A056673 Number of unitary and squarefree divisors of binomial(n, floor(n/2)). Also the number of divisors of the powerfree part of A001405(n), A056060(n).

1, 2, 2, 4, 4, 2, 4, 8, 4, 2, 16, 8, 8, 8, 8, 16, 32, 16, 32, 16, 32, 32, 64, 32, 16, 16, 8, 8, 32, 32, 64, 128, 128, 64, 256, 128, 128, 128, 512, 256, 512, 512, 512, 512, 64, 64, 256, 128, 128, 128, 128, 128, 256, 256, 2048, 2048, 4096, 4096, 2048, 2048, 2048, 2048

Offset: 1

Author

Labos Elemer, Aug 10 2000

Keywords

• nonn

Examples

			n = 14: binomial(15,7) = 3432 = 2*2*2*3*11*13, which has 32 divisors. Of those divisors, 16 are unitary: {1, 3, 8, 11, 13, 24, 33, 39, 88, 104, 143, 264, 312, 429, 1144, 3432}; 16 are squarefree: {1, 2, 3, 6, 11, 13, 22, 26, 33, 39, 66, 78, 143, 286, 429, 858}. Only 8 of the divisors belong to both classes: {1, 3, 11, 13, 33, 39, 143, 429}. Thus, a(14) = 8.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001405, A007913, A055229, A055231, A056060, A056671.

Programs

Mathematica

Table[With[{m = Binomial[n, Floor[n/2]]}, DivisorSum[m, 1 &, And[CoprimeQ[#, m/#], SquareFreeQ@ #] &]], {n, 62}] (* Michael De Vlieger, Sep 05 2017 *)
f[p_, e_] := If[e == 1, 2, 1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[ Binomial[n, Floor[n/2]]]); Array[a, 60] (* Amiram Eldar, Sep 06 2020 *)

PARI

a(n) = my(b=binomial(n, n\2)); sumdiv(b, d, issquarefree(d) && (gcd(d, b/d) == 1)); \\ Michel Marcus, Sep 05 2017

Formula

a(n) = A000005(A055231(x)) = A000005(A007913(x)/A055229(x)), where x = A001405(n) = binomial(n, floor(n/2)).

a(n) = A056671(A001405(n)). - Amiram Eldar, Sep 06 2020