A118562 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.

1, 3, 5, 15, 0, 23, 89, 95, 0, 123, 0, 215, 0, 0, 1117, 943, 0, 2003, 0, 0, 0, 3455, 0, 1439, 0, 7846, 0, 7916, 0, 14735, 13103, 0, 0, 0, 0, 23711, 0, 0, 0, 24049, 0, 44857, 0, 0, 0, 44711, 0, 47594, 0, 0, 0, 77021, 0, 0, 0, 0, 0, 195765, 0, 381398, 0, 0, 374435, 0, 0

Offset: 1

Author

Robert G. Wilson v, Nov 23 2005

Keywords

• nonn

Comments

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

Examples

			a(3)=5 because binomial(10,5)=252 which is divisible by the squares of 1, 2 & 3 but not 4 squared.
a(70)=385823.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A059097, A111869, A080765, A024619.

Programs

Mathematica

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 *)
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

Formula

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

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