login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) is the smallest integer of the form a*b*c.../p*q*r..., where the numerator and the denominator contain n numbers each and a,b,c,...p,q,r... are all the integers from 1 to 2n.
1

%I #23 Aug 29 2018 10:34:59

%S 2,6,5,70,7,231,858,1430,12155,46189,176358,676039,104006,44574,

%T 1077205,66786710,64822395,90751353,353452638,3829070245,134564468610,

%U 526024740930,2287064091,35830670759,71661341518,281132955186

%N a(n) is the smallest integer of the form a*b*c.../p*q*r..., where the numerator and the denominator contain n numbers each and a,b,c,...p,q,r... are all the integers from 1 to 2n.

%C The largest value of the ratio is C(2n, n). a(4) = 70 is the smallest as well as the largest such integer. The smallest number can arise in more than one ways. i.e. a(3) = (2*5*6)/(1*3*4) = (3*4*5)/(1*2*6) = 5.

%C a(n) is bounded below by the squarefree part of (2n)! and above by C(2n, n). When those bounds are equal, that is a(n); for example, a(4) = 70. When a(n) equals that lower bound, it is fairly easy to compute. That happens for all of the first 3400 terms, anyway.

%C Conjecture: a(n) always equals that lower bound. - _Don Reble_, Jul 01 2003

%e a(3) = (2*5*6)/(1*3*4) = 5. a(5) = (1*7*8*9*10)/(2*3*4*5*6) = 7.

%Y This is conjectured to be the same as A069113; they agree for at least the first 3400 terms.

%K nonn

%O 1,1

%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 26 2003

%E More terms from _Ray Chandler_ and _Don Reble_, Jul 01 2003