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 1-10 of 11 results. Next

A174826 Hankel transform of median Euler numbers A000657.

Original entry on oeis.org

1, 3, 1260, 471517200, 550527639742080000, 5190428913555819988454400000000, 850897011688550259860209488988778496000000000000, 4613713285873452716084822982464795709127827429408768000000000000000000
Offset: 0

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Author

Paul Barry, Mar 30 2010

Keywords

Programs

  • Mathematica
    Table[Product[((k+1)*(4*k+1)*(4*k+3)*Floor[(2*k+3)/2])^(n-k), {k,0,n}], {n,0,10}] (* Vaclav Kotesovec, Feb 24 2019 *)

Formula

a(n) = Product_{k=0..n} ((k+1)*(4k+1)*(4k+3)*floor((2k+3)/2))^(n-k).
a(n) ~ sqrt(Gamma(1/4)) * 2^(2*n^2 + 7*n/2 + 9/8) * n^(n*(2*n+3) + 47/48) * Pi^(n + 3/4) / (A^(7/4) * exp(3*n*(n+1) - 7/48)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Feb 24 2019

Extensions

a(5) and following corrected by Georg Fischer, Jan 18 2024

A008280 Boustrophedon version of triangle of Euler-Bernoulli or Entringer numbers read by rows.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 0, 1, 2, 2, 5, 5, 4, 2, 0, 0, 5, 10, 14, 16, 16, 61, 61, 56, 46, 32, 16, 0, 0, 61, 122, 178, 224, 256, 272, 272, 1385, 1385, 1324, 1202, 1024, 800, 544, 272, 0, 0, 1385, 2770, 4094, 5296, 6320, 7120, 7664, 7936, 7936
Offset: 0

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Author

Keywords

Comments

The earliest known reference for this triangle is Seidel (1877). - Don Knuth, Jul 13 2007
Sum of row n = A000111(n+1). - Reinhard Zumkeller, Nov 01 2013

Examples

			This version of the triangle begins:
  [0] [   1]
  [1] [   0,    1]
  [2] [   1,    1,    0]
  [3] [   0,    1,    2,    2]
  [4] [   5,    5,    4,    2,    0]
  [5] [   0,    5,   10,   14,   16,   16]
  [6] [  61,   61,   56,   46,   32,   16,    0]
  [7] [   0,   61,  122,  178,  224,  256,  272,  272]
  [8] [1385, 1385, 1324, 1202, 1024,  800,  544,  272,    0]
  [9] [   0, 1385, 2770, 4094, 5296, 6320, 7120, 7664, 7936, 7936]
See A008281 and A108040 for other versions.
		

References

  • M. D. Atkinson: Partial orders and comparison problems, Sixteenth Southeastern Conference on Combinatorics, Graph Theory and Computing, (Boca Raton, Feb 1985), Congressus Numerantium 47, 77-88.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 110.
  • A. J. Kempner, On the shape of polynomial curves, Tohoku Math. J., 37 (1933), 347-362.
  • A. A. Kirillov, Variations on the triangular theme, Amer. Math. Soc. Transl., (2), Vol. 169, 1995, pp. 43-73, see p. 53.
  • R. P. Stanley, Enumerative Combinatorics, volume 1, second edition, chapter 1, exercise 141, Cambridge University Press (2012), p. 128, 174, 175.

Crossrefs

Cf. A000657 (central terms); A227862.

Programs

  • Haskell
    a008280 n k = a008280_tabl !! n !! k
    a008280_row n = a008280_tabl !! n
    a008280_tabl = ox True a008281_tabl where
      ox turn (xs:xss) = (if turn then reverse xs else xs) : ox (not turn) xss
    -- Reinhard Zumkeller, Nov 01 2013
    
  • Mathematica
    max = 9; t[0, 0] = 1; t[n_, m_] /; n < m || m < 0 = 0; t[n_, m_] := t[n, m] = Sum[t[n-1, n-k], {k, m}]; tri = Table[t[n, m], {n, 0, max}, {m, 0, n}]; Flatten[ {Reverse[#[[1]]], #[[2]]} & /@ Partition[tri, 2]] (* Jean-François Alcover, Oct 24 2011 *)
    T[0,0] := 1; T[n_?OddQ,k_]/;0<=k<=n := T[n,k]=T[n,k-1]+T[n-1,k-1]; T[n_?EvenQ,k_]/;0<= k<=n := T[n,k]=T[n,k+1]+T[n-1,k]; T[n_,k_] := 0; Flatten@Table[T[n,k], {n,0,9}, {k,0,n}] (* Oliver Seipel, Nov 24 2024 *)
  • Maxima
    T(n, m):=abs(sum(binomial(m, k)*euler(n-m+k), k, 0, m)); /* Vladimir Kruchinin, Apr 06 2015 */
  • Python
    # Python 3.2 or higher required.
    from itertools import accumulate
    A008280_list = blist = [1]
    for n in range(10):
        blist = list(reversed(list(accumulate(reversed(blist))))) + [0] if n % 2 else [0]+list(accumulate(blist))
        A008280_list.extend(blist)
    print(A008280_list) # Chai Wah Wu, Sep 20 2014
    
  • Python
    # Uses function seidel from A008281.
    def A008280row(n): return seidel(n) if n % 2 else seidel(n)[::-1]
    for n in range(8): print(A008280row(n)) # Peter Luschny, Jun 01 2022
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle.
    def A008280_triangle(n) :
        A = {-1:0, 0:1}
        k = 0; e = 1
        for i in range(n) :
            Am = 0
            A[k + e] = 0
            e = -e
            for j in (0..i) :
                Am += A[k]
                A[k] = Am
                k += e
            print([A[z] for z in (-i//2..i//2)])
    A008280_triangle(10) # Peter Luschny, Jun 02 2012
    

Formula

T(n,m) = abs( Sum_{k=0..n} C(m,k)*Euler(n-m+k) ). - Vladimir Kruchinin, Apr 06 2015
E.g.f.: (cos(x) + sin(x))/cos(x+y). - Ira M. Gessel, Nov 18 2024

A005799 Generalized Euler numbers of type 2^n.

Original entry on oeis.org

1, 1, 2, 10, 104, 1816, 47312, 1714000, 82285184, 5052370816, 386051862272, 35917232669440, 3996998043812864, 524203898507631616, 80011968856686405632, 14061403972845412526080, 2818858067801804443910144
Offset: 0

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Author

Keywords

Comments

Also, a(n) equals the number of alternating permutations (p(1),...,p(2n)) of the multiset {1,1,2,2,...,n,n} satisfying p(1) <= p(2) > p(3) <= p(4) > p(5) <= ... <= p(2n). Hence, A275801(n) <= a(n) <= A275829(n). - Max Alekseyev, Aug 10 2016
This is the BinomialMean transform of A000364 (see A075271 for definition of transform). - John W. Layman, Dec 04 2002
This sequence appears to be middle column in Poupard's triangle A008301.

References

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

Crossrefs

Right edge of triangle A210108.

Programs

  • Maple
    T := proc(n, k) option remember;
    if n < 0 or k < 0 then 0
    elif n = 0 then euler(k, 1)
    else T(n-1, k+1) - T(n-1, k) fi end:
    a := n -> (-2)^n*T(n, n); seq(a(n), n=0..16); # Peter Luschny, Aug 23 2017
  • Mathematica
    a[n_] := Sum[Binomial[n, i]Abs[EulerE[2i]], {i, 0, n}]/2^n

Formula

a(n) = (1/2^n) * Sum_{i=0..n} binomial(n, i) * A000364(i).
From Sergei N. Gladkovskii, Dec 27 2012, Oct 11 2013, Oct 27 2013, Jan 08 2014: (Start) Continued fractions:
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)*(2*k+1)/(1 - x*(k+1)*(2*k+1)/G(k+1)).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)^2*(2*k+1)^2/(x^2*(k+1)^2*(2*k+1)^2 - (4*x*k^2 + 2*x*k + x - 1)*( 4*x*k^2 + 10*x*k + 7*x - 1)/Q(k+1)).
G.f.: R(0), where R(k) = 1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/(1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/R(k+1)))).
G.f.: 2/(x*Q(0)), where Q(k) = 2/x - 1 - (2*k+1)^2/(1 - (2*k+2)^2/Q(k+1)). (End)
a(n) ~ 2^(3*n+3) * n^(2*n+1/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, May 30 2015
a(n) = 2^n * Sum_{k=0..n} (-1)^k*binomial(n, k)*euler(n+k, 1). - Peter Luschny, Aug 23 2017
From Peter Bala, Dec 21 2019: (Start)
O.g.f. as a continued fraction: 1/(1 - x/(1 - x/(1 - 6*x/(1 - 6*x/(1 - 15*x/(1 - 15*x/(1 - ... - n*(2*n-1)*x/(1 - n*(2*n-1)*x/(1 - ...))))))))) - apply Bala, Proposition 3, with a = 0, b = 1 and replace x with x/2.
Conjectures:
E.g.f. as a continued fraction: 2/(2 - (1-exp(-4*t))/(2 - (1-exp(-8*t))/(2 - (1-exp(-12*t))/(2 - ... )))) = 1 + t + 2*t^2/2! + 10*t^3/3! + 104*t^4/4! + ....
Cf. A000657. [added April 18 2024: for a proof of this conjecture see Fu et al., Section 4.3.]
a(n) = (-2)^(n+1)*Sum_{k = 0..floor((n-1)/2)} binomial(n,2*k+1)*(2^(2*n-2*k) - 1)*Bernoulli(2*n-2*k)/(2*n-2*k) for n >= 1. (End)

Extensions

Edited by Dean Hickerson, Dec 10 2002

A002832 Median Euler numbers.

Original entry on oeis.org

1, 3, 24, 402, 11616, 514608, 32394624, 2748340752, 302234850816, 41811782731008, 7106160248346624, 1455425220196234752, 353536812021243273216, 100492698847094242603008, 33045185784774350171111424
Offset: 1

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

There are two kinds of Euler median numbers, the 'right' median numbers (this sequence), and the 'left' median numbers (A000657).
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021

Crossrefs

Cf. A000657.
See related polynomials in A098277.
A diagonal of A323833.

Programs

  • Maple
    rr := array(1..40,1..40):rr[1,1] := 0:for i from 1 to 39 do rr[i+1,1] := (subs(x=0,diff((exp(x)-1)/cosh(x),x$i))):od: for i from 2 to 40 do for j from 2 to i do rr[i,j] := rr[i,j-1]-rr[i-1,j-1]:od:od: seq(rr[2*i-1,i-1],i=2..20); # Barbara Haas Margolius (margolius(AT)math.csuohio.edu) Feb 16 2001, corrected by R. J. Mathar, Dec 22 2010
    # alternative
    A002832 := proc(n)
        abs(A323833(n-1,n)) ;
    end proc:
    seq(A002832(n),n=1..40) ; # R. J. Mathar, Jun 11 2025
  • Mathematica
    max = 20; rr[1, 1] = 0; For[i = 1, i <= 2*max - 1, i++, rr[i + 1, 1] = D[(Exp[x] - 1)/Cosh[x], {x, i}] /. x -> 0]; For[i = 2, i <= 2*max, i++, For[j = 2, j <= i, j++, rr[i, j] = rr[i, j - 1] - rr[i - 1, j - 1]]]; Table[(-1)^i*rr[2*i - 1, i - 1], {i, 2, max}] (* Jean-François Alcover, Jul 10 2012, after Maple *)

Formula

G.f.: Sum_{n>=0} a(n)*x^n = 1/(1-1*3x/(1-1*5x/(1-2*7x/(1-2*9x/(1-3*11x/...))))).
G.f.: -1/G(0) where G(k)= x*(8*k^2+8*k+3) - 1 - (4*k+5)*(4*k+3)*(k+1)^2*x^2/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 08 2012
a(n) ~ 2^(4*n+3/2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Apr 23 2015

Extensions

More terms from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Feb 16 2001
Terms corrected by R. J. Mathar, Dec 22 2010

A099023 Diagonal of Euler-Seidel matrix with start sequence e.g.f. 1-tanh(x).

Original entry on oeis.org

1, -1, 4, -46, 1024, -36976, 1965664, -144361456, 13997185024, -1731678144256, 266182076161024, -49763143319190016, 11118629668610842624, -2925890822304510631936, 895658946905031792553984
Offset: 0

Views

Author

Ralf Stephan, Sep 23 2004

Keywords

Comments

T(2n,n), where T is A008280 (signed).

Crossrefs

Programs

  • Mathematica
    A099023List[n_] := Module[{e, dim, m, k}, dim = 2 n; e[0, 0] = 1; For[m = 1, m <= dim - 1, m++, If[EvenQ[m], e[m, 0] = 1; For[k = m - 1, k >= -1, k--, e[k, m - k] = e[k + 1, m - k - 1] - e[k, m - k - 1]], e[0, m] = 1; For[k = 1, k <= m + 1, k++, e[k, m - k] = e[k - 1, m - k + 1] + e[k - 1, m - k]]]]; Table[e[k, k], {k, 0, (dim + 1)/2 - 1}]];
    A099023List[15] (* Jean-François Alcover, Jun 11 2019, after Peter Luschny *)
  • Sage
    # Variant of an algorithm of L. Seidel (1877).
    def A099023_list(n) :
        dim = 2*n; E = matrix(ZZ, dim); E[0,0] = 1
        for m in (1..dim-1) :
            if m % 2 == 0 :
                E[m,0] = 1;
                for k in range(m-1,-1,-1) :
                    E[k,m-k] = E[k+1,m-k-1] - E[k,m-k-1]
            else :
                E[0,m] = 1;
                for k in range(1,m+1,1) :
                    E[k,m-k] = E[k-1,m-k+1] + E[k-1,m-k]
        return [E[k,k] for k in range((dim+1)//2)]
    # Peter Luschny, Jul 14 2012

Formula

|a(n)| = A000657(n) - Sean A. Irvine, Dec 22 2010
G.f.: 1/G(0) where G(k) = 1 + x*(k+1)*(4*k+1)/(1 + x*(k+1)*(4*k+3)/G(k+1) ) ; (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 05 2013
G.f.: G(0)/(1+x), where G(k) = 1 - x^2*(k+1)^2*(4*k+1)*(4*k+3)/( x^2*(k+1)^2*(4*k+1)*(4*k+3) - (1 + x*(8*k^2+4*k+1))*(1 + x*(8*k^2+20*k+13))/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 01 2014

A098277 Coefficients of polynomials D(n,x) related to median Euler numbers.

Original entry on oeis.org

1, 2, 2, 8, 20, 12, 48, 224, 344, 168, 384, 2880, 8096, 9872, 4272, 3840, 42240, 186816, 407936, 430688, 171168, 46080, 698880, 4451328, 15030528, 27944576, 26627648, 9915072, 645120, 12902400, 111605760, 535271424, 1519126272
Offset: 0

Views

Author

Ralf Stephan, Sep 07 2004

Keywords

Comments

2^n(x+1) divides D(n,x).

Examples

			D(0,x) = 1,
D(1,x) = 2*x + 2,
D(2,x) = 8*x^2 + 20*x + 12,
D(3,x) = 48*x^3 + 224*x^2 + 344*x + 168,
D(4,x) = 384*x^4 + 2880*x^3 + 8096*x^2 + 9872*x + 4272.
		

Crossrefs

D(n, 1/2) = A002832(n+1), D(n, -1/2) = A000657(n).
D(n, 0)/2^n = A098278(n), D(n, 1)/2^n = A098279(n).
Leading coefficients are A000165. Constant terms are in A098431.

Programs

  • Mathematica
    d[0, ] = 1; d[n, x_] := d[n, x] = (x+1)(x+2)d[n-1, x+2] - x(x+1)d[n-1, x];
    Table[CoefficientList[d[n, x], x] // Reverse, {n, 0, 8}] // Flatten (* Jean-François Alcover, Jul 27 2018 *)
  • PARI
    D(n,x)=if(n<1,1,(x+1)*(x+2)*D(n-1,x+2)-x*(x+1)*D(n-1,x))
    
  • PARI
    T(n,k)=local(A=sum(m=0,n,m!*(2*x)^m*prod(j=1,m,(j+y)/(1+j*(j+1)*x +x*O(x^n)))));polcoeff(polcoeff(A,n,x),n-k,y)
    {for(n=0,8,for(k=0,n,print1(T(n,k),", "));print())} \\ Paul D. Hanna, Sep 05 2012

Formula

Recurrence: D(0, x)=1, D(n, x) = (x+1)(x+2)D(n-1, x+2) - x(x+1)D(n-1, x).
G.f.: Sum[n>=0, D(n, x)t^n] = 1/(1-2(x+1)t/(1-2(x+2)t/(1-4(x+3)t/(1-4(x+4)t/...)))).
G.f.: Sum_{n>=0} D(n,y)*x^n = Sum_{n>=0} n!*(2*x)^n*Product_{k=1..n} (k+y)/(1+k*(k+1)*x). - Paul D. Hanna, Sep 05 2012

A241209 a(n) = E(n) - E(n+1), where E(n) are the Euler numbers A122045(n).

Original entry on oeis.org

1, 1, -1, -5, 5, 61, -61, -1385, 1385, 50521, -50521, -2702765, 2702765, 199360981, -199360981, -19391512145, 19391512145, 2404879675441, -2404879675441, -370371188237525, 370371188237525, 69348874393137901, -69348874393137901, -15514534163557086905
Offset: 0

Views

Author

Paul Curtz, Apr 17 2014

Keywords

Comments

A version of the Seidel triangle (1877) for the integer Euler numbers is
1
1 1
2 2 1
2 4 5 5
16 16 14 10 5
16 32 46 56 61 61
etc.
It is not in the OEIS. See A008282.
The first diagonal, Es(n) = 1, 1, 1, 5, 5, 61, 61, 1385, 1385, ..., comes from essentially A000364(n) repeated.
a(n) is Es(n) signed two by two.
Difference table of a(n):
1, 1, -1, -5, 5, 61, -61, -1385, ...
0, -2, -4, 10, 56, -122, -1324, ...
-2, -2, 14, 46, -178, -1202, ...
0, 16, 32, -224, -1024, ...
16, 16, -256, -800, ...
0, -272, -544, ...
-272, -272, ...
0, ...
etc.
Sum of the antidiagonals: 1, 1, -5, -11, 61, 211, -385, ... = A239322(n+1).
Main diagonal interleaved with the first upper diagonal: 1, 1, -2, -4, 14, 46, ... = signed A214267(n+1).
Inverse binomial transform (first column): A155585(n+1).
The Akiyama-Tanigawa transform applied to A046978(n+1)/A016116(n) gives
1, 1, 1/2, 0, -1/4, -1/4, -1/8, 0, ...
0, 1, 3/2, 1, 0, -3/4, -7/8, ...
-1, -1, 3/2, 4, 15/4, 3/4, ...
0, -5, -15/2, 1, 15, ...
5, 5, -51/2, -56, ...
0, 61, 183/2, ...
-61, -61, ...
0, ...
etc.
A122045(n) and A239005(n) are reciprocal sequences by their inverse binomial transform. In their respective difference table, two different signed versions of A214247(n) appear: 1) interleaved main diagonal and first under diagonal (1, -1, -1, 2, 4, -14, ...) and 2) interleaved main diagonal and first upper diagonal (1, 1, -1, -2, 4, 14, ...).

Crossrefs

Programs

  • Magma
    EulerPoly:= func< n,x | (&+[ (&+[ (-1)^j*Binomial(k,j)*(x+j)^n : j in [0..k]])/2^k: k in [0..n]]) >;
    Euler:= func< n | 2^n*EulerPoly(n, 1/2) >; // A122045
    [Euler(n) - Euler(n+1): n in [0..40]]; // G. C. Greubel, Jun 07 2023
    
  • Maple
    A241209 := proc(n) local v, k, h, m; m := `if`(n mod 2 = 0, n, n+1);
    h := k -> `if`(k mod 4 = 0, 0, (-1)^iquo(k,4));
    (-1)^n*add(2^iquo(-k,2)*h(k+1)*add((-1)^v*binomial(k,v)*(v+1)^m, v=0..k)
    ,k=0..m) end: seq(A241209(n),n=0..24); # Peter Luschny, Apr 17 2014
  • Mathematica
    skp[n_, x_]:= Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}];
    a[n_]:= skp[n, x] - skp[n+1, x]/. x->0; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Apr 17 2014, after Peter Luschny *)
    Table[EulerE[n] - EulerE[n+1], {n,0,30}] (* Vincenzo Librandi, Jan 24 2016 *)
    -Differences/@Partition[EulerE[Range[0,30]],2,1]//Flatten (* Harvey P. Dale, Apr 16 2019 *)
  • SageMath
    [euler_number(n) - euler_number(n+1) for n in range(41)] # G. C. Greubel, Jun 07 2023

Formula

a(n) = A119880(n+1) - A119880(n).
a(n) is the second column of the fractional array.
a(n) = (-1)^n*second column of the array in A239005(n).
a(n) = skp(n, 0) - skp(n+1, 0), where skp(n, x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 17 2014
E.g.f.: exp(x)/cosh(x)^2. - Sergei N. Gladkovskii, Jan 23 2016
G.f. T(0)/x-1/x, where T(k) = 1 - x*(k+1)/(x*(k+1)-(1-x)/(1-x*(k+1)/(x*(k+1)+(1-x)/T(k+1)))). - Sergei N. Gladkovskii, Jan 23 2016

A323834 A Seidel matrix A(n,k) read by antidiagonals downwards.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, -2, -1, 1, 4, -5, -7, -8, -7, -3, 16, 11, 4, -4, -11, -14, 61, 77, 88, 92, 88, 77, 63, -272, -211, -134, -46, 46, 134, 211, 274, -1385, -1657, -1868, -2002, -2048, -2002, -1868, -1657, -1383, 7936, 6551, 4894, 3026, 1024, -1024, -3026, -4894, -6551, -7934, 50521, 58457, 65008, 69902, 72928, 73952, 72928, 69902, 65008, 58457, 50523
Offset: 0

Views

Author

N. J. A. Sloane, Feb 03 2019

Keywords

Comments

The first row is a signed version of the Euler numbers A000111.
Other rows are defined by A(n+1,k) = A(n,k) + A(n,k+1).

Examples

			Read as triangle T(n,k) = A(k, n-k) (n >= 0, k = 0..n), the first few antidiagonals of the square array A are:
     0,
     1,    1,
     1,    2,    3,
    -2,   -1,    1,   4,
    -5,   -7,   -8,  -7,  -3,
    16,   11,    4,  -4, -11, -14,
    61,   77,   88,  92,  88,  77,  63,
  -272, -211, -134, -46,  46, 134, 211, 274,
  ...
From _Petros Hadjicostas_, Mar 02 2021: (Start)
Square array A(n,k) (with rows n >= 0 and columns k >= 0) begins:
     0,   1,   1,    -2,    -5,    16,     61,     -272,    -1385, ...
     1,   2,  -1,    -7,    11,    77,   -211,    -1657,     6551, ...
     3,   1,  -8,     4,    88,  -134,  -1868,     4894,    65008, ...
     4,  -7,  -4,    92,   -46, -2002,   3026,    69902,  -179806, ...
    -3, -11,  88,    46, -2048,  1024,  72928,  -109904, -3784448, ...
   -14,  77, 134, -2002, -1024, 73952, -36976, -3894352,  5860016, ...
   ... (End)
		

Crossrefs

Cf. A000111, A000657 (next-to-main diagonal), A323833.

Programs

  • PARI
    {b(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos's PARI program for A000111.
    c(n) = if(n==0, 0, (-1)^floor((n-1)/2)*b(n))
    A(n, k) = sum(i=0, n, binomial(n, i)*c(k+i)) \\ Petros Hadjicostas, Mar 02 2021

Formula

From Petros Hadjicostas, Mar 02 2021: (Start)
Formulas for the square array A(n,k) (n, k >= 0):
A(0,k) = (-1)^floor((k-1)/2)*A000111(k) for k > 0 with A(0,0) = 0.
A(n,k) = Sum_{i=0..n} binomial(n, i)*A(0,k+i) for n, k >= 0.
A(n,n)/2 = A(n+1,n) = +/- A000657(n) for n > 0.
Bivariate e.g.f.: Sum_{n,k >= 0} A(n,k)*(x^n/n!)*(y^k/k!) = (-sech(x + y) + tanh(x + y) + 1)*exp(x).
Formulas for the triangular array T(n,k) = A(k,n-k) (n >= 0, 0 <= k <= n):
T(n,k) = T(n-1,k-1) + T(n,k-1) for 1 <= k <= n with T(n,0) = (-1)^floor((n-1)/2) * A000111(n) for n > 0 and T(0,0) = 0.
T(n,k) = Sum_{i=0..k} binomial(k,i)*T(n-k+i,0) for 0 <= k <= n. (End)

Extensions

Typo corrected by and more terms from Petros Hadjicostas, Mar 02 2021

A135594 a(n) = (1/2^n) * Sum_{i=0..n} (-1)^(n-i) * binomial(n,i) * A000364(i).

Original entry on oeis.org

1, 0, 1, 6, 73, 1380, 37801, 1417626, 69802993, 4369750440, 339034806001, 31935510092046, 3590398569115513, 474937566660074700, 73024143791301120601, 12914495107705743175266, 2603190607000627341985633, 593297406341867021292734160
Offset: 0

Views

Author

Vladeta Jovovic, Feb 25 2008

Keywords

Comments

Let k be a positive integer. It appears that reducing the sequence {a(n): n >= 1} modulo k produces a periodic sequence with period a divisor of phi(k) unless k is of the form 2^j, when the period equals k. For example, modulo 7 the sequence becomes [0, 1, 6, 3, 1, 1, 0, 1, 6, 3, 1, 1, 0, 1, 6, 3, 1, 1, ...], with an apparent period of 6 = phi(7), while modulo 8 the sequence becomes [0, 1, 6, 1, 4, 1, 2, 1, 0, 1, 6, 1, 4, 1, 2, 1, 0, 1, 6, 1, 4, 1, 2, 1, ...] with an apparent period of 8. - Peter Bala, May 07 2023

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, Exercise 4.2.2.(b).

Crossrefs

Programs

  • Maple
    A000364 := proc(n) option remember ; (2*n)!*coeftayl(sec(x),x=0,2*n) ; end: A135594 := proc(n) add((-1)^(n-i)*binomial(n,i)*A000364(i),i=0..n)/2^n ; end: seq(A135594(n),n=0..20) ; # R. J. Mathar, Mar 14 2008
    f:=sec(z): fser:=series(f,z=0,63): for n from 0 to 60 do b[n]:=factorial(n)*coeff(fser,z,n) end do: a:= proc(n) options operator, arrow: add((-1)^(n-k)*binomial(n,k)*b[2*k],k=0..n)/2^n end proc: seq(a(n),n=0..16); # Emeric Deutsch, Mar 17 2008
  • Mathematica
    Table[(-1)^n*Sum[Binomial[n, k]*EulerE[2*k], {k, 0, n}]/2^n, {n, 0, 20}] (* Vaclav Kotesovec, Jun 08 2019 *)

Formula

G.f.: 1/Q(0), where Q(k)= 1 + x - x*(2*k+1)*(k+1)/(1 - x*(2*k+1)*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 04 2013
a(n) ~ 2^(3*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Jun 08 2019
Conjecture: e.g.f. as a continued fraction: 2*exp(-t)/(2 - (1-exp(-4*t))/(2 - (1-exp(-8*t))/(2 - (1-exp(-12*t))/(2 - ... )))) = 1 + t^2/2! + 6*t^3/3! + 73*t^4/4! + .... Cf. A000657 and A005799. - Peter Bala, Dec 21 2019

Extensions

More terms from R. J. Mathar and Emeric Deutsch, Mar 03 2008

A088969 Triangle T(n,k), read by rows, given by A000290 DELTA [1, 2, 6, 5, 11, 8, 16, 11, 21, 14, 26, 17, 31, 20, 36, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 5, 20, 21, 0, 61, 287, 445, 231, 0, 1385, 7244, 13688, 11206, 3453, 0, 50521, 285295, 612556, 631862, 320159, 65271, 0, 2702765, 16195816, 38143429, 45667208, 29860381, 10300476, 1491381
Offset: 0

Views

Author

Philippe Deléham, Oct 29 2003, Nov 07 2007

Keywords

Examples

			Triangle begins:
1;
0, 1;
0, 1, 3;
0, 5, 20, 21;
0, 61, 287, 445, 231;
0, 1385, 7244, 13688, 11206, 3453;
0, 50521, 285295, 612556, 631862, 320159, 65271;
0, 2702765, 16195816, 38143429, 45667208, 29860381, 10300476, 1491381;
		

Crossrefs

Formula

Row sums : A000657; median Euler numbers (the middle of Arnold's shuttle triangle) Columns : A000007; A000364 (Euler numbers)
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