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A100967 Least k such that binomial(2k + 1, k - n) >= binomial(2k, k). 3
3, 9, 18, 29, 44, 61, 81, 104, 130, 159, 191, 225, 263, 303, 347, 393, 442, 494, 549, 606, 667, 730, 797, 866, 938, 1013, 1091, 1172, 1255, 1342, 1431, 1524, 1619, 1717, 1818, 1922, 2029, 2138, 2251, 2366, 2485, 2606, 2730, 2857, 2987, 3119, 3255, 3394, 3535 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
From the formula, if we know k, we can estimate n as approximately 0.83 sqrt(k).
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
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..500
FORMULA
Round(0.3807 + 1.43869 n + 1.44276 n^2) is an exact fit for the first 50 terms.
As n -> infinity, we have a(n) = (n^2+n)/log(2) + o(n). - Robert Israel, Mar 16 2016
MAPLE
F:= proc(n) local Q, LQ, k, k0;
LQ:= -ln(GAMMA(k-n+1))-ln(GAMMA(k+1+n))-ln(k+1+n)+ln(2*k+1)+2*ln(GAMMA(k+1));
k0:= floor(fsolve(LQ, k=n..max(2*n^2, 9)));
if (2*k0+1)*binomial(k0, n) >= (n+1)*binomial(k0+1+n, n+1) then
while (2*k0-1)*binomial(k0-1, n) >= (n+1)*binomial(k0+n, n+1) do k0:= k0-1 od
else
while (2*k0+1)*binomial(k0, n) < (n+1)*binomial(k0+1+n, n+1) do k0:= k0+1 od;
fi;
k0;
end proc:
map(F, [$1..100]); # Robert Israel, Mar 16 2016
MATHEMATICA
k=1; Table[While[Binomial[2k+1, k-n] < Binomial[2k, k], k++ ]; k, {n, 50}]
PROG
(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
CROSSREFS
Cf. A000984, A003015 (numbers that occur 5 or more times in Pascal's triangle).
Sequence in context: A325729 A159794 A352643 * A193567 A045943 A134479
KEYWORD
nonn
AUTHOR
T. D. Noe, Nov 23 2004
STATUS
approved

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