A265388 a(n) = gcd{k=1..n-1} binomial(2*n, 2*k), a(1) = 0.
0, 6, 15, 14, 15, 33, 91, 2, 51, 19, 11, 23, 65, 3, 435, 62, 17, 3, 703, 1, 41, 43, 23, 47, 35, 1, 159, 7, 29, 59, 1891, 2, 1, 67, 1, 71, 2701, 1, 1, 79, 123, 249, 43, 1, 267, 1, 47, 1, 679, 1, 101, 103, 53, 321, 109, 1, 113, 1, 59, 1, 671, 1, 5, 254, 5, 1441
Offset: 1
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Carl McTague, On the Greatest Common Divisor of C(q*n,n), C(q*n,2*n), ...C(q*n,q*n-q), arXiv:1510.06696 [math.CO], 2015.
Crossrefs
Programs
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Mathematica
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 *) -
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 for(n=1, 10000, write("b265388.txt", n, " ", A265388(n)));
-
Scheme
(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)))))) ;; A version using fold. Instead of fold-left we could as well use fold-right: (define (A265388 n) (fold-left gcd 0 (map (lambda (k) (A007318tr (* 2 n) (* 2 k))) (range1-n (- n 1))))) (define (range1-n n) (let loop ((n n) (result (list))) (cond ((zero? n) result) (else (loop (- n 1) (cons n result)))))) ;; In above code A007318tr(n,k) computes the binomial coefficient C(n,k), i.e., Pascal's triangle A007318. - Antti Karttunen, Dec 11 2015
Formula
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).