OFFSET
0,4
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).
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
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 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Vladeta Jovovic, Feb 25 2008
EXTENSIONS
More terms from R. J. Mathar and Emeric Deutsch, Mar 03 2008
STATUS
approved