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%I #38 Aug 26 2017 03:50:19
%S 6,9,16,27,28,95,96,121,122,123,124,125,126,537,538,539,540,905,906,
%T 1149,1150,1349,1350,1351,1352,1353,1354,1355,1356,1357,1358,1359,
%U 1360,9585,9586,15719,15720,15721,15722,15723,15724,15725,15726,19653,19654,19655
%N a(n) is the smallest integer k > n such that (k+1)(k+2)...(2k-2n+1)/(k(k-1)...(k-n+1)) is an integer.
%C This sequence comes from a small change of an exercise proposed by Paul Erdős for Crux Mathematicorum (see link). In the solution, they show that for n >= 3, the fraction is always an integer for k = (n+1)! - 2. Be careful, n and k are swapped between Crux Mathematicorum and this sequence.
%H Chai Wah Wu, <a href="/A290791/b290791.txt">Table of n, a(n) for n = 1..100</a>
%H Wojcich Komornicki, <a href="https://cms.math.ca/crux/backfile/Crux_v8n02_Feb.pdf">Problem 556</a>, Crux Mathematicorum, page 49, Vol. 8, Feb. 82.
%e If n = 1, for k = 2, 3, 4, 5, the fraction is respectively equal to 3/2, (4*5)/3, (5*6*7)/4, (6*7*8*9)/5 but for k = 6, the quotient is (7*8*9*10*11)/6 = 9240 and so a(1) = 6.
%t a[n_] := Block[{k = n+1}, While[! IntegerQ[(1 + 2*k - 2*n)! (k-n)! / (k!)^2], k++]; k]; Array[a, 30] (* _Giovanni Resta_, Aug 11 2017 *)
%K nonn
%O 1,1
%A _Bernard Schott_, Aug 10 2017
%E a(6)-a(46) from _Giovanni Resta_, Aug 11 2017