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%I #7 Dec 05 2013 19:56:49
%S 2,3,8,39,52,187,204,863,773,6621,34038,2404,34440,223097,11976,
%T 1106290,1980047,85119892,15308072,496820597,2590416388,1087065675,
%U 4736428784,1128909067,242793786666,2791304683100,273924845940
%N Let f(k, n) be the product of n consecutive numbers beginning with k. Then a(n) is the least k > 1+n*(n-1)/2 such that f(k, n) is a multiple of f(1+n*(n-1)/2, n).
%C f(k, n) = A008279(n+k-1, n). 1+n*(n-1)/2 = A000124(n-1). f(1+n*(n-1)/2, n) = A057003(n).
%C a(28) > 88*10^12.
%e a(4) = 39 because 39*40*41*42 is divisible by 7*8*9*10. No
%e smaller set gives a product that is a multiple of 7*8*9*10.
%Y Cf. A000124, A008279, A057003, A093909.
%K nonn
%O 1,1
%A _Amarnath Murthy_, Apr 24 2004
%E Edited and extended by _David Wasserman_, Apr 25 2007
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