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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A027612 Numerator of 1/n + 2/(n-1) + 3/(n-2) + ... + (n-1)/2 + n.

Original entry on oeis.org

1, 5, 13, 77, 87, 223, 481, 4609, 4861, 55991, 58301, 785633, 811373, 835397, 1715839, 29889983, 30570663, 197698279, 201578155, 41054655, 13920029, 325333835, 990874363, 25128807667, 25472027467, 232222818803, 235091155703, 6897956948587, 6975593267347
Offset: 1

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Author

Glen Burch (gburch(AT)erols.com)

Keywords

Comments

Numerator of a second-order harmonic number H(n, (2)) = Sum_{k=1..n} HarmonicNumber(k). - Alexander Adamchuk, Apr 12 2006
p divides a(p-3) for prime p > 3. - Alexander Adamchuk, Jul 06 2006
Denominator is A027611(n+1). p divides a(p-3) for prime p > 3. - Alexander Adamchuk, Jul 26 2006
a(n) = A213998(n,n-2) for n > 1. - Reinhard Zumkeller, Jul 03 2012

Crossrefs

Programs

  • Haskell
    import Data.Ratio ((%), numerator)
    a027612 n = numerator $ sum $ zipWith (%) [1 .. n] [n, n-1 .. 1]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Numerator((&+[j/(n-j+1): j in [1..n]])): n in [1..30]]; // G. C. Greubel, Aug 23 2022
    
  • Maple
    a := n -> numer(add((n+1-j)/j, j=1..n));
    seq(a(n), n = 1..29); # Peter Luschny, May 12 2023
  • Mathematica
    Numerator[Table[Sum[Sum[1/i,{i,1,k}],{k,1,n}],{n,1,30}]] (* Alexander Adamchuk, Apr 12 2006 *)
    Numerator[Table[Sum[k/(n-k+1),{k,1,n}],{n,1,50}]] (* Alexander Adamchuk, Jul 26 2006 *)
  • PARI
    a(n) = numerator(sum(k=1, n, k/(n-k+1))); \\ Michel Marcus, Jul 14 2018
    
  • SageMath
    [numerator(n*(harmonic_number(n+1) - 1)) for n in (1..30)] # G. C. Greubel, Aug 23 2022

Formula

From Vladeta Jovovic, Sep 02 2002: (Start)
a(n) = numerators of coefficients in expansion of -log(1-x)/(1-x)^2.
a(n) = numerators of (n+1)*(harmonic(n+1) - 1).
a(n) = numerators of (n+1)*(Psi(n+2) + Euler-gamma - 1). (End)
a(n) = numerator( Sum_{k=1..n} Sum_{i=1..k} 1/i ). - Alexander Adamchuk, Apr 12 2006
a(n) = numerator( Sum_{k=1..n} k/(n-k+1) ). - Alexander Adamchuk, Jul 26 2006
a(n) = numerator of integral_{x=1..n+1} floor((n+1)/x). - Jean-François Alcover, Jun 18 2013

A028421 Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 6, 22, 18, 4, 24, 100, 105, 40, 5, 120, 548, 675, 340, 75, 6, 720, 3528, 4872, 2940, 875, 126, 7, 5040, 26136, 39396, 27076, 9800, 1932, 196, 8, 40320, 219168, 354372, 269136, 112245, 27216, 3822, 288, 9
Offset: 0

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Author

Peter Wiggen (wiggen(AT)math.psu.edu)

Keywords

Comments

Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.
We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2 + 6*n + 3*n^2)/x^2 - (6 + 22*n + 18*n^2 + 4*n^3)/x^3 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.
The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).
(End)
For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - Tom Copeland, Feb 28 2014
From Wolfdieter Lang, Nov 25 2018: (Start)
The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).
The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).
The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)
The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
n\k       0        1        2        3        4       5       6      7     8   9 10
------------------------------------------------------------------------------------
0:        1
1:        1        2
2:        2        6        3
3:        6       22       18        4
4:       24      100      105       40        5
5:      120      548      675      340       75       6
6:      720     3528     4872     2940      875     126       7
7:     5040    26136    39396    27076     9800    1932     196      8
8:    40320   219168   354372   269136   112245   27216    3822    288     9
9:   362880  2053152  3518100  2894720  1346625  379638   66150   6960   405  10
10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11
... - _Wolfdieter Lang_, Nov 23 2018
		

Crossrefs

Row sums give A000254(n+1), n >= 0.
Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.
From Johannes W. Meijer, Oct 07 2009: (Start)
A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.
The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).
Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

Programs

  • Maple
    A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:
    seq(seq(A028421(n,k), k=0..n), n=0..8);
    # Johannes W. Meijer, Oct 07 2009, Revised Sep 09 2012
    egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):
    ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):
    seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # Peter Luschny, Jun 12 2022
  • Mathematica
    f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(-ln(1 - x), 10, True) # Peter Luschny, Jan 03 2019

Formula

E.g.f.: d/dt(-log(1-t)/(1-t)^x). - Vladeta Jovovic, Oct 12 2003
The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - Peter Bala, Jul 18 2013
Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - Wolfdieter Lang, Nov 25 2018

Extensions

Edited by Wolfdieter Lang, Nov 23 2018

A001711 Generalized Stirling numbers.

Original entry on oeis.org

1, 7, 47, 342, 2754, 24552, 241128, 2592720, 30334320, 383970240, 5231113920, 76349105280, 1188825724800, 19675048780800, 344937224217600, 6386713749964800, 124548748102195200, 2551797512248320000, 54804198761303040000, 1231237843834521600000
Offset: 0

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Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=3) ~ exp(-x)/x^2*(1 - 7/x + 47/x^2 - 342/x^3 + 2754/x^4 - 24552/x^5 + 241128/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
For n > 4, a(n) mod n = 0 for n composite, = n-3 for n prime. - Gary Detlefs, Jul 18 2011
From Petros Hadjicostas, Jun 11 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) using slightly different notation. They were further examined by Mitrinovic and Mitrinovic (1962).
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m. (Because an empty product is by definition 1, we may let R_0^0(a,b) = 1.)
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m). (Array A008275 is the same as array A048994 but with no zero row and no zero column.)
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+1}^1(a=-3, b=-1) for n >= 0. (End)

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Maple
    a := n-> add(1/2*((n+3)!/(k+3)), k=0..n): seq(a(n), n=0..19); # Zerinvary Lajos, Jan 22 2008
    a := n -> (n+1)!*hs2(n+1): hs2 := n-> add(hs(k), k=0..n): hs := n-> add(h(k), k=0..n): h := n-> add(1/k, k=1..n): seq(a(n), n=0..19); # Gary Detlefs, Jan 01 2011
  • Mathematica
    f[k_] := k + 2; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}]; (* Clark Kimberling, Dec 29 2011 *)
    Table[(n + 3)!*Sum[1/(2*k + 4), {k, 1, n + 1}], {n,0,100}] (* G. C. Greubel, Jan 15 2017 *)
  • PARI
    for(n=0, 19, print1((n+1)! * sum(k=0, n, binomial(k + 2, 2) / (n + 1 - k)),", ")) \\ Indranil Ghosh, Mar 13 2017
    
  • PARI
    R(n,m,a,b) =  sum(k=0, n-m, (-1)^k*a^k*b^(n-m-k)*binomial(m+k,k)*stirling(n, m+k,1));
    aa(n) = R(n+1,1,-3,-1);
    for(n=0, 19, print1(aa(n), ",")) \\ Petros Hadjicostas, Jun 11 2020

Formula

E.g.f.: -log(1 - x)/(1 - x)^3 if offset 1. With offset 0: (d/dx)(-log(1 - x)/(1 - x)^3) = (1 - 3*log(1 - x))/(1 - x)^4.
a(n) = Sum_{k=0..n} ((-1)^(n+k)*(k+1)*3^k*Stirling1(n+1, k+1)). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-3,k)/(n-k)). - Milan Janjic, Dec 14 2008
a(n) = ( A000254(n+3) - 3*A001710(n+3) )/2. - Gary Detlefs, May 24 2010
a(n) = ((n+3)!/4) * (2*h(n+3) - 3), where h(n) = Sum_{k=1..n} (1/k) is the n-th harmonic number. - Gary Detlefs, Aug 15 2010
a(n) = n!*[2]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. With offset 1. - Gary Detlefs, Jan 04 2011
a(n) = (n+3)! * Sum_{k=1..n+1} (1/(2*k+4)). - Gary Detlefs, Sep 14 2011
a(n) = (n+1)! * Sum_{k=0..n} (binomial(k+2,2)/(n+1-k)). - Gary Detlefs, Dec 01 2011
a(n) = A001705(n+2) - A182541(n+4). - Anton Zakharov, Jul 02 2016
a(n) ~ n^(n+7/2) * exp(-n) * sqrt(Pi/2) * log(n) * (1 + (gamma - 3/2)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 12 2016
Conjectural D-finite with recurrence: a(n) + (-2*n-5)*a(n-1) + (n+2)^2*a(n-2)=0. - R. J. Mathar, Feb 16 2020
From Petros Hadjicostas, Jun 11 2020: (Start)
Since a(n) = R_{n+1}^1(a=-3, b=-1), it follows from Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) that:
a(n) = [x] Product_{r=0}^n (x + 3 + r) = (Product_{r=0}^n (3 + r)) * Sum_{s=0}^n 1/(3 + s).
a(n) = (n + 2)!/2 + (n + 3)*a(n-1) for n >= 1. [This can be used to prove R. J. Mathar's recurrence above.] (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
Maple programs corrected and edited by Johannes W. Meijer, Nov 28 2012

A027611 Denominator of n * n-th harmonic number.

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 20, 35, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 680680, 4084080, 3879876, 739024, 235144, 5173168, 14872858, 356948592, 343219800, 2974571600, 2868336900, 80313433200, 77636318760
Offset: 1

Views

Author

Glen Burch (gburch(AT)erols.com)

Keywords

Comments

This is very similar to A128438, which is a different sequence. They differ at n=6 (and nowhere else?). - N. J. A. Sloane, Nov 21 2008
Denominator of 1/n + 2/(n-1) + 3/(n-2) + ... + (n-1)/2 + n.
Denominator of Sum_{k=1..n} frac(n/k) where frac(x/y) denotes the fractional part of x/y. - Benoit Cloitre, Oct 03 2002
Denominator of Sum_{d=2..n-1, n mod d > 0} n/d. Numerator = A079076. - Reinhard Zumkeller, Dec 21 2002
a(n) is odd iff n is a power of 2. - Benoit Cloitre, Oct 03 2002
Indices where a(n) differs from A128438 are terms of A074791. - Gary Detlefs, Sep 03 2011

Crossrefs

Programs

  • Haskell
    import Data.Ratio ((%), denominator)
    a027611 n = denominator $ sum $ map (n %) [1..n]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Denominator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014
    
  • Maple
    a := n -> denom(add((n-j)/j, j=1..n));
    seq(a(n), n = 1..30); # Peter Luschny, May 12 2023
  • Mathematica
    a[n_]:=Denominator[n*HarmonicNumber[n]]; Array[a,100] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *)
  • PARI
    a(n) = denominator(n*sum(k=1, n, 1/k)); \\ Michel Marcus, Feb 15 2015
    
  • Python
    from sympy import harmonic
    def A027611(n): return (n*harmonic(n)).q # Chai Wah Wu, Sep 26 2021
    
  • SageMath
    [denominator(n*harmonic_number(n)) for n in (1..40)] # G. C. Greubel, Aug 24 2022

Formula

From Vladeta Jovovic, Sep 02 2002: (Start)
a(n) = denominators of coefficients in expansion of -log(1-x)/(1-x)^2.
a(n) = denominators of (n+1)*(harmonic(n+1) - 1).
a(n) = denominators of (n+1)*(Psi(n+2) + Euler-gamma - 1). (End)
a(n) = numerator(h(n)/h(n-1)) - denominator(h(n)/h(n-1)), n > 1, where h(n) is the n-th harmonic number. - Gary Detlefs, Sep 03 2011
a(n) = A213999(n, n-2) for n > 1. - Reinhard Zumkeller, Jul 03 2012
a(n) = denominators of coefficients of e.g.f. -1 + exp(x)*(1 + Sum_{j >= 0} (-x)^(j+1)/(j * j!)). - G. C. Greubel, Aug 24 2022

Extensions

Entry revised by N. J. A. Sloane following a suggestion of Eric W. Weisstein, Jul 02 2004

A052750 a(n) = (2*n + 1)^(n - 1).

Original entry on oeis.org

1, 1, 5, 49, 729, 14641, 371293, 11390625, 410338673, 16983563041, 794280046581, 41426511213649, 2384185791015625, 150094635296999121, 10260628712958602189, 756943935220796320321, 59938945498865420543457
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left child has a larger label than its parent. - Brian Drake, Jul 28 2008
Put a(0) = 1. For n > 0, let x(n,k) = 2*cos((2*k-1)*Pi/(2*n+1)), k=1..n. Define the recurrences S(n;0,x(n,k)) = 1, S(n;1,x(n,k)) = x(n,k), S(n;r,x(n,k)) = x(n,k)*S(n;r-1,x(n,k)) - S(n;r-2,x(n,k)), r > 1 an integer, k=1..n. CONJECTURE: For n > 0, a(n) = Product_{k=1..n} (Sum_{m=0..n-1} S(n;2*m,x(n,k))^2). - L. Edson Jeffery, Sep 11 2013
From Wolfdieter Lang, Dec 16 2013: (Start)
Discriminants of the first difference of Chebyshev S-polynomials.
The coefficient table for the first difference polynomials P(n, x) = S(n, x) - S(n-1, x), n >= 0, S(-1, x) = 0, with the Chebyshev S polynomials (see A049310), is given in A130777.
For the discriminant of a polynomial in terms of the square of a determinant of a Vandermonde matrix build from the zeros of the polynomial see, e.g., A127670.
For the proof that D(n) := discriminant(P(n,x)) = (2*n + 1)^(n - 1), n >= 1, use the formula given e.g., in the Rivlin reference, p. 218, Theorem 5.13, eq. (5.3), namely D(n) = (-1)^(n*(n-1)/2)*Product_{j=1..n} P'(n, x(n,j)), with the zeros x(n,j) = -2*cos(2*Pi*j/(2*n+1)) of P(n, x) (see A130777). P'(n, x(n,j)) = (2*n+1)*P(n-1, x(n,j))/(2*sin(Pi*j/(2*n+1))*2*cos(Pi*j/(2*n+1)))^2. P(n-1, x(n,j)) = (-1)^(n+j)*2*cos(Pi*j/(2*n+1)). Product_{j=1..n} 2*sin(Pi*j/(2*n+1)) = 2*n+1 (see the Oct 10 2013 formula in A005408. Product_{j=1..n} 2*cos(Pi*j/(2*n+1)) = 1, because S(2*n, 0) = (-1)^n.
(End)
a(n) is the number of labeled 2-trees with n+2 vertices, rooted at a given edge. - Nikos Apostolakis, Nov 30 2018
a(n) is also the number of 2-trees with n labeled triangles and with a distinguished oriented edge. - Nikos Apostolakis, Dec 14 2018

Examples

			Discriminant: n=4: P(4, x) = 1 + 2*x - 3*x^2 - x^3 + x^4 with the zeros x[1] = -2*cos((2/9)*Pi), x[2] = -2*cos((4/9)*Pi), x[3] = 1, x[4] = 2*cos((1/9)*Pi). D(4) = (Det(Vandermonde(4,[x[1],x[2],x[3],x[4]])))^2 = 729 = a(4). - _Wolfdieter Lang_, Dec 16 2013
		

References

  • L. W. Beineke, and J. W. Moon, Several proofs of the number of labelled 2-dimensional trees, In "Proof Techniques in Graph Theory" (F. Harary editor). Academic Press, New York, 1969, pp. 11-20.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.

Crossrefs

Programs

  • GAP
    List([0..20],n->(2*n+1)^(n-1)); # Muniru A Asiru, Dec 05 2018
  • Magma
    [(2*n+1)^(n-1) : n in [0..20]]; // Wesley Ivan Hurt, Jan 20 2017
    
  • Maple
    spec := [S,{B=Prod(Z,S,S),S=Set(B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    max = 16; (Series[Exp[-1/2*ProductLog[-2*x]], {x, 0, max}] // CoefficientList[#, x] & ) * Range[0, max]! (* Jean-François Alcover, Jun 20 2013 *)
    Table[(2*n+1)^(n-1),{n,0,20}] (* Harvey P. Dale, Jul 14 2025 *)
  • PARI
    a(n)=(2*n+1)^(n-1) \\ Charles R Greathouse IV, Nov 20 2011
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,21,A=sqrt(1+2*sum(n=1,21,x^(2*n-1)/(2*n-1)!*A^(4*n-1))+x*O(x^n)));n!*polcoeff(A,n)} \\ Paul D. Hanna, Sep 07 2012
    
  • Python
    for n in range(0, 20): print((2*n + 1)**(n - 1), end=', ') # Stefano Spezia, Dec 01 2018
    

Formula

E.g.f.: exp(-1/2*W(-2*x)), where W is Lambert's W function.
E.g.f. satisfies: A(x) = sqrt(1 + 2*Sum_{n>=1} x^(2*n-1)/(2*n-1)! * A(x)^(4*n-1)). - Paul D. Hanna, Sep 07 2012
E.g.f. satisfies: A(x) = 1/A(-x*A(x)^4). - Paul D. Hanna, Sep 07 2012
a(n) = discriminant of P(n,x) = S(n,x) - S(n-1,x), n >= 1, with the Chebyshev S polynomials from A049310. For the proof see the comment above. a(n) is also the discriminant of S(n,x) + S(n-1,x) = (-1)^n*(S(n,-x) - S(n-1,-x)). - Wolfdieter Lang, Dec 16 2013
From Peter Bala, Dec 19 2013: (Start)
The e.g.f. A(x) = 1 + x + 5*x^2/2! + 49*x^3/3! + 729*x^4/4! + ... satisfies:
1) A(x*exp(-2*x)) = exp(x) = 1/A(-x*exp(2*x));
2) A^2(x) = 1/x*series reversion(x*exp(-2*x));
3) A(x^2) = 1/x*series reversion(x*exp(-x^2));
4) A(x) = exp(x*A(x)^2). (End)
E.g.f.: sqrt(-LambertW(-2*x)/(2*x)). - Vaclav Kotesovec, Dec 07 2014
Related to A001705 by Sum_{n >= 1} a(n)*x^n/n! = series reversion( 1/(1 + x)^2*log(1 + x) ) = series reversion(x - 5*x^2/2! + 26*x^3/3! - 154*x^4/4! + ...). Cf. A000272, A052752, A052774, A052782. - Peter Bala, Jun 15 2016
From Peter Bala, Dec 13 2022: (Start)
The e.g.f. A(x) = 1/x * series reversion of x^2/T(x), where the tree function T(x) = Sum_{n >= 1} n^(n-1)*x^n/n!. See A000169.
For c in C, A(x)^c = 1 + Sum_{n >= 1} c*(2*n + c)^(n-1)*x^n/n!.
First derivative A'(x) = A(x)^3/(1 - 2*x*A(x)^2).
Series reversion of (1 - A(-z)) = -log(1 - z)/(1 - z)^2 is the e.g.f. of A001705.
1/z * series reversion of z/A(z) = 1 + z + 7*z^2/2! + (10^2)*z^3/3! + (13^3)*z^4/4! + ... is the e.g.f. of A052752.
1/z * series reversion of z/A(z^2) = 1 + z^2 + 9*z^4/2! + (13^2)*z^6/3! + (17^3)*z^8/4! + ... = Sum_{n >= 0} A052774(n)*z^(2*n)/n!.
1/z * series reversion of z/A(z^3) = 1 + z^3 + 11*z^6/2! + (16^2)*z^9/3! + (21^3)*z^12/4! + ... = Sum_{n >= 0} A052782(n)*z^(3*n)/n!.
1/z * series reversion of z/A(z)^2 = A(2*z) = 2*Sum_{n >= 0} (4*n + 2)^(n-1)*z^n/n!.
1/z * series reversion of z/A(z)^k = k*Sum_{n >= 0} ((k+2)*n + k)^(n-1)*z^n/n!. (End)
a(n) = Sum_{k=1..n} (-1)^(n-k)*(n+k)^(n-1)*binomial(n,k-1), a(0)=1. - Vladimir Kruchinin, Aug 14 2025

Extensions

Better description from Vladeta Jovovic, Sep 02 2003

A001716 Generalized Stirling numbers.

Original entry on oeis.org

1, 9, 74, 638, 5944, 60216, 662640, 7893840, 101378880, 1397759040, 20606463360, 323626665600, 5395972377600, 95218662067200, 1773217155225600, 34758188233574400, 715437948072960000, 15429680577561600000, 347968129734973440000, 8190600438533990400000
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=4) ~ exp(-x)/x^2*(1 - 9/x + 74/x^2 - 638/x^3 + 5944/x^4 - 60216/x^5 + 662640/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 23 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) using slightly different notation.
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_0^0(a,b) = 1, R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m) for n, m >= 0.
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+1}^1(a=-4, b=-1) for n >= 0. (End)

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Mathematica
    f[k_] := k + 3; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}] (* Clark Kimberling, Dec 29 2011 *)
    Rest[CoefficientList[Series[(1-x)^(-4)*Log[1/(1-x)],{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Jan 19 2014 *)
  • PARI
    R(n, m, a, b) =  sum(k=0, n-m, (-1)^k*a^k*b^(n-m-k)*binomial(m+k, k)*stirling(n, m+k, 1));
    aa(n) = R(n+1, 1, -4, -1);
    for(n=0, 19, print1(aa(n), ", ")) \\ Petros Hadjicostas, Jun 23 2020

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k) * (k+1) * 4^k * stirling1(n+1, k+1). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n-1) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-4,k)/(n-k) for n >= 1. [Milan Janjic, Dec 14 2008] [Edited by Petros Hadjicostas, Jun 23 2020]
a(n)= n! * [3]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n (with offset 1). [Gary Detlefs, Jan 04 2011]
a(n) = (n+1)! * Sum_{k=0..n} (-1)^k*binomial(-4,k)/(n+1-k). [Gary Detlefs, Jul 16 2011]
a(n) = (n+4)! * Sum_{k=1..n+1} 1/(k+3)/6. [Gary Detlefs, Sep 14 2011]
E.g.f. (for offset 1): 1/(1-x)^4 * log(1/(1-x)). - Vaclav Kotesovec, Jan 19 2014
E.g.f.: (1 + 4*log(1/(1 - x)))/(1 - x)^5. - Ilya Gutkovskiy, Jan 23 2017
From Petros Hadjicostas, Jun 23 2020: (Start)
a(n) = [x] Product_{r=0..n} (x + 4 + r) = (Product_{r=0..n} (4 + r)) * Sum_{i=0..n} 1/(4 + i).
Since a(n) = R_{n+1}^1(a=-4, b=-1) and R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b), we conclude that:
(i) a(n) = (n+3)!/6 + (n+4)*a(n-1) for n >= 1;
(ii) a(n) = (2*n+7)*a(n-1) - (n+3)^2*a(n-2) for n >= 2. (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A001721 Generalized Stirling numbers.

Original entry on oeis.org

1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, 284574960, 4243508640, 67285058400, 1131047366400, 20099588140800, 376612896038400, 7422410595801600, 153516757766400000, 3325222830101760000, 75283691519393280000, 1778358268603445760000
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=5) ~ exp(-x)/x^2*(1 - 11/x + 107/x^2 - 1066/x^3 + 11274/x^4 - 127860/x^5 + 1557660/x^6 - ... ) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Related to n!*(the k-th successive summation of the harmonic numbers): k=0..A000254, k=1..A001705,k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Mathematica
    f[k_] := k + 4; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}] (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+1, 1)*5^k*Stirling1(n+1, k+1). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-5,k)/(n-k). - Milan Janjic, Dec 14 2008
a(n) = n!*[4]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. With offset 1. - Gary Detlefs, Jan 04 2011
E.g.f.: (1 + 5*log(1/(1-x)))/(1 - x)^6. - Ilya Gutkovskiy, Jan 23 2017

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A051524 Second unsigned column of triangle A051338.

Original entry on oeis.org

0, 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, 703404576, 11277554400, 191338156800, 3427105248000, 64651956364800, 1281740285145600, 26648514872985600, 579892995734169600, 13183403757582643200
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=6) ~ exp(-x)/x^2*(1 - 13/x + 146/x^2 - 1650/x^3 + 19524/x^4 - 245004/x^5 + 3272688/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S.: see reference given for triangle A051338.

Crossrefs

Cf. A001725 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs, Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 5; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051338(n, 1)*(-1)^(n-1);
E.g.f.: -log(1-x)/(1-x)^6.
For n>=1, a(n) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-6,k)/(n-k). - Milan Janjic, Dec 14 2008
a(n) = n!*[5]h(n), where [k]h(n) denotes the k-th successive summation of h(n) from 0 to n. - Gary Detlefs, Jan 04 2011
Conjecture: a(n) +(-2*n-9)*a(n-1) +(n+4)^2*a(n-2)=0. - R. J. Mathar, Aug 04 2013

A051545 Second unsigned column of triangle A051339.

Original entry on oeis.org

0, 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, 1576890000, 27046454400, 488849155200, 9293295110400, 185464792800000, 3878247384345600, 84822225638169600, 1937048605944883200, 46113230058645657600
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=7) ~ exp(-x)/x^2*(1 - 15/x + 191/x^2 - 2414/x^3 + 31594/x^4 - 434568/x^5 + 6314664/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051339.

Crossrefs

Cf. A001730 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..(this sequence), k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs, Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 6; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051339(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^7.
a(n) = n!*Sum_{k=0,..,n-1}((-1)^k*binomial(-7,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[6]h(n), where [k]h(n) denotes the k-th successive summation of The harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A051560 Second unsigned column of triangle A051379.

Original entry on oeis.org

0, 1, 17, 242, 3382, 48504, 725592, 11393808, 188204400, 3270729600, 59753750400, 1146140409600, 23046980025600, 485075533132800, 10669304848204800, 244861798361241600, 5854837379724748800
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=8) ~ exp(-x)/x^2*(1 - 17/x + 242/x^2 - 3382/x^3 + 48504/x^4 - 725592/x^5 + 11393808/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051379.

Crossrefs

Cf. A049388 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 7; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051379(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^8.
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-8,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[7]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011
Conjecture: a(n) +(-2*n-13)*a(n-1) +(n+6)^2*a(n-2)=0. - R. J. Mathar, Aug 04 2013
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