OFFSET
1,2
COMMENTS
The 15th all-Russia Mathematical Olympiad competition in 1989 asked competitors to prove that a(1001) is an integer: 2002*2003*...*4002 = (4003 - 2001)*(4003 - 2000)*...*(4003 - 1) = 4003*k - 2001! for some large integer k, so 2001! + 2002*2003*...*4002 = 4003*k, which can be divided by 4*1001 - 1 = 4003.
a(n)/((2n - 1)!/((4n - 1)*(n - 1)!)) is an integer.
For 5k+2 < n < 5k+8, k is a natural number, a(n)/10^(k - [k/5]) is an integer, [k/5] is the integral part of k/5.
LINKS
The IMO Compendium Group, 15-st All-Russian Mathematical Olympiad 1989: Grade 9 Problem 1
FORMULA
a(n) = ((2n - 1)! + (4n - 2)!/(2n - 1)!)/(4n - 1) for n > 0.
a(n) = Sum_{k=0..2*(n-1)} (4*n-1)^(2*(n-1)- k)*(-1)^k*sigma(k,[2*n-1]), with the elementary symmetric functions sigma(k,[n]) with [n] := [1, 2, ..., n], and k = 0..n, with sigma(0, [n]) := 1. Generalized from the example for n = 1001 given above. - Wolfdieter Lang, Oct 02 2018
EXAMPLE
a(2) = 18 from 7^2*1 - 7^1*(1 + 2 + 3) + 7^0*(1*2 + 1*3 + 2*3) = 18, from the elementary symmetric functions sigma[k, [3]], k = 0..2. - Wolfdieter Lang, Oct 02 2018
MATHEMATICA
Table[(2n - 1)!/(4n - 1) + (4n - 2)!/(4n - 1)(2n - 1)!, {n, 0, 50}]
PROG
(Magma) [(Factorial(2*n-1)+Factorial(4*n-2)/Factorial(2*n-1))/(4*n-1): n in [1..15]] // (adapted by Vincenzo Librandi, Aug 27 2018)
(PARI) a(n) = ( (2*n - 1)! + (4*n - 2)!/(2*n - 1)!)/(4*n - 1);
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Aug 23 2018
STATUS
approved