 2787318255464814752951533
Offset: 0

Views

Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

See A285061 for details. These are generalized Bell numbers (A000110) because A285061 is a generalized Stirling2 triangle.
For the alternating row sums of A285061 see A285065.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]*BellB[k]*4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 19 2017 *)
  • Python
    from sympy import binomial, bell
    def a(n): return sum([binomial(n, k)*bell(k)*4**k for k in range(n + 1)]) # Indranil Ghosh, Apr 19 2017

Formula

a(n) = Sum_{m=0..n} A285061(n, m), n >= 0.
E.g.f.: exp(x)*exp(exp(4*x) - 1).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(1+4*m)^n, n >= 0. (Dobiński type formula from the A285061(n,m) sum formula, after interchange of summations).
a(n) = Sum_{k=0..n} binomial(n, k)*A000110(k)*4^k, n >= 0. From the Vaclav Kotesovec program. This follows from the S2[4,1] formula in terms of Stirling2. - Wolfdieter Lang, Apr 24 2017
a(0) = 1; a(n) = a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ Bell(n) * (4 + LambertW(n)/n)^n. - Vaclav Kotesovec, Jun 22 2022
a(n) ~ 4^n * n^(n + 1/4) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 1/4)). - Vaclav Kotesovec, Jun 27 2022

A285065 Alternating row sums of Sheffer triangle S2[4,1] = A285061.

Original entry on oeis.org

1, -3, -7, 53, 497, -147, -44055, -437339, 971745, 90858205, 1254551513, -56188139, -361749699119, -7793811482035, -47717641321527, 2053219888651909, 77548473901557697, 1171383881442334141, -8155337883596701767
Offset: 0

Views

Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

See A285061 for details. This is a generalization of A000587.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]*BellB[k, -1]*4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 19 2017 *)
  • Python
    from sympy import bell, binomial
    def a(n): return sum([binomial(n, k)*bell(k, -1)*4**k for k in range(n + 1)]) # Indranil Ghosh, May 06 2017

Formula

a(n) = Sum_{m=0..n} (-1)^m*A285061(n, m), n >= 0.
E.g.f.: exp(x)*exp(1 - exp(4*x)).
a(n) = e*Sum_{m>=0} ((-1)^m / m!)*(1 + 4*m)^n, n >= 0, (Dobiński type formula).
a(n) = Sum_{k=0..n} binomial(n, k) * 4^k * A000587(k), n >= 0. - Vaclav Kotesovec, Apr 23 2017
a(0) = 1; a(n) = a(n-1) - Sum_{k=1..n} binomial(n-1,k-1) * 4^k * a(n-k). - Ilya Gutkovskiy, Nov 30 2023

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.

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

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

Views

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)

A157866 Numerator of Bernoulli(n, 1/5).

Original entry on oeis.org

1, -3, 1, 6, -29, -74, 4537, 1946, -23789, -88434, 15034541, 6154786, -10417027559, -607884394, 13199705071, 80834386026, -34108052679853, -13923204233954, 51709981061257363, 3015393801263666, -1029159167703800359, -801997872697905114, 629565265428734672873
Offset: 0

Views

Author

N. J. A. Sloane, Nov 08 2009

Keywords

Comments

From Wolfdieter Lang, Jul 05 2017: (Start)
a(n) gives also the numerators of the generalized Bernoulli numbers B[5,1](n) = 5^n*B(n, 1/5) with the Bernoulli polynomials B(n, x) = Bernoulli(n, x) from A196838/A196839 or A053382/A053383. For the denominators see A288872(n) = A157867(n)/5^n.
(-1)^n*a(n) gives the numerators of the generalized Bernoulli numbers B[5,4](n). The denominators are also A288872(n).
The generalized Bernoulli numbers B[d,a](n), for d >= 1, a >= 0, with gcd(d, a) = 1 are defined in terms of generalized Stirling2 numbers by B[d,a](n) = Sum_{k=0..n} ((-1)^k / (k+1))*S2[d,a](n, k)*k!, n >= 0. See A285061 for more details.
(End)

Crossrefs

For denominators see A157867, and also A288872.

Programs

  • Mathematica
    Table[Numerator[BernoulliB[n, 1/5]], {n, 0, 50}] (* Vincenzo Librandi, Mar 16 2014 *)
  • PARI
    a(n)=numerator(subst(bernpol(n, x), x, 1/5)); \\ Michel Marcus, Jul 06 2017

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

A285062 Numerators of the exponential expansion of (4/(log(1+x)))*(1-1/(1+x)^(1/4)).

Original entry on oeis.org

1, -1, 7, -81, 3853, -25721, 1862773, -52571875, 2828694491, -20554196553, 2489317910533, -36843139557745, 187344440646279463, -200535626786994961, 15853768141768274581, -319644021424695652161, 927777140067161706072467, -1412565248386878259675625, 2151379749437782936765977859
Offset: 0

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Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

This gives the numerators of the z-sequence for the Sheffer triangle (exp(x), exp(4*x) - 1) shown in A285061. 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 4*A006232/A006233.
For the nontrivial recurrence for the sequence {1^n} of column m=0 of A285061 by the z-sequence see the example n=4 below.

Examples

			The rationals r(n) = a(n)/A285063(n),  n >= 0,  start: 1, -1/8, 7/48, -81/256, 3853/3840, -25721/6144, 1862773/86016, -52571875/393216, 2828694491/2949120, -20554196553/2621440, ...
The z-Recurrence for A285061(4, 0) = 1 is  1 = 4*(1*1 + 124*(-1/8) + 240*(7/48) + 64*(-81/256)).
		

Crossrefs

Formula

The e.g.f. of the rationals r(n) = a(n)/A285063(n) is (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)).

A285063 Denominators of the exponential expansion of (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)).

Original entry on oeis.org

1, 8, 48, 256, 3840, 6144, 86016, 393216, 2949120, 2621440, 34603008, 50331648, 22900899840, 2013265920, 12079595520, 17179869184, 3285649981440, 309237645312, 27419071217664, 23089744183296
Offset: 0

Views

Author

Wolfdieter Lang, Apr 13 2017

Keywords

Comments

For the numerators see A285062, also for the rationals r(n) = A285062(n)/a(n), the z-sequence of the Sheffer triangle S2[4,1].

Examples

			See A285062.
		

Crossrefs

Formula

E.g.f. for sequence {r(n) = A285062(n)/a(n)}: (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)).
Showing 1-9 of 9 results.