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Final terms of rows of A077553.
5

%I #10 Dec 02 2015 03:55:38

%S 4,6,9,10,15,25,21,25,35,49,35,49,55,77,121,65,77,91,121,143,169,119,

%T 121,143,169,187,221,289,169,187,209,221,247,289,323,361,247,253,289,

%U 299,323,361,391,437,529,323,361,377,391,437,493,529,551,667,841,437

%N Final terms of rows of A077553.

%C If there are two sets of distinct composite numbers satisfying the above condition then the set with lesser product is chosen irrespective of the number of prime divisors. Perhaps the ambiguity may not arise. E.g., row 6 is 4,6,9,10,15,25. This row cannot be extended to get the next row without bringing in another prime because every number divisible by 2,3 or 5 will be a multiple of one of the previous terms. Hence in row 7, prime 7 has to be brought in and then we get a new set of numbers: 4,6,9,10,14,15,21.

%Y Cf. A002024.

%Y Cf. A001358, A005843, A077553, A077555, A087112.

%K nonn

%O 0,1

%A _Amarnath Murthy_, Nov 10 2002

%E More terms from _Ray Chandler_, Aug 24 2003