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%I #27 Oct 24 2019 11:20:39
%S 1,1,4,5,5,21,84,42,84,132,264,6435,6435,715,2860,4862,9724,352716,
%T 705432,58786,117572,1040060,2080120,6686100,13372200,2674440,5348880,
%U 9694845,9694845,583401555
%N a(n) = (binomial(n,floor(n/2)))/(greatest common divisor of all numbers in n-th row of Pascal's triangle excluding 1 and n).
%C For all values of a(n), where a(n) is not equal to A001405(n), n is either: a prime, a power of a prime, a prime +1 or a power of a prime +1.
%F a(n) = A001405(n)/A328202(n).
%e For n = 17, a(17) = A001405(17)/A328202(17) = 24310/34 = 715.
%t a[n_] := Binomial[n, Floor[n/2]]/GCD @@ Binomial[n, Range[2, n/2]]; Array[a, 30, 4] (* _Amiram Eldar_, Oct 24 2019 *)
%o (PARI) a(n) = binomial(n, n\2)/gcd(vector((n+1)\2-1, k, binomial(n, k+1))); \\ _Michel Marcus_, Oct 24 2019
%Y Cf. A001405, A328202.
%K nonn
%O 4,3
%A _Joel Kaufmann_, Oct 24 2019
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