%I #5 Mar 17 2018 21:22:03

%S 1,3,5,15,0,23,89,95,0,123,0,215,0,0,1117,943,0,2003,0,0,0,3455,0,

%T 1439,0,7846,0,7916,0,14735,13103,0,0,0,0,23711,0,0,0,24049,0,44857,0,

%U 0,0,44711,0,47594,0,0,0,77021,0,0,0,0,0,195765,0,381398,0,0,374435,0,0

%N Least number k such that binomial(2k,k) is divisible by all squares to n squared but not (n+1) squared, or 0 if impossible.

%C a(5)=0 because any number squared which would divide binomial(2k,k) would also be divided by 6^2 since 6=2*3.

%F a(n)=0 iff n is a member of A080765: m such that m+1 divides lcm(1..m).

%F a(n-1)=0 iff n-1 is a member of A024619: Numbers that are not powers of primes.

%e a(3)=5 because binomial(10,5)=252 which is divisible by the squares of 1, 2 & 3 but not 4 squared.

%e a(70)=385823.

%t f[n_] := Block[{k = 1, b = Binomial[2n, n]}, While[Mod[b, k^2] == 0, k++ ]; k - 1]; t = Table[0, {100}]; Do[ a = f[n]; If[a < 101 &t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 38000}] (* or *)

%t expoPF[k_, n_] := Module[{s = 0, x = n}, While[x > 0, x = Floor[x/k]; s += x]; s]; expoCF[k_, n_] := Min[expoPF[ #[[1]], n]/#[[2]] & /@ FactorInteger@k]; f[n_] := Module[{k = 2}, While[ expoCF[k, 2n] >= 2(1 + expoCF[k, n]), k++ ]; k-1]; t = Table[0, {100}]; Do[ a = f[n]; If[a < 101 &t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 400000}]; t

%Y Cf. A059097, A111869, A080765, A024619.

%K nonn

%O 1,2

%A _Robert G. Wilson v_, Nov 23 2005