cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A001900 Successive numerators of Wallis's approximation to Pi/2 (unreduced).

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%I A001900 #72 Jul 30 2024 04:22:52
%S A001900 1,2,4,16,64,384,2304,18432,147456,1474560,14745600,176947200,
%T A001900 2123366400,29727129600,416179814400,6658877030400,106542032486400,
%U A001900 1917756584755200,34519618525593600,690392370511872000,13807847410237440000,303772643025223680000
%N A001900 Successive numerators of Wallis's approximation to Pi/2 (unreduced).
%C A001900 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
%C A001900 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
%C A001900 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
%D A001900 H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146.
%H A001900 Alois P. Heinz, <a href="/A001900/b001900.txt">Table of n, a(n) for n = 0..449</a>
%H A001900 John Derbyshire, <a href="http://tomlr.free.fr/Math%E9matiques/Fichiers%20Claude/Nombres/Derbyshire%20-%20Prime%20Obsession%20-%20Bernhard%20Riemann%20and%20the%20Greatest%20Unsolved%20Problem%20in%20MathematicsAAA.pdf ">Prime Obsession</a>, Plume books, p. 16, 2003.
%H A001900 Jonathan Sondow, <a href="https://arxiv.org/abs/math/0401406">A faster product for Pi and a new integral for ln(Pi/2)</a>, arXiv:math/0401406 [math.NT], 2004.
%H A001900 Jonathan Sondow, <a href="http://www.jstor.org/stable/30037575">A faster product for Pi and a new integral for ln(Pi/2)</a>, Amer. Math. Monthly 112 (2005), 729-734 and 113 (2006), 670.
%H A001900 <a href="/index/Di#divseq">Index to divisibility sequences</a>
%F A001900 2.2.4.4.6.6....2n.2n.../1.3.3.5.5.7.7....(2n-1).(2n+1) ...for n >= 1.
%F A001900 a(n) = 2^n * A010551(n) = 2^n * [n/2]! * [(n+1)/2]!. - _Ralf Stephan_, Mar 11 2004
%F A001900 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]
%F A001900 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
%F A001900 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
%F A001900 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
%F A001900 From _Mathias Zechmeister_, Jul 24 2022: (Start)
%F A001900 Let J_0(x) and J_1(x) denote the Bessel functions and i = sqrt(-1).
%F A001900 a(n) = denominator([x^n](J_0(x) + J_1(x))).
%F A001900 a(n) = denominator([x^n](J_0(i*x) - i*J_1(i*x))).
%F A001900 Bisection: a(2*k) = A002454(k), a(2*k+1) = A002474(k).
%F A001900 G.f. for 1/a(n): J_0(i*x) - i*J_1(i*x).  (End)
%e A001900 From _Wolfdieter Lang_, Dec 06 2017: (Start)
%e A001900 Partial products of the rows N (for numerators a(n)) and D (for denominators b(n) = A000246(n+1)) begin:
%e A001900 n:    0  1  2  3  4   5    6     7      8       9       10 ...
%e A001900 N:    1  2  2  4  4   6    6     8      8      10       10 ...
%e A001900 D:    1  1  3  3  5   5    7     7      9       9       11 ...
%e A001900 a(n): 1  2  4 16 64 384 2304 18432 147456 14745601 4745600 ...
%e A001900 b(n): 1  1  3  9 45 225 1575 11025  99225   893025 9823275 ...  (End)
%t A001900 a[n_] := a[n] = If[n==0, 1, (n+Mod[n, 2]) a[n-1]];
%t A001900 a /@ Range[0, 21] (* _Jean-François Alcover_, Jan 31 2020 *)
%o A001900 (PARI) a(n)=if(n<0,0,prod(k=1,n,if(k%2,k+1,k)))
%Y A001900 Cf. A000246, A284230.
%Y A001900 For the reduced form see A001901(n)/A001902(n), n >= 0.
%K A001900 nonn,frac,easy
%O A001900 0,2
%A A001900 _N. J. A. Sloane_