OFFSET
1,2
COMMENTS
From Chai Wah Wu, May 04 2026: (Start)
Since for k=1, binomial(2*n,2*k) = n*(2*n-1), this implies that valuation(a(n),p) = 0 if p is prime and not a prime factor of n*(2*n-1).
Let nu_2(m) denote the 2-adic valuation of m.
Theorem: For n>1, nu_2(a(n)) = 1 if n is a power of 2 and nu_2(a(n)) = 0 otherwise.
Proof: First note that nu_2(a(n)) = min_{k=1..n-1} nu_2(binomial(2*n,2*k)). If n is not a power of 2, then there exists j such that 2^j < n and the j-th bit of n is 1. Then by setting k = 2^j, it is clear by Kummer's theorem that nu_2(binomial(2*n,2*k)) = 0 and thus nu_2(a(n))=0. If n=2^i is a power of 2, then i>0 since n>1. For each 1<=k<n, there will be a carry when adding 2*k to 2*(n-k) in binary and thus nu_2(binomial(2*n,2*k))>0. For k = n/2 = 2^(i-1), since 2*(n-k) = 2*k, adding 2*k to 2*k in binary will result in one carry, and Kummer's theorem shows that nu_2(binomial(2*n,2*k)) = 1. QED
This can be written in a similar form as the formula for odd prime p (assuming n>1):
For prime p=2, valuation(a(n), p) = 1 if 2*n = p^i+p^j for some 0<i=j, 0 otherwise.
A consequence is that all terms except for a(1) are squarefree. (End)
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], 2015-2018.
Chai Wah Wu, Computing the Greatest Common Divisor of Binomial Coefficients C(m*n,m*k), arXiv:2606.20940 [math.NT], 2026.
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).
For k > 1, a(2^(k-1)) <> 2 if and only if k is a Mersenne exponent (A000043) if and only if a(2^(k-1))/2 = 2^k-1 is a Mersenne prime (A000668). - Chai Wah Wu, Jun 22 2026
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 *)
PROG
(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
(Python)
from math import prod
from sympy import primefactors
from sympy.ntheory.factor_ import digits
def A265388(n): return prod((p if sum(digits(n<<1, p)[1:])==2 else 1) for p in primefactors(n*((n<<1)-1)) if p>2)<<(not(n&-n)^n) if n>1 else 0 # Chai Wah Wu, May 04 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Dec 08 2015
STATUS
approved