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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A000051 a(n) = 2^n + 1.

Original entry on oeis.org

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049, 4097, 8193, 16385, 32769, 65537, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649, 4294967297, 8589934593
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequence L(2,3).
Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003
See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004
From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)* charpoly(A,3). - Milan Janjic, Jan 27 2010
First differences of A006127. - Reinhard Zumkeller, Apr 14 2011
The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011
Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012
Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012
Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012
For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014
Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014
a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014
For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017
Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017
Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017
Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

References

  • Paul Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 60, 244.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 141.

Crossrefs

Apart from the initial 1, identical to A094373.
See A008776 for definitions of Pisot sequences.
Column 2 of array A103438.
Cf. A007583 (a((n-1)/2)/3 for odd n).

Programs

  • Haskell
    a000051 = (+ 1) . a000079
    a000051_list = iterate ((subtract 1) . (* 2)) 2
    -- Reinhard Zumkeller, May 03 2012
    
  • Magma
    [2^n+1: n in [0..40]]; // G. C. Greubel, Jan 18 2025
  • Maple
    A000051:=-(-2+3*z)/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
    a := n -> add(binomial(n,k)*bernoulli(n-k,1)*2^(k+1)/(k+1),k=0..n); # Peter Luschny, Apr 20 2009
  • Mathematica
    Table[2^n + 1, {n,0,40}]
    2^Range[0,40] + 1 (* Eric W. Weisstein, Jul 17 2017 *)
    LinearRecurrence[{3, -2}, {2, 3}, 40] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    a(n)=2^n+1
    
  • PARI
    first(n) = Vec((2 - 3*x)/((1 - x)*(1 - 2*x)) + O(x^n)) \\ Iain Fox, Dec 31 2017
    
  • Python
    def A000051(n): return (1<Chai Wah Wu, Dec 21 2022
    

Formula

a(n) = 2*a(n-1) - 1 = 3*a(n-1) - 2*a(n-2).
G.f.: (2-3*x)/((1-x)*(1-2*x)).
First differences of A052944. - Emeric Deutsch, Mar 04 2004
a(0) = 1, then a(n) = (Sum_{i=0..n-1} a(i)) - (n-2). - Gerald McGarvey, Jul 10 2004
Inverse binomial transform of A007689. Also, V sequence in Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005
a(n) = A127904(n+1) for n>0. - Reinhard Zumkeller, Feb 05 2007
Equals binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = A000079(n)+1. - Omar E. Pol, May 18 2008
E.g.f.: exp(x) + exp(2*x). - Mohammad K. Azarian, Jan 02 2009
a(n) = A024036(n)/A000225(n). - Reinhard Zumkeller, Feb 14 2009
From Peter Luschny, Apr 20 2009: (Start)
A weighted binomial sum of the Bernoulli numbers A027641/A027642 with A027641(1)=1 (which amounts to the definition B_{n} = B_{n}(1)).
a(n) = Sum_{k=0..n} C(n,k)*B_{n-k}*2^(k+1)/(k+1). (See also A052584.) (End)
a(n) is the a(n-1)-th odd number for n >= 1. - Jaroslav Krizek, Apr 25 2009
From Reinhard Zumkeller, Feb 28 2010: (Start)
a(n)*A000225(n) = A000225(2*n).
a(n) = A173786(n,0). (End)
If p[i]=Fibonacci(i-4) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise, then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
a(n+2) = a(n) + a(n+1) + A000225(n). - Ivan N. Ianakiev, Jun 24 2012
a(A006521(n)) mod A006521(n) = 0. - Reinhard Zumkeller, Jul 17 2014
a(n) = 3*A007583((n-1)/2) for n odd. - Eric W. Weisstein, Jul 17 2017
Sum_{n>=0} 1/a(n) = A323482. - Amiram Eldar, Nov 11 2020

A153641 Nonzero coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent, and Bernoulli numbers (triangle read by rows).

Original entry on oeis.org

1, 1, 1, -1, 1, -3, 1, -6, 5, 1, -10, 25, 1, -15, 75, -61, 1, -21, 175, -427, 1, -28, 350, -1708, 1385, 1, -36, 630, -5124, 12465, 1, -45, 1050, -12810, 62325, -50521, 1, -55, 1650, -28182, 228525, -555731, 1, -66, 2475, -56364, 685575, -3334386, 2702765, 1
Offset: 0

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Author

Peter Luschny, Dec 29 2008

Keywords

Comments

In the following the expression [n odd] is 1 if n is odd, 0 otherwise.
(+) W_n(0) = E_n are the Euler (or secant) numbers A122045.
(+) W_n(1) = T_n are the signed tangent numbers, see A009006.
(+) W_{n-1}(1) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli number A027641/A027642.
(+) W_n(-1) 2^{-n}(n+1) = G_n the Genocchi number A036968.
(+) W_n(1/2) 2^{n} are the signed generalized Euler (Springer) number, see A001586.
(+) | W_n([n odd]) | the number of alternating permutations A000111.
(+) | W_n([n odd]) / n! | for 0<=n the Euler zeta number A099612/A099617 (see Wikipedia on Bernoulli number). - Peter Luschny, Dec 29 2008
The diagonals in the full triangle (with zero coefficients) of the polynomials have the general form E(k)*binomial(n+k,k) (k>=0 fixed, n=0,1,...) where E(n) are the Euler numbers in the enumeration A122045. For k=2 we find the triangular numbers A000217 and for k=4 A154286. - Peter Luschny, Jan 06 2009
From Peter Bala, Jun 10 2009: (Start)
The Swiss-Knife polynomials W_n(x) may be expressed in terms of the Bernoulli polynomials B(n,x) as
... W_n(x) = 4^(n+1)/(2*n+2)*[B(n+1,(x+3)/4) - B(n+1,(x+1)/4)].
The Swiss-Knife polynomials are, apart from a multiplying factor, examples of generalized Bernoulli polynomials.
Let X be the Dirichlet character modulus 4 defined by X(4*n+1) = 1, X(4*n+3) = -1 and X(2*n) = 0. The generalized Bernoulli polynomials B(X;n,x), n = 1,2,..., associated with the character X are defined by means of the generating function
... t*exp(x*t)*(exp(t)-exp(3*t))/(exp(4*t)-1) = sum {n = 1..inf} B(X;n,x)*t^n/n!.
The first few values are B(X;1,x) = -1/2, B(X;2,x) = -x, B(X,3,x) = -3/2*(x^2-1) and B(X;4,x) = -2*(x^3-3*x).
In general, W_n(x) = -2/(n+1)*B(X;n+1,x).
For the theory of generalized Bernoulli polynomials associated to a periodic arithmetical function see [Cohen, Section 9.4].
The generalized Bernoulli polynomials may be used to evaluate twisted sums of k-th powers. For the present case the result is
sum{n = 0..4*N-1} X(n)*n^k = 1^k - 3^k + 5^k - 7^k + ... - (4*N-1)^k
= [B(X;k+1,4*N) - B(X;k+1,0)]/(k+1) = [W_k(0) - W_k(4*N)]/2.
For the proof apply [Cohen, Corollary 9.4.17 with m = 4 and x = 0].
The generalized Bernoulli polynomials and the Swiss-Knife polynomials are also related to infinite sums of powers through their Fourier series - see the formula section below. For a table of the coefficients of generalized Bernoulli polynomials attached to a Dirichlet character modulus 8 see A151751.
(End)
The Swiss-Knife polynomials provide a general formula for alternating sums of powers similar to the formula which are provided by the Bernoulli polynomials for non-alternating sums of powers (see the Luschny link). Sequences covered by this formula include A001057, A062393, A062392, A011934, A144129, A077221, A137501, A046092. - Peter Luschny, Jul 12 2009
The greatest common divisor of the nonzero coefficients of the decapitated Swiss-Knife polynomials is exp(Lambda(n)), where Lambda(n) is the von Mangoldt function for odd primes, symbolically:
gcd(coeffs(SKP_{n}(x) - x^n)) = A155457(n) (n>1). - Peter Luschny, Dec 16 2009
Another version is at A119879. - Philippe Deléham, Oct 26 2013

Examples

			1
x
x^2  -1
x^3  -3x
x^4  -6x^2   +5
x^5 -10x^3  +25x
x^6 -15x^4  +75x^2  -61
x^7 -21x^5 +175x^3 -427x
		

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag. [From Peter Bala, Jun 10 2009]

Crossrefs

W_n(k), k=0,1,...
W_0: 1, 1, 1, 1, 1, 1, ........ A000012
W_1: 0, 1, 2, 3, 4, 5, ........ A001477
W_2: -1, 0, 3, 8, 15, 24, ........ A067998
W_3: 0, -2, 2, 18, 52, 110, ........ A121670
W_4: 5, 0, -3, 32, 165, 480, ........
W_n(k), n=0,1,...
k=0: 1, 0, -1, 0, 5, 0, -61, ... A122045
k=1: 1, 1, 0, -2, 0, 16, 0, ... A155585
k=2: 1, 2, 3, 2, -3, 2, 63, ... A119880
k=3: 1, 3, 8, 18, 32, 48, 128, ... A119881
k=4: 1, 4, 15, 52, 165, 484, ........ [Peter Luschny, Jul 07 2009]

Programs

  • Maple
    w := proc(n,x) local v,k,pow,chen; pow := (a,b) -> if a = 0 and b = 0 then 1 else a^b fi; chen := proc(m) if irem(m+1,4) = 0 then RETURN(0) fi; 1/((-1)^iquo(m+1,4) *2^iquo(m,2)) end; add(add((-1)^v*binomial(k,v)*pow(v+x+1,n)*chen(k),v=0..k), k=0..n) end:
    # Coefficients with zeros:
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)*sech(t),t,16),t,i),x,i-n),n=0..i)), i=0..8);
    # Recursion
    W := proc(n,z) option remember; local k,p;
    if n = 0 then 1 else p := irem(n+1,2);
    z^n - p + add(`if`(irem(k,2)=1,0,
    W(k,0)*binomial(n,k)*(power(z,n-k)-p)),k=2..n-1) fi end:
    # Peter Luschny, edited and additions Jul 07 2009, May 13 2010, Oct 24 2011
  • Mathematica
    max = 9; rows = (Reverse[ CoefficientList[ #, x]] & ) /@ CoefficientList[ Series[ Exp[x*t]*Sech[t], {t, 0, max}], t]*Range[0, max]!; par[coefs_] := (p = Partition[ coefs, 2][[All, 1]]; If[ EvenQ[ Length[ coefs]], p, Append[ p, Last[ coefs]]]); Flatten[ par /@ rows] (* Jean-François Alcover, Oct 03 2011, after g.f. *)
    sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; Table[CoefficientList[sk[n, x], x] // Reverse // Select[#, # =!= 0 &] &, {n, 0, 13}] // Flatten (* Jean-François Alcover, May 21 2013 *)
    Flatten@Table[Binomial[n, 2k] EulerE[2k], {n, 0, 12}, {k, 0, n/2}](* Oliver Seipel, Jan 14 2025 *)
  • Sage
    def A046978(k):
        if k % 4 == 0:
            return 0
        return (-1)**(k // 4)
    def A153641_poly(n, x):
        return expand(add(2**(-(k // 2))*A046978(k+1)*add((-1)**v*binomial(k,v)*(v+x+1)**n for v in (0..k)) for k in (0..n)))
    for n in (0..7): print(A153641_poly(n, x))  # Peter Luschny, Oct 24 2011

Formula

W_n(x) = Sum_{k=0..n}{v=0..k} (-1)^v binomial(k,v)*c_k*(x+v+1)^n where c_k = frac((-1)^(floor(k/4))/2^(floor(k/2))) [4 not div k] (Iverson notation).
From Peter Bala, Jun 10 2009: (Start)
E.g.f.: 2*exp(x*t)*(exp(t)-exp(3*t))/(1-exp(4*t))= 1 + x*t + (x^2-1)*t^2/2! + (x^3-3*x)*t^3/3! + ....
W_n(x) = 1/(2*n+2)*Sum_{k=0..n+1} 1/(k+1)*Sum_{i=0..k} (-1)^i*binomial(k,i)*((x+4*i+3)^(n+1) - (x+4*i+1)^(n+1)).
Fourier series expansion for the generalized Bernoulli polynomials:
B(X;2*n,x) = (-1)^n*(2/Pi)^(2*n)*(2*n)! * {sin(Pi*x/2)/1^(2*n) - sin(3*Pi*x/2)/3^(2*n) + sin(5*Pi*x/2)/5^(2*n) - ...}, valid for 0 <= x <= 1 when n >= 1.
B(X;2*n+1,x) = (-1)^(n+1)*(2/Pi)^(2*n+1)*(2*n+1)! * {cos(Pi*x/2)/1^(2*n+1) - cos(3*Pi*x/2)/3^(2*n+1) + cos(5*Pi*x/2)/5^(2*n+1) - ...}, valid for 0 <= x <= 1 when n >= 1 and for 0 <= x < 1 when n = 0.
(End)
E.g.f.: exp(x*t) * sech(t). - Peter Luschny, Jul 07 2009
O.g.f. as a J-fraction: z/(1-x*z+z^2/(1-x*z+4*z^2/(1-x*z+9*z^2/(1-x*z+...)))) = z + x*z^2 + (x^2-1)*z^3 + (x^3-3*x)*z^4 + .... - Peter Bala, Mar 11 2012
Conjectural o.g.f.: Sum_{n >= 0} (1/2^((n-1)/2))*cos((n+1)*Pi/4)*( Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - (k + x)*t) ) = 1 + x*t + (x^2 - 1)*t^2 + (x^3 - 3*x)*t^3 + ... (checked up to O(t^13)), which leads to W_n(x) = Sum_{k = 0..n} 1/2^((k - 1)/2)*cos((k + 1)*Pi/4)*( Sum_{j = 0..k} (-1)^j*binomial(k, j)*(j + x)^n ). - Peter Bala, Oct 03 2016

A110501 Unsigned Genocchi numbers (of first kind) of even index.

Original entry on oeis.org

1, 1, 3, 17, 155, 2073, 38227, 929569, 28820619, 1109652905, 51943281731, 2905151042481, 191329672483963, 14655626154768697, 1291885088448017715, 129848163681107301953, 14761446733784164001387, 1884515541728818675112649, 268463531464165471482681379
Offset: 1

Views

Author

Michael Somos, Jul 23 2005

Keywords

Comments

The Genocchi numbers satisfy Seidel's recurrence: for n > 1, 0 = Sum_{j=0..floor(n/2)} (-1)^j*binomial(n, 2*j)*a(n-j). - Ralf Stephan, Apr 17 2004
The (n+1)-st Genocchi number is the number of Dumont permutations of the first kind on 2n letters. In a Dumont permutation of the first kind, each even integer must be followed by a smaller integer and each odd integer is either followed by a larger integer or is the last element. - Ralf Stephan, Apr 26 2004
The (n+1)-st Genocchi number is also the number of ways to place n rooks (attacking along planes; also called super rooks of power 2 by Golomb and Posner) on the three-dimensional Genocchi boards of size n. The Genocchi board of size n consists of cells of the form (i, j, k) where min{i, j} <= k and 1 <= k <= n. A rook placement on this board can also be realized as a pair of permutations of n the smallest number in the i-th position of the two permutations is not larger than i. - Feryal Alayont, Nov 03 2012
The (n+1)-st Genocchi number is also the number of Dumont permutations of the second kind, third kind, and fourth kind on 2n letters. In a Dumont permutation of the second kind, all odd positions are weak excedances and all even positions are deficiencies. In a Dumont permutation of the third kind, all descents are from an even value to an even value. In a Dumont permutation of the fourth kind, all deficiencies are even values at even positions. - Alexander Burstein, Jun 21 2019
The (n+1)-st Genocchi number is also the number of semistandard Young tableaux of skew shape (n+1,n,...,1)/(n-1,n-2,...,1) such that the entries in row i are at most i for i=1,...,n+1. - Alejandro H. Morales, Jul 26 2020
The (n+1)-st Genocchi number is also the number of positive terms of the Okounkov-Olshanski formula for the number of standard tableaux of skew shape (n+1,n,n-1,...,1)/(n-1,n-2,...,1), given by the (2n+1)-st Euler number A000111. - Alejandro H. Morales, Jul 26 2020
The (n+1)-st Genocchi number is also the number of collapsed permutations in (2n-1) letters. A permutation pi of size 2n-1 is said to be collapsed if ceil(k/2) <= pi^{-1}(k) <= n + floor(k/2). There are 3 collapsed permutations of size 3, namely 123, 132 and 213. - Arvind Ayyer, Oct 23 2020

Examples

			E.g.f.: x*tan(x/2) = x^2/2! + x^4/4! + 3*x^6/6! + 17*x^8/8! + 155*x^10/10! + ...
O.g.f.: A(x) = x + x^2 + 3*x^3 + 17*x^4 + 155*x^5 + 2073*x^6 + ...
where A(x) = x + x^2/(1+x) + 2!^2*x^3/((1+x)*(1+4*x)) + 3!^2*x^4/((1+x)*(1+4*x)*(1+9*x)) + 4!^2*x^5/((1+x)*(1+4*x)*(1+9*x)*(1+16*x)) + ... . - _Paul D. Hanna_, Jul 21 2011
From _Gary W. Adamson_, Jul 19 2011: (Start)
The first few rows of production matrix M are:
  1, 2,  0,  0,  0, 0, ...
  1, 3,  3,  0,  0, 0, ...
  1, 4,  6,  4,  0, 0, ...
  1, 5, 10, 10,  5, 0, ...
  1, 6, 15, 20, 15, 6, ... (End)
		

References

  • L. Carlitz, A conjecture concerning Genocchi numbers. Norske Vid. Selsk. Skr. (Trondheim) 1971, no. 9, 4 pp. MR0297697 (45 #6749)
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • Leonhard Euler, Institutionum Calculi Differentialis, volume 2 (1755), para. 181.
  • A. Genocchi, Intorno all'espressione generale de'numeri Bernulliani, Ann. Sci. Mat. Fis., 3 (1852), 395-405.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2 (1999) p. 74; see Problem 5.8.

Crossrefs

Programs

  • Magma
    [Abs(2*(4^n-1)*Bernoulli(2*n)): n in [1..20]]; // Vincenzo Librandi, Jul 28 2017
    
  • Maple
    A110501 := proc(n)
        2*(-1)^n*(1-4^n)*bernoulli(2*n) ;
    end proc:
    seq(A110501(n),n=0..10) ; # R. J. Mathar, Aug 02 2013
  • Mathematica
    a[n_] := 2*(4^n - 1) * BernoulliB[2n] // Abs; Table[a[n], {n, 19}] (* Jean-François Alcover, May 23 2013 *)
  • PARI
    {a(n) = if( n<1, 0, 2 * (-1)^n * (1 - 4^n) * bernfrac( 2*n))};
    
  • PARI
    {a(n) = if( n<1, 0, (2*n)! * polcoeff( x * tan(x/2 + x * O(x^(2*n))), 2*n))};
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m!^2*x^(m+1)/prod(k=1,m, 1+k^2*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 21 2011 */
    
  • PARI
    upto(n) = my(v1, v2, v3); v1 = vector(n, i, 0); v1[1] = 1; v2 = vector(n-1, i, ((i+1)^2)\4); v3 = v1; for(i=2, n, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 28 2025
    
  • Python
    from sympy import bernoulli
    def A110501(n): return ((2<<(m:=n<<1))-2)*abs(bernoulli(m)) # Chai Wah Wu, Apr 14 2023
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A110501_list(n) :
        D = []; [D.append(0) for i in (0..n+2)]; D[1] = 1
        R = [] ; b = True
        for i in(0..2*n-1) :
            h = i//2 + 1
            if b :
                for k in range(h-1,0,-1) : D[k] += D[k+1]
            else :
                for k in range(1,h+1,1) :  D[k] += D[k-1]
            b = not b
            if b : R.append(D[h])
        return R
    A110501_list(19) # Peter Luschny, Apr 01 2012
    
  • Sage
    [2*(-1)^n*(1-4^n)*bernoulli(2*n) for n in (1..20)] # G. C. Greubel, Nov 28 2018
    

Formula

(-1)^n * a(n) = A036968(2*n) = A001469(n).
a(n) = 2*(-1)^n*(1-4^n)*B_{2*n} (B = A027641/A027642 are Bernoulli numbers).
A002105(n) = 2^(n-1)/n * a(n). - Don Knuth, Jan 16 2007
A000111(2*n-1) = a(n)*2^(2*n-2)/n. - Alejandro H. Morales, Jul 26 2020
E.g.f.: x * tan(x/2) = Sum_{k > 0} a(k) * x^(2*k) / (2*k)!.
E.g.f.: x * tan(x/2) = x^2 / (2 - x^2 / (6 - x^2 / (... 4*k+2 - x^2 / (...)))). - Michael Somos, Mar 13 2014
O.g.f.: Sum_{n >= 0} n!^2 * x^(n+1) / Product_{k = 1..n} (1 + k^2*x). - Paul D. Hanna, Jul 21 2011
a(n) = Sum_{k = 0..2*n} (-1)^(n-k+1)*Stirling2(2*n, k)*A059371(k). - Vladeta Jovovic, Feb 07 2004
O.g.f.: A(x) = x/(1-x/(1-2*x/(1-4*x/(1-6*x/(1-9*x/(1-12*x/(... -[(n+1)/2]*[(n+2)/2]*x/(1- ...)))))))) (continued fraction). - Paul D. Hanna, Jan 16 2006
a(n) = Pi^(-2*n)*integral(log(t/(1-t))^(2*n)-log(1-1/t)^(2*n) dt,t=0,1). - Gerry Martens, May 25 2011
a(n) = the upper left term of M^(n-1); M is an infinite square production matrix with M[i,j] = C(i+1,j-1), i.e., Pascal's triangle without the first two rows and right border, see the examples and Maple program. - Gary W. Adamson, Jul 19 2011
G.f.: 1/U(0) where U(k) = 1 + 2*(k^2)*x - x*((k+1)^2)*(x*(k^2)+1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 15 2012
a(n+1) = Sum_{k=0..n} A211183(n, k)*2^(n-k). - Philippe Deléham, Feb 03 2013
G.f.: 1 + x/(G(0)-x) where G(k) = 2*x*(k+1)^2 + 1 - x*(k+2)^2*(x*k^2+2*x*k+x+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 10 2013
G.f.: G(0) where G(k) = 1 + x*(2*k+1)^2/( 1 + x + 4*x*k + 4*x*k^2 - 4*x*(k+1)^2*(1 + x + 4*x*k + 4*x*k^2)/(4*x*(k+1)^2 + (1 + 4*x + 8*x*k + 4*x*k^2)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 11 2013
G.f.: R(0), where R(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/(1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/R(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 27 2013
E.g.f. (offset 1): sqrt(x)*tan(sqrt(x)/2) = Q(0)*x/2, where Q(k) = 1 - x/(x - 4*(2*k+1)*(2*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 06 2014
Pi^2/6 = 2*Sum_{k=1..N} (-1)^(k-1)/k^2 + (-1)^N/N^2(1 - 1/N + 1/N^3 - 3/N^5 + 17/N^7 - 155/N^9 +- ...), where the terms in the parenthesis are (-1)^n*a(n)/N^(2n-1). - M. F. Hasler, Mar 11 2015
a(n) = 2*n*|euler(2*n-1, 0)|. - Peter Luschny, Jun 09 2016
a(n) = 4^(1-n) * (4^n-1) * Pi^(-2*n) * (2*n)! * zeta(2*n). - Daniel Suteu, Oct 14 2016
a(n) ~ 8*Pi*(2^(2*n)-1)*(n/(Pi*exp(1)))^(2*n+1/2)*exp(1/2+(1/24)/n-(1/2880)/n^3+(1/40320)/n^5+...). [Given in A001469 by Peter Luschny, Jul 24 2013, copied May 14 2022.]
a(n) = A000182(n) * n / 4^(n-1) (Han and Liu, 2018). - Amiram Eldar, May 17 2024

Extensions

Edited by M. F. Hasler, Mar 22 2015

A005439 Genocchi medians (or Genocchi numbers of second kind).

Original entry on oeis.org

1, 1, 2, 8, 56, 608, 9440, 198272, 5410688, 186043904, 7867739648, 401293838336, 24290513745920, 1721379917619200, 141174819474169856, 13266093250285568000, 1415974941618255921152, 170361620874699124637696, 22948071824232932086513664, 3439933090471867097102680064
Offset: 0

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Author

Keywords

Comments

a(n) is the number of Boolean functions of n variables whose ROBDD (reduced ordered binary decision diagram) contains exactly n branch nodes, one for each variable. - Don Knuth, Jul 11 2007
The earliest known reference for these numbers is Seidel (1877, pages 185 and 186). - Don Knuth, Jul 13 2007
Hankel transform of 1,1,2,8,... is A168488. - Paul Barry, Nov 27 2009
According to Hetyei [2017], alternation acyclic tournaments "are counted by the median Genocchi numbers"; an alternation acyclic tournament "does not contain a cycle in which descents and ascents alternate." - Danny Rorabaugh, Apr 25 2017
The n-th Genocchi number of the second kind is also the number of collapsed permutations in (2n) letters. A permutation pi of size 2n is said to be collapsed if 1+floor(k/2) <= pi^{-1}(k) <= n + floor(k/2). There are 2 collapsed permutations of size 4, namely 1234 and 1324. - Arvind Ayyer, Oct 23 2020
For any positive integer n, a(n) is (-1)^n times the permanent of the 2n X 2n matrix M with M(j, k) = floor((2*j-k-1)/(2*n)). This former conjecture of Luschny, inspired by a conjecture of Zhi-Wei Sun in A036968, was proven by Fu, Lin and Sun (see link). - Peter Luschny, Sep 07 2021 [updated Sep 24 2021]

References

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

Crossrefs

Programs

  • GAP
    List([1..20],n->2*(-1)^n*Sum([0..n],k->Binomial(n,k)*(1-2^(n+k+1))*Bernoulli(n+k+1))); # Muniru A Asiru, Nov 29 2018
    
  • Magma
    [2*(-1)^n*(&+[Binomial(n, k)*(1-2^(n+k+1))*Bernoulli(n+k+1): k in [0..n]]): n in [1..20]]; // G. C. Greubel, Nov 28 2018
    
  • Maple
    seq(2*(-1)^n*add(binomial(n,k)*(1 - 2^(n+k+1))*bernoulli(n+k+1), k=0..n), n=0..20); # G. C. Greubel, Oct 18 2019
  • Mathematica
    a[n_]:= 2*(-1)^(n-2)*Sum[Binomial[n, k]*(1 -2^(n+k+1))*BernoulliB[n+k+1], {k, 0, n}]; Table[a[n], {n,16}] (* Jean-François Alcover, Jul 18 2011, after PARI prog. *)
  • PARI
    a(n)=2*(-1)^n*sum(k=0,n,binomial(n,k)*(1-2^(n+k+1))* bernfrac(n+k+1))
    
  • PARI
    a(n)=local(CF=1+x*O(x^(n+2)));if(n<0,return(0), for(k=1,n+1,CF=1/(1-((n-k+1)\2+1)^2*x*CF));return(Vec(CF)[n+2])) \\ Paul D. Hanna
    
  • Python
    from math import comb
    from sympy import bernoulli
    def A005439(n): return (-2 if n&1 else 2)*sum(comb(n,k)*(1-(1<Chai Wah Wu, Apr 14 2023
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A005439_list(n) :
        D = []; [D.append(0) for i in (0..n+2)]; D[1] = 1
        R = [] ; b = True
        for i in(0..2*n-1) :
            h = i//2 + 1
            if b :
                for k in range(h-1,0,-1) : D[k] += D[k+1]
            else :
                for k in range(1,h+1,1) :  D[k] += D[k-1]
            if b : R.append(D[1])
            b = not b
        return R
    A005439_list(18) # Peter Luschny, Apr 01 2012
    
  • Sage
    [2*(-1)^n*sum(binomial(n,k)*(1-2^(n+k+1))*bernoulli(n+k+1) for k in (0..n)) for n in (1..20)] # G. C. Greubel, Oct 18 2019
    

Formula

a(n) = T(n, 1) where T(1, x) = 1; T(n, x) = (x+1)*((x+1)*T(n-1, x+1)-x*T(n-1, x)); see A058942.
a(n) = A000366(n)*2^(n-1).
a(n) = 2 * (-1)^n * Sum_{k=0..n} binomial(n, k)*(1-2^(n+k+1))*B(n+k+1), with B(n) the Bernoulli numbers. - Ralf Stephan, Apr 17 2004
O.g.f.: 1 + x*A(x) = 1/(1-x/(1-x/(1-4*x/(1-4*x/(1-9*x/(1-9*x/(... -[(n+1)/2]^2*x/(1-...)))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
G.f.: (of 1,1,2,8,...) 1/(1-x-x^2/(1-5*x-16*x^2/(1-13*x-81*x^2/(1-25*x-256*x^2/(1-41*x-625*x^2/(1-... (continued fraction). - Paul Barry, Nov 27 2009
O.g.f.: Sum_{n>=0} n!*(n+1)! * x^(n+1) / Product_{k=1..n} (1 + k*(k+1)*x). - Paul D. Hanna, May 10 2012
From Sergei N. Gladkovskii, Dec 14 2011, Dec 27 2012, May 29 2013, Oct 09 2013, Oct 24 2013, Oct 27 2013: (Start)
Continued fractions:
G.f.: A(x) = 1/S(0), S(k) = 1 - x*(k+1)*(k+2)/(1 - x*(k+1)*(k+2)/S(k+1)).
G.f.: A(x) = -1/S(0), S(k) = 2*x*(k+1)^2 - 1 - x^2*(k+1)^2*(k+2)^2/S(k+1).
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)^2/(1 - x*(k+1)^2/G(k+1)).
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(4*x*(k+1)) + 1/G(k+1))).
G.f.: Q(0)/x - 1/x, where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/(1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)))).
G.f.: T(0)/(1-2*x), where T(k) = 1 - x^2*((k + 2)*(k+1))^2/(x^2*((k + 2)*(k+1))^2 - (1 - 2*x*k^2 - 4*x*k - 2*x)*(1 - 2*x*k^2 - 8*x*k - 8*x)/T(k+1)).
G.f.: R(0), where R(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/(1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/R(k+1) ))). (End)
a(n) ~ 2^(2*n+4) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, Oct 28 2014
Rewriting the above: a(n) ~ 4*(2*n+1)! / Pi^(2*n+1). Compare to Genocchi numbers A110501(n) = g_n ~ 4*(2*n)! / Pi^(2*n). So these are indeed like "Genocchi medians" g_{n + 1/2}. - Alan Sokal, May 13 2022
Asymptotic expansion: a(n) ~ 4*(2*n+1)! * Pi^(-(2*n+1)) * (1 + (Pi^2/16)/n + (Pi^2 (Pi^2 - 16)/512)/n^2 + (Pi^2 (Pi^4 + 384)/24576)/n^3 + (Pi^2 (Pi^6 + 96*Pi^4 + 768*Pi^2 - 12288)/1572864)/n^4 + (Pi^2 (Pi^8 + 320*Pi^6 + 12800*Pi^4 + 491520)/125829120)/n^5 + ...) --- Proof uses binomial sum for Genocchi medians in terms of Genocchi or Bernoulli numbers, combined with leading term of convergent sum (with exponentially small corrections) for the latter. Can also check against the 10000 term a-file. - Alan Sokal, May 23 2022.
a(n) = n!^2 * [x^n*y^n] exp(x)*f(x-y), where f(x) is the derivative of the Genocchi number generating function 2*x/(exp(x)+1). - Ira M. Gessel, Jul 23 2024

Extensions

More terms and additional comments from David W. Wilson, Jan 11 2001
a(0)=1 prepended by Peter Luschny, Apr 14 2023

A226158 a(n) = 2*n*(2^n - 1)*zeta(1-n) where in the case n=0 the limit is understood, zeta(s) the Riemann zeta function.

Original entry on oeis.org

0, -1, -1, 0, 1, 0, -3, 0, 17, 0, -155, 0, 2073, 0, -38227, 0, 929569, 0, -28820619, 0, 1109652905, 0, -51943281731, 0, 2905151042481, 0, -191329672483963, 0, 14655626154768697, 0, -1291885088448017715, 0, 129848163681107301953
Offset: 0

Views

Author

Peter Luschny, Jun 28 2013

Keywords

Comments

Also known as the Genocchi numbers, apart from a(0) and a(1) same as A036968.
Consider the difference table of a(n), which is a variant of Seidel's Genocchi table A014781:
0 -1 -1 0 1 0 -3 0 17
-1 0 1 1 -1 -3 3 17 -17
1 1 0 -2 -2 6 14 -34 -138
0 -1 -2 0 8 8 -48 -104 448
-1 -1 2 8 0 -56 -56 552 1160
0 3 6 -8 -56 0 608 608 -8832
3 3 -14 -48 56 608 0 -9440 -9440
0 -17 -34 104 552 -608 -9440 0 198272
-17 -17 138 448 -1160 -8832 9440 198272 0
a(n) is an autosequence: its inverse binomial transform is the sequence signed (see A181722). The first column (inverse binomial transform) is 0, followed by -A036968. - Paul Curtz, Jul 22 2013
a(n+1) = p(0) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 1, ..., n+1. - Michael Somos, Apr 23 2014

Examples

			G.f. = - x - x^2 + x^4 - 3*x^6 + 17*x^8 - 155*x^10 + 2073*x^12 - 38227*x^14 + ...
		

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 50); [0] cat Coefficients(R!(Laplace( -2*x/(1+Exp(-x)) ))); // G. C. Greubel, Apr 22 2023
  • Maple
    seq(n!*coeff(series(-2*x/(1+exp(-x)), x, 34), x, n), n=0..32);
    # Second program:
    A226158 := proc(n) local f; f := z -> Zeta(1-z)*2*z*(2^z-1);
    if n=0 then limit(f(z), z=0) else f(n) fi end: seq(A226158(n), n=0..32);
  • Mathematica
    a[0]=0; a[1]= -1; a[n_]:= n*EulerE[n-1, 0]; Table[a[n], {n,0,32}] (* Jean-François Alcover, Sep 12 2013 *)
    (* Programs from Michael Somos, Apr 23 2014 *)
    a[n_]:= If[n<1, 0, -n*EulerE[n-1, 1]];
    a[n_]:= If[n<0, 0, 2*(1-2^n)*BernoulliB[n,1]]; (* End *)
    Table[2*n*PolyLog[1-n, -1], {n,0,32}] (* Peter Luschny, Aug 17 2021 *)
  • PARI
    my(x='x+O('x^40)); concat([0], Vec(serlaplace(-2*x/(1+exp(-x))))) \\ G. C. Greubel, Jan 19 2018
    
  • Sage
    def A226158(n): return -2*n*zeta(1-n)*(1-2^n) if n != 0 else 0
    [A226158(n) for n in (0..32)]
    # Alternatively:
    def A226158_list(len):
        e, f, R, C = 4, 1, [0], [1]+[0]*(len-1)
        for n in (2..len-1):
            for k in range(n, 0, -1):
                C[k] = -C[k-1] / (k+1)
            C[0] = -sum(C[k] for k in (1..n))
            R.append((2-e)*f*C[0])
            f *= n; e *= 2
        return R
    print(A226158_list(34)) # Peter Luschny, Feb 22 2016
    

Formula

E.g.f.: -2*x/(1+exp(-x)).
a(2n) = -A000367(n)*A090648(n). - Paul Curtz, Jul 22 2013
E.g.f.: -2*x/(1+exp(-x))= -2 - 2*T(0), where T(k) = 4*k-1 + x/( 2 - x/( 4*k+1 + x/( 2 - x/T(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2013
G.f.: conjecture: -x/Q(0),where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/(1 - x*(k+1)/(1 + x*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2013
a(n) = 2*(1 - 2^n)*Bernoulli(n, 1). - Peter Luschny, Apr 16 2014
a(n) = -n*Euler(n - 1, 1). - Michael Somos, Apr 23 2014
a(n) = 2^n*(Bernoulli(n, 1/2) - Bernoulli(n, 1)). - Peter Luschny, Jul 10 2020
a(n) = 2*n*PolyLog[1 - n, -1] - Peter Luschny, Aug 17 2021

A083007 a(n) = Sum_{k=0..n-1} 3^k*B(k)*C(n,k) where B(k) is the k-th Bernoulli number and C(n,k)=binomial(n,k).

Original entry on oeis.org

0, 1, -2, 1, 4, -5, -26, 49, 328, -809, -6710, 20317, 201772, -722813, -8370194, 34607305, 457941136, -2145998417, -31945440878, 167317266613, 2767413231220, -16020403322021, -291473080313162, 1848020950359841, 36679231132772824, -252778977216700025, -5435210060467425446
Offset: 0

Views

Author

Benoit Cloitre, May 31 2003

Keywords

Crossrefs

Programs

  • Maple
    A083007 := proc(n)
        3*x/(1+exp(x)+exp(2*x)) ;
        coeftayl(%,x=0,n) ;
        %*n! ;
    end proc:
    seq(A083007(n),n=0..30) ; # R. J. Mathar, Jul 13 2023
  • Mathematica
    Range[0, 15]! CoefficientList[ Series[ 3x/(1 + Exp[x] + Exp[ 2x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
    Table[Sum[3^k BernoulliB[k]Binomial[n,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, May 26 2014 *)
  • PARI
    a(n)=sum(k=0,n-1,3^k*binomial(n,k)*bernfrac(k))

Formula

E.g.f.: 3x/(1+e^x+e^(2x)). - Ira M. Gessel, Jan 28 2012
From Peter Bala, Mar 01 2015: (Start)
a(2*n+1) = (-1)^(n+1)*A002111(n) for n >= 1.
a(n) = 3^n * ( B(n,1/3) - B(n,0) ), where B(n,x) denotes the n-th Bernoulli polynomial. More generally, Almkvist and Meurman show that k^n * ( B(n, 1/k) - B(n, 0) ) is an integer sequence for k = 2,3,4,..., which proves the integrality of A083008 through A083014.
a(0) = 1 and for n >= 1, a(n) = 1 - 1/(n + 1)*Sum_{k = 1..n-1} 3^(n-k)*binomial(n+1,k)*a(k) (Sury, Section 1). (End)

A083008 a(n) = Sum_{k=0..n-1} 4^k*B(k)*C(n,k) where B(k) is the k-th Bernoulli number and C(n,k) = binomial(n,k).

Original entry on oeis.org

0, 1, -3, 3, 9, -25, -99, 427, 2193, -12465, -79515, 555731, 4247577, -35135945, -313193811, 2990414715, 30461046561, -329655706465, -3777604994187, 45692713833379, 581778811909545, -7777794952988025, -108933009112011843, 1595024111042171723, 24370176181315498929
Offset: 0

Views

Author

Benoit Cloitre, May 31 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Range[0, 15]! CoefficientList[ Series[ 4x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
    Table[Sum[4^k*BernoulliB[k] Binomial[n, k], {k, 0, n - 1}], {n, 0, 24}] (* Michael De Vlieger, Sep 28 2016 *)
  • PARI
    a(n)=sum(k=0,n-1,4^k*binomial(n,k)*bernfrac(k))

Formula

E.g.f.: 4*x/(1+exp(x)+exp(2*x)+exp(3*x)). - Ira M. Gessel, Feb 23 2012
a(n) ~ n! * (cos(n*Pi/2)-sin(n*Pi/2)) * 2^(n+1) / Pi^n. - Vaclav Kotesovec, Mar 02 2014

Extensions

Offset changed to 0 by Seiichi Manyama, Sep 28 2016

A083009 a(n) = Sum_{k=0,n-1} 5^k*B(k)*binomial(n,k) where B(k) is the k-th Bernoulli number.

Original entry on oeis.org

0, 1, -4, 6, 16, -74, -264, 1946, 9056, -88434, -512024, 6154786, 42716496, -607884394, -4920817384, 80834386026, 747784582336, -13923204233954, -144898927180344, 3015393801263666, 34867899296006576, -801997872697905114, -10201104981227536904, 256982712667627683706
Offset: 0

Views

Author

Benoit Cloitre, May 31 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Range[0, 15]! CoefficientList[ Series[ 5x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x] + Exp[ 4x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
    Table[Sum[5^k*BernoulliB[k] Binomial[n, k], {k, 0, n - 1}], {n, 0, 23}] (* Michael De Vlieger, Sep 28 2016 *)
  • PARI
    a(n)=sum(k=0,n-1,5^k*binomial(n,k)*bernfrac(k))

Formula

E.g.f.: 5x/(1+exp(x)+exp(2x)+exp(3x)+exp(4x)). - Benoit Cloitre, Oct 26 2012 (following I. Gessel).

Extensions

Offset changed to 0 by Seiichi Manyama, Sep 28 2016

A083010 a(n) = 6^n(B_n(1/6)-B_n(0)) where B_n(x) is the n-th Bernoulli polynomial.

Original entry on oeis.org

0, 1, -5, 10, 25, -170, -575, 6370, 28225, -415826, -2294975, 41649850, 275622625, -5922729722, -45718037855, 1134081384850, 10004182986625, -281284596509858, -2791456543622015, 87722769712529770, 967282878165054625, -33597252908389628234, -407509096583935700255
Offset: 0

Views

Author

Benoit Cloitre, May 31 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Range[0, 15]! CoefficientList[ Series[ 6x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x] + Exp[ 4x] + Exp[ 5x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
  • PARI
    a(n)=sum(k=0,n-1,6^k*binomial(n,k)*bernfrac(k))
    
  • PARI
    {a(n)=if(n<1, 0, n!*polcoeff( 6*x*(exp(x+x*O(x^n))-1)/(exp(6*x +x*O(x^n))-1), n))} /* Michael Somos, Aug 02 2006 */

Formula

E.g.f.: 6x(exp(x)-1)/(exp(6x)-1). - Michael Somos, Aug 02 2006
a(n) = Sum_{k=0..n-1} 6^k*B(k)*C(n,k) where B(k) is the k-th Bernoulli number and C(n,k) = binomial(n,k).

A083011 a(n) = Sum_{k=0..n-1} 7^k*B(k)*binomial(n,k) where B(k) is the k-th Bernoulli number.

Original entry on oeis.org

0, 1, -6, 15, 36, -335, -1098, 16955, 73032, -1503963, -8075430, 204957775, 1319806188, -39666688711, -297958666242, 10337889346275, 88743928066704, -3489994294713779, -33703905982634334, 1481439997178305655, 15896303102840841780, -772269573963075710367
Offset: 0

Views

Author

Benoit Cloitre, May 31 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Range[0, 15]! CoefficientList[ Series[ 7x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x] + Exp[ 4x] + Exp[ 5x] + Exp[ 6x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
  • PARI
    a(n)=sum(k=0,n-1,7^k*binomial(n,k)*bernfrac(k))

Extensions

Offset changed to 0 by Seiichi Manyama, Sep 28 2016
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