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A005799
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Generalized Euler numbers of type 2^n.
(Formerly M1979)
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11
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1, 1, 2, 10, 104, 1816, 47312, 1714000, 82285184, 5052370816, 386051862272, 35917232669440, 3996998043812864, 524203898507631616, 80011968856686405632, 14061403972845412526080, 2818858067801804443910144
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OFFSET
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0,3
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COMMENTS
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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 sequence appears to be middle column in Poupard's triangle A008301.
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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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
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)
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MAPLE
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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
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MATHEMATICA
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a[n_] := Sum[Binomial[n, i]Abs[EulerE[2i]], {i, 0, n}]/2^n
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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