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Bisection of A001790.
5

%I #39 May 18 2020 04:03:09

%S 1,5,63,429,12155,88179,1300075,9694845,583401555,4418157975,

%T 67282234305,514589420475,15801325804719,121683714103007,

%U 1879204156221315,14544636039226909,1804857108504066435

%N Bisection of A001790.

%H Vincenzo Librandi, <a href="/A063079/b063079.txt">Table of n, a(n) for n = 1..200</a>

%H Petros Hadjicostas, <a href="/A334907/a334907.pdf">Proof of the claim A334907(n)/n! = a(n+1)/A060818(n)</a>, 2020.

%F Numerators of binomial(2*n-3/2, -1/2).

%F Because A334907(n)/n! = a(n+1)/A060818(n) for n >= 0, the o.g.f. of a(n+1)/A060818(n), for n >= 0, is (sqrt(1 + sqrt(8*s)) - sqrt(1 - sqrt(8*s)))/sqrt(8*s * (1 - 8*s)), which is the e.g.f. of A334907 (see the link above for a proof). - _Petros Hadjicostas_, May 16 2020

%p seq(numer(binomial(2*n-3/2,-1/2)), n=1..20);

%t Numerator[Binomial[2Range[20]-3/2,-(1/2)]] (* _Harvey P. Dale_, Feb 27 2012 *)

%Y Cf. A001790, A060818, A334907. Other bisection gives A061548.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Aug 07 2001

%E More terms from _Vladeta Jovovic_, Aug 07 2001