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 A001900 Successive numerators of Wallis's approximation to Pi/2 (unreduced). 8
 1, 2, 4, 16, 64, 384, 2304, 18432, 147456, 1474560, 14745600, 176947200, 2123366400, 29727129600, 416179814400, 6658877030400, 106542032486400, 1917756584755200, 34519618525593600, 690392370511872000, 13807847410237440000, 303772643025223680000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) = number of permutations of [n+1] all of whose non-initial left-to-right minima are at even positions in the permutation. For example, a(2) = 4 counts 123, 132, 213, 312. - David Callan, Jul 22 2008 Number of self-avoiding planar walks starting at (0,0), ending at (n,0), remaining in the first quadrant and using steps (0,1), (1,0), (-1,1), and (1,-1) with the restriction that (0,1) is never used below the diagonal and (1,0) is never used above the diagonal. a(2) = 4: [(0,0),(1,0),(2,0)], [(0,0),(0,1),(1,0),(2,0)], [(0,0),(0,1),(0,2),(1,1),(2,0)], [(0,0),(1,0),(0,1),(0,2),(1,1),(2,0)]. - Alois P. Heinz, Mar 23 2017 REFERENCES H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..449 John Derbyshire, Prime Obsession, Plume books, p. 16, 2003. J. Sondow, A faster product for Pi and a new integral for ln(Pi/2), arXiv:math/0401406 [math.NT], 2004. J. Sondow, A faster product for Pi and a new integral for ln(Pi/2), Amer. Math. Monthly 112 (2005), 729-734 and 113 (2006), 670. FORMULA 2.2.4.4.6.6....2n.2n.../1.3.3.5.5.7.7....(2n-1).(2n+1) ...for n >= 1. a(n) = 2^n * A010551(n) = 2^n * [n/2]! * [(n+1)/2]!. - Ralf Stephan, Mar 11 2004 Conjecture: a(n) - a(n-1) - n*(n-1)*a(n-2) = 0. - R. J. Mathar, Jun 07 2013 [The proof, for n >= 2, follows from the bisection recurrence given below. - Wolfdieter Lang, Dec 07 2017] E.g.f.: E(0), where E(k)= 1 + 2*x*(k+1)/((2*k+1) - x*(2*k+1)/(x + 1/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 08 2013 G.f.: G(0), where G(k)= 1 + 2*x*(k+1)/(1 - 2*x*(k+1)/(2*x*(k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 08 2013 Bisection: a(2*k+1) = ((2*k+1)+1)*a(2*k), a(2*k) = 2*k*a(2*k-1), k >= 0, with a(0) = 1. The proof is obvious from the numbers in the numerator (see the row N in the example). From a proposal by David James Sycamore, Nov 02 2017 based on the fractions 4/1, 8/3, 32/9, 128/45, ... converging very slowly to Pi, given on p. 16 of the Derbyshire link. - Wolfdieter Lang, Dec 06 2017 EXAMPLE From Wolfdieter Lang, Dec 06 2017: (Start) Partial products of the rows N (for numerators a(n)) and D (for denominators b(n) = A000246(n+1)) begin: n:    0  1  2  3  4   5    6     7      8       9       10 ... N:    1  2  2  4  4   6    6     8      8      10       10 ... D:    1  1  3  3  5   5    7     7      9       9       11 ... a(n): 1  2  4 16 64 384 2304 18432 147456 14745601 4745600 ... b(n): 1  1  3  9 45 225 1575 11025  99225   893025 9823275 ...  (End) PROG (PARI) a(n)=if(n<0, 0, prod(k=1, n, if(k%2, k+1, k))) CROSSREFS Cf. A000246, A284230. For the reduced form see A001901(n)/A001902(n), n >= 0. Sequence in context: A106186 A155543 A151371 * A113247 A280132 A138870 Adjacent sequences:  A001897 A001898 A001899 * A001901 A001902 A001903 KEYWORD nonn,frac,easy AUTHOR STATUS approved

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Last modified April 21 20:03 EDT 2019. Contains 322328 sequences. (Running on oeis4.)