OFFSET
0,2
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..500
D. Doster, Problem 1318, Three Term Recurrence, Math. Magazine, 63 (1990), 127-128.
FORMULA
f(n) = Sum_{k=0..n} binomial(-1/2,k)*(-1/2)^k.
f(n) -> sqrt(2) as n -> oo.
G.f.: (sqrt(-x)*arccsc(1-x)/sqrt(2)-(Pi*i*sqrt(x))/sqrt(2)^3)/x. - Vladimir Kruchinin, Oct 10 2012
a(n) = numerator( Sum_{k=0..n} binomial(2*k, k)/8^k ). - G. C. Greubel, Jan 29 2020
MAPLE
seq( numer( add(binomial(2*k, k)/8^k, k=0..n) ), n=0..25); # G. C. Greubel, Jan 29 2020
MATHEMATICA
a[n_] := Sqrt[2](1-(Gamma[1/2+n] Hypergeometric2F1[n, 1/2+n, 1+n, -1])/(Sqrt[Pi] Gamma[1+n])); Table[Numerator[FullSimplify[a[n]]], {n, 20}] (* Gerry Martens, Aug 09 2015 *)
f[n_]:= If[n==0, 1, If[n==1, 5/4, ((6*n-1)*f[n-1]-(2*n-1)*f[n-2])/(4*n)]];
Table[Numerator[f[n]], {n, 0, 25}] (* G. C. Greubel, Jan 29 2020 *)
PROG
(PARI) A126963(n)=numerator(sum(k=0, n, binomial(-1/2, k)/(-2)^k)) \\ f(n)=if(n>1, ((6*n-1)*f(n-1)-(2*n-1)*f(n-2))/(4*n), (5/4)^n) yields the same results. - M. F. Hasler, Aug 11 2015
(Magma) [Numerator( &+[Binomial(2*k, k)/8^k: k in [0..n]] ): n in [0..25]]; // G. C. Greubel, Jan 29 2020
(Sage) [numerator( sum(binomial(2*k, k)/8^k for k in (0..n)) ) for n in (0..25)] # G. C. Greubel, Jan 29 2020
(GAP) List([0..25], n-> NumeratorRat( Sum([0..n], k-> Binomial(2*k, k)/8^k) )); # G. C. Greubel, Jan 29 2020
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
N. J. A. Sloane, Mar 20 2007
STATUS
approved