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%I #39 Sep 08 2022 08:46:22
%S 1,18,2760,1153488,928609920,1224288172800,2398694768870400,
%T 6543329233529088000,23715268426751879577600,
%U 110245593949982051033088000,639537254337962130647777280000,4528740191242360945670704005120000,38446695454134018174768929636352000000
%N a(n) = ((2n - 1)! + (4n - 2)!/(2n - 1)!)/(4n - 1).
%C 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.
%C a(n)/((2n - 1)!/((4n - 1)*(n - 1)!)) is an integer.
%C 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.
%H The IMO Compendium Group, <a href="https://imomath.com/othercomp/Rus/RusMO89.pdf">15-st All-Russian Mathematical Olympiad 1989: Grade 9 Problem 1</a>
%F a(n) = ((2n - 1)! + (4n - 2)!/(2n - 1)!)/(4n - 1) for n > 0.
%F 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
%e 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
%t Table[(2n - 1)!/(4n - 1) + (4n - 2)!/(4n - 1)(2n - 1)!, {n, 0, 50}]
%o (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)
%o (PARI) a(n) = ( (2*n - 1)! + (4*n - 2)!/(2*n - 1)!)/(4*n - 1);
%K nonn
%O 1,2
%A _Jinyuan Wang_, Aug 23 2018
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