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H(n) = 3* A000436(n)/2^(2n+1) = 3*A002114(n). - Philippe Deléham, Jan 17 2004
E.g.f.: E(x) = 3*x^2/(G(0)-x^2) ; G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 03 2012
If E(x) = Sum(_{k>=0,1,..., } a(k+1)*x^(2k+2 )), then A002112(k) = a(k+1)*(2*k+2)!. - Sergei N. Gladkovskii, Jan 09 2012
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J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
J. W. L. Glaisher, <a href="httphttps://doi.org/10.1112/plms.oxfordjournals.org/content/s1-31/.1/.216.extract">On a set of coefficients analogous to the Eulerian numbers</a>, Proc. London Math. Soc., 31 (1899), 216-235.
Michael E. Hoffman, <a href="https://doi.org/10.37236/1453">Derivative polynomials, Euler polynomials, and associated integer sequences</a>, The Electronic Journal of Combinatorics [electronic only] 6.1 (1999).
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