A267825 Index of largest primorial factor of binomial(2n,n).

0, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 3, 3, 3, 3, 5, 5, 6, 3, 3, 3, 3, 2, 2, 1, 1, 5, 1, 1, 2, 4, 4, 2, 1, 1, 4, 1, 1, 5, 5, 5, 4, 4, 4, 4, 4, 3, 2, 2, 2, 5, 5, 2, 2, 2, 3, 3, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 3, 3, 3, 3, 6, 6, 6, 7, 5, 5, 5, 1, 1, 5, 1, 1, 6, 6, 6, 6, 1, 1, 6, 1, 1, 7, 7, 7, 3, 3, 3

Offset: 0

Author

Jonathan Sondow, Jan 27 2016

Keywords

• nonn

Comments

For n > 0, binomial(2n,n) is even, so a(n) >= 1.

Is a(n) unbounded? (The largest value for n <= 100000 is a(45416) = 43.)

From Robert Israel, Jan 28 2016: (Start)

a(n) = A000720(p)-1 where p is the least prime that does not divide A000984(n).

Equivalently, p is the least prime such that the base-p representation of n has all digits < p/2.

a(primorial(k)-1) >= k. In particular the sequence is unbounded. (End)

Examples

			Binomial(16,8) = 12870 is divisible by primorial(3) = 2*3*5 = 30, but not by prime(4) = 7, so a(8) = 3.
		

Links

• Wikipedia, Lucas' theorem

Crossrefs

Cf. A000720, A000984, A002110, A226078, A267823.

Programs

Mathematica

PrimorialFactor[n_] := (k = 0; While[Mod[n, Prime[k + 1]] == 0, k++]; k);
Table[PrimorialFactor[Binomial[2 n, n]], {n, 0, 100}]

PARI

pf(n) = {my(k = 0); while (n % prime(k+1) == 0, k++); k;}
a(n) = pf(binomial(2*n, n)); \\ adapted from Mathematica; Michel Marcus, Jan 29 2016

Formula

a(A267823(n)) >= n.

min{k : a(k) >= n} = A267823(n).