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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
a(n+1) is the number of 0-1 square matrices of order n+1 with 2n+1 nonzero entries where the cell (i,j) is 1 for all i+j=n+2 and every diagonal, parallel to the main diagonal, has exactly one 1. For example, a(2) = 4: [(0,1,1), (1,1,0), (1,0,0)], [(0,1,1), (0,1,0), (1,1,0)], [(0,0,1), (1,1,1), (1,0,0)], [(0,0,1), (0,1,1), (1,1,0)]. - Christian Barrientos, Jul 17 2021
REFERENCES
H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146.
LINKS
John Derbyshire, Prime Obsession, Plume books, p. 16, 2003.
Jonathan Sondow, A faster product for Pi and a new integral for ln(Pi/2), arXiv:math/0401406 [math.NT], 2004.
Jonathan 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
From Mathias Zechmeister, Jul 24 2022: (Start)
Let J_0(x) and J_1(x) denote the Bessel functions and i = sqrt(-1).
a(n) = denominator([x^n](J_0(x) + J_1(x))).
a(n) = denominator([x^n](J_0(i*x) - i*J_1(i*x))).
Bisection: a(2*k) = A002454(k), a(2*k+1) = A002474(k).
G.f. for 1/a(n): J_0(i*x) - i*J_1(i*x). (End)
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)
MATHEMATICA
a[n_] := a[n] = If[n==0, 1, (n+Mod[n, 2]) a[n-1]];
a /@ Range[0, 21] (* Jean-François Alcover, Jan 31 2020 *)
PROG
(PARI) a(n)=if(n<0, 0, prod(k=1, n, if(k%2, k+1, k)))
CROSSREFS
For the reduced form see A001901(n)/A001902(n), n >= 0.
Sequence in context: A106186 A155543 A151371 * A113247 A280132 A138870
KEYWORD
nonn,frac,easy
AUTHOR
STATUS
approved

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Last modified September 6 00:26 EDT 2024. Contains 375701 sequences. (Running on oeis4.)