cp's OEIS Frontend

A100967 Least k such that binomial(2k + 1, k - n) >= binomial(2k, k).

%I A100967 #24 Jul 04 2017 14:37:50
%S A100967 3,9,18,29,44,61,81,104,130,159,191,225,263,303,347,393,442,494,549,
%T A100967 606,667,730,797,866,938,1013,1091,1172,1255,1342,1431,1524,1619,1717,
%U A100967 1818,1922,2029,2138,2251,2366,2485,2606,2730,2857,2987,3119,3255,3394,3535
%N A100967 Least k such that binomial(2k + 1, k - n) >= binomial(2k, k).
%C A100967 From the formula, if we know k, we can estimate n as approximately 0.83 sqrt(k).
%C A100967 Open question: Does binomial(2*a(n) + 1, a(n) - n) = binomial(2*a(n), a(n)) for any n? An affirmative answer would settle whether there exists an odd term greater than 3 in A003016. - _Danny Rorabaugh_, Mar 16 2016
%H A100967 Charles R Greathouse IV, <a href="/A100967/b100967.txt">Table of n, a(n) for n = 1..500</a>
%F A100967 Round(0.3807 + 1.43869 n + 1.44276 n^2) is an exact fit for the first 50 terms.
%F A100967 As n -> infinity, we have a(n) = (n^2+n)/log(2) + o(n). - _Robert Israel_, Mar 16 2016
%p A100967 F:= proc(n) local Q, LQ, k, k0;
%p A100967      LQ:= -ln(GAMMA(k-n+1))-ln(GAMMA(k+1+n))-ln(k+1+n)+ln(2*k+1)+2*ln(GAMMA(k+1));
%p A100967       k0:= floor(fsolve(LQ, k=n..max(2*n^2, 9)));
%p A100967       if (2*k0+1)*binomial(k0, n) >= (n+1)*binomial(k0+1+n, n+1)  then
%p A100967         while (2*k0-1)*binomial(k0-1, n) >= (n+1)*binomial(k0+n, n+1) do k0:= k0-1 od
%p A100967       else
%p A100967         while (2*k0+1)*binomial(k0, n) < (n+1)*binomial(k0+1+n, n+1) do k0:= k0+1 od;
%p A100967       fi;
%p A100967       k0;
%p A100967 end proc:
%p A100967 map(F, [$1..100]); # _Robert Israel_, Mar 16 2016
%t A100967 k=1; Table[While[Binomial[2k+1, k-n] < Binomial[2k, k], k++ ]; k, {n, 50}]
%o A100967 (PARI) a(n,k=n+1)=while((2*k+1)*k!^2<(n+k+2)!*(k-n-1)!,k++);k \\ _Charles R Greathouse IV_, Sep 09 2013
%Y A100967 Cf. A000984, A003015 (numbers that occur 5 or more times in Pascal's triangle).
%K A100967 nonn
%O A100967 1,1
%A A100967 _T. D. Noe_, Nov 23 2004