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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).
2
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
1,2
LINKS
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
EXAMPLE
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.
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 *)
PROG
(PARI) a(n) = my(b=binomial(n, n\2)); sumdiv(b, d, issquarefree(d) && (gcd(d, b/d) == 1)); \\ Michel Marcus, Sep 05 2017
KEYWORD
nonn
AUTHOR
Labos Elemer, Aug 10 2000
STATUS
approved