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%I #9 Jun 30 2024 22:08:57
%S 29,76,141,172,193,197,205,211,245,181,162,85,86,69,71,61,62,67,52,53,
%T 58,58,58,59,62,45,46,49,20,51,7,22,10,7,10,7,7,7,7,15,12,13,7,7,7,7,
%U 7,7,7,7,7,7,7,7,10,7,7,7,7,7,7,10,7,7,7,7,7,7,7,7,7
%N a(n) is the smallest integer >=0 we cannot obtain from n, n+1, n+2, n+3, n+4 and the operators +, -, *, /, using each number only once.
%C It seems that a(n)=7 for all n>=62, but this needs to be proved.
%H Gilles Bannay, <a href="https://web.archive.org/web/20061201125224/http://gilles.bannay.free.fr/jeux_us.html">Countdown Problem</a>
%e a(61)=10 because by using 61, 62, 63, 64, 65 we can get 62-61=1, 63-61=2, ..., 63/((64-61)*(65-62))=7, (65-61)*(64+62)/63=8, (65-62)*(64-61)=9 but we cannot obtain 10 in the same way.
%e a(32) != 7 because 7 = 35 / (34 - (32 - (36 - 33))). - _Sean A. Irvine_, Jun 30 2024
%Y Cf. A060316.
%K nonn
%O 0,1
%A Koksal Karakus (karakusk(AT)hotmail.com), May 27 2002
%E a(32) and a(40) corrected by _Sean A. Irvine_, Jun 30 2024