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 1-3 of 3 results.

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

A275784 Number A(n,k) of up-down sequences with k copies each of 1,2,...,n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 4, 5, 1, 1, 0, 1, 12, 53, 16, 1, 1, 0, 1, 36, 761, 936, 61, 1, 1, 0, 1, 120, 12661, 87336, 25325, 272, 1, 1, 0, 1, 400, 229705, 9929000, 18528505, 933980, 1385, 1, 1, 0, 1, 1400, 4410665, 1267945800, 17504311533, 6376563600, 45504649, 7936, 1
Offset: 0

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Author

Alois P. Heinz, Aug 12 2016

Keywords

Examples

			A(4,1) = 5: 1324, 1423, 2314, 2413, 3412.
A(3,2) = 4: 121323, 132312, 231213, 231312.
A(3,3) = 12: 121313232, 121323132, 121323231, 131213232, 132312132, 132323121, 231213132, 231213231, 231312132, 231323121, 232312131, 232313121.
A(2,4) = 1: 12121212.
Square array A(n,k) begins:
  1,   1,      1,          1,              1,              1, ...
  1,   1,      0,          0,              0,              0, ...
  1,   1,      1,          1,              1,              1, ...
  1,   2,      4,         12,             36,            120, ...
  1,   5,     53,        761,          12661,         229705, ...
  1,  16,    936,      87336,        9929000,     1267945800, ...
  1,  61,  25325,   18528505,    17504311533, 19126165462061, ...
  1, 272, 933980, 6376563600, 59163289699260, ...
		

Crossrefs

Columns k=0-3 give: A000012, A000111, A275801, A276636.
Rows n=2-5 give: A000012, A241530, A036916, A276637.

Programs

  • Maple
    b:= proc(n, l) option remember; `if`(l=[], 1, `if`(irem(add(i,
          i=l), 2)=0, add(b(i, subsop(i=`if`(l[i]=1, [][], l[i]-1),
          l)), i=n+1..nops(l)), add(b(i-`if`(l[i]=1, 1, 0), subsop(
          i=`if`(l[i]=1, [][], l[i]-1), l)), i=1..n-1)))
        end:
    A:= (n, k)->`if`(k=0, 1, b(`if`(irem(k*n, 2)=0, 0, n+1), [k$n])):
    seq(seq(A(n, d-n), n=0..d), d=0..10);
  • Mathematica
    b[n_, l_List] := b[n, l] = If[l == {}, 1, If[EvenQ[Total[l]], Sum[b[i, ReplacePart[l, i -> If[l[[i]] == 1, Nothing, l[[i]]-1]]], {i, n+1, Length[l]}], Sum[b[i - If[l[[i]] == 1, 1, 0], ReplacePart[l, i -> If[l[[i]] == 1, Nothing, l[[i]]-1]]], {i, 1, n-1}]]]; A[n_, k_] := If[k == 0, 1, b[If[EvenQ[k*n], 0, n+1], Array[k&, n]]]; Table[A[n, d-n], {d, 0, 10}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 23 2017, adapted from Maple *)

A275829 Number of weakly alternating permutations of the multiset {1,1,2,2,...,n,n}.

Original entry on oeis.org

1, 1, 2, 12, 140, 2564, 68728, 2539632, 123686800, 7677924688, 591741636128, 55438330474944, 6204888219697856, 817697605612952384, 125322509904814743424, 22102340129003429880576, 4444468680409243484516608, 1010802175212828388101900544, 258152577318424951261637001728
Offset: 0

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Author

Max Alekseyev, Aug 11 2016

Keywords

Comments

Number of permutations (p(1),...,p(2n)) of {1,1,2,2,...,n,n} satisfying p(1) <= p(2) >= p(3) <= p(4) >= p(5) <= ... <= p(2n).
a(n) >= A005799(n) >= A275801(n).

Crossrefs

Showing 1-3 of 3 results.