%I #40 Dec 17 2015 10:37:46
%S 0,6,15,14,15,33,91,2,51,19,11,23,65,3,435,62,17,3,703,1,41,43,23,47,
%T 35,1,159,7,29,59,1891,2,1,67,1,71,2701,1,1,79,123,249,43,1,267,1,47,
%U 1,679,1,101,103,53,321,109,1,113,1,59,1,671,1,5,254,5,1441
%N a(n) = gcd{k=1..n-1} binomial(2*n, 2*k), a(1) = 0.
%H Antti Karttunen, <a href="/A265388/b265388.txt">Table of n, a(n) for n = 1..10000</a>
%H Carl McTague, <a href="http://arxiv.org/abs/1510.06696">On the Greatest Common Divisor of C(q*n,n), C(q*n,2*n), ...C(q*n,q*n-q)</a>, arXiv:1510.06696 [math.CO], 2015.
%F For prime p>2, valuation(a(n), p) = 1 if 2*n = p^i+p^j for some i<=j, 0 otherwise (see Theorem 2 in McTague).
%t Table[GCD @@ Array[Binomial[2 n, 2 #] &, {n - 1}], {n, 1, 66}] (* _Michael De Vlieger_, Dec 09 2015, modified to match the new corrected data by _Antti Karttunen_, Dec 11 2015 *)
%o (PARI) allocatemem(2^30); A265388(n) = if(n<=1, 0, gcd(vector(n-1, k, binomial(2*n, 2*k)))) \\ PARI versions before 2.8 return an erroneous value 1 for gcd of an empty vector/set. - _Michel Marcus_, Dec 08 2015 and _Antti Karttunen_, Dec 11 2015
%o for(n=1, 10000, write("b265388.txt", n, " ", A265388(n)));
%o (Scheme)
%o (define (A265388 n) (let loop ((z 0) (k 1)) (cond ((>= k n) z) ((= 1 z) z) (else (loop (gcd z (A007318tr (* 2 n) (* 2 k))) (+ k 1))))))
%o ;; A version using fold. Instead of fold-left we could as well use fold-right:
%o (define (A265388 n) (fold-left gcd 0 (map (lambda (k) (A007318tr (* 2 n) (* 2 k))) (range1-n (- n 1)))))
%o (define (range1-n n) (let loop ((n n) (result (list))) (cond ((zero? n) result) (else (loop (- n 1) (cons n result))))))
%o ;; In above code A007318tr(n,k) computes the binomial coefficient C(n,k), i.e., Pascal's triangle A007318. - _Antti Karttunen_, Dec 11 2015
%Y Cf. A007318, A014410, A014963.
%Y Cf. A265394 (positions of records), A265395 (record values), A265401 (positions of ones), A265402 (fixed points), A265403 (positions where a(n) = 2n-1).
%K nonn
%O 1,2
%A _Michel Marcus_, Dec 08 2015