A284016 - OEIS

%I #88 Jan 19 2025 01:27:44

%S -1,2,2,4,10,28,84,264,858,2860,9724,33592,117572,416024,1485800,

%T 5348880,19389690,70715340,259289580,955277400,3534526380,13128240840,

%U 48932534040,182965127280,686119227300,2579808294648,9723892802904,36734706144304,139067101832008,527495903500720,2004484433302736

%N a(-1)=-1; a(n) = 2*A000108(n) for n >= 0.

%C There exists a set of Ramanujan-Sato series using this sequence.

%C a(n-1) = binomial(2n,n)/(2n-1) is the number of walks on a line that start and end at origin after 2n steps, not touching origin at intermediate stages. - _Robert FERREOL_, Aug 24 2019

%C From _Bernard Schott_, Apr 18 2020: (Start)

%C If E_n is the set of sequences with 2n digits composed with n digits 0 and n digits 1, then #E_n = binomial(2n,n). Now, if F_n is the subset of E_n composed of sequences where number of 0's = number of 1's for the first time at index 2n, then #F_n = a(n-1) = binomial(2n,n)/(2n-1) [see reference for proof].

%C Example: For n = 2, E_2 = {(0,0,1,1), (0,1,0,1), (1,0,1,0), (1,1,0,0), (0,1,1,0), (1,0,0,1)} with #E_2 = 6, but, F_2 = {(0,0,1,1), (1,1,0,0)} and #F_2 = a(1) = 2. (End)

%D Patrice Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 11, p. 296.

%H G. C. Greubel, <a href="/A284016/b284016.txt">Table of n, a(n) for n = -1..1000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series#Similar_series">Ramanujan-Sato series</a>.

%F a(n) = -A002420(n+1).

%F a(n) = (2*4^n*Gamma(n+1/2))/(sqrt(Pi)*Gamma(n+2)) for n >= -1. - _Ralf Steiner_, Apr 02 2017

%F a(n) = binomial(2*n+2, n+1) / (2*n+1) = 4*binomial(2*n, n) - binomial(2*n+2, n+1) for all n in Z. - _Michael Somos_, Jan 26 2018

%F a(n) = A228403(n) for n > 1 (essentially twice the Catalan numbers). - _Georg Fischer_, Oct 23 2018

%F From _Stefano Spezia_, Aug 24 2019: (Start)

%F G.f. for n >= 0: (1 - sqrt(1 - 4*x))/x.

%F E.g.f. for n >= 0: 2*(exp(2*x))*(I_{0}(2*x) - I_{1}(2*x)) where I_{k}(x) is the modified Bessel function of the first kind. (End)

%F Sum_{n>=-1} 1/a(n) = 2*Pi/(9*sqrt(3)). - _Amiram Eldar_, Jan 18 2025

%e The a(3)=10 8-step walks starting from and ending at the origin are [0, -1, -2, -3, -4, -3, -2, -1, 0], [0, -1, -2, -3, -2, -3, -2, -1, 0], [0, -1, -2, -3, -2, -1, -2, -1, 0], [0, -1, -2, -1, -2, -3, -2, -1, 0], [0, -1, -2, -1, -2, -1, -2, -1, 0], [0, 1, 2, 1, 2, 1, 2, 1, 0], [0, 1, 2, 1, 2, 3, 2, 1, 0], [0, 1, 2, 3, 2, 1, 2, 1, 0], [0, 1, 2, 3, 2, 3, 2, 1, 0], [0, 1, 2, 3, 4, 3, 2, 1, 0]. - _Robert FERREOL_, Aug 24 2019

%p seq(binomial(2*n+2, n+1) / (2*n+1),n=-1..20); # _Robert FERREOL_, Aug 24 2019

%t Prepend[Table[(2*CatalanNumber[n]), {n, 0, 20}], -1]

%o (PARI) for(n=-1, 25, print1(round((2*4^n*gamma(n+1/2))/(sqrt(Pi)*gamma(n+2))), ", ")) \\ _G. C. Greubel_, Apr 11 2017

%o (PARI) {a(n) = binomial(2*n+2, n+1) / (2*n+1)}; /* _Michael Somos_, Jan 26 2018 */

%o (Magma) [Binomial(2*n+2, n+1) / (2*n+1): n in [-1..30]]; // _Vincenzo Librandi_, Jan 27 2018

%Y Cf. A000108, A002420, A228403.

%Y Essentially the same as A262543 and A068875.

%K sign,changed

%O -1,2

%A _Ralf Steiner_, Mar 28 2017