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%I #7 Apr 09 2014 10:15:19
%S 3,5,6,10,12,15,18,20,27,30,36,40,42,45,48,54,60,63,70,72,77,80,88,90,
%T 100,102,105,108,110,120,130,135,138,140,150,153,160,165,170,176,180,
%U 187,192,195,198,204,221,225,228,230,238,240,242,252,259,264,270,272
%N a(n) = the multiple of A142972(n) such that the n-th prime <= a(n) <= the (n+1)th prime.
%C There is always only one multiple of A142972(n) that is between the n-th prime and the (n+1)th prime.
%e The 15th prime is 47 and the 16th prime is 53. So we will examine the integers 47,48,49,50,51,52,53. Now, 1 divides each of these integers. 2 divides 48,50,52. 3 divides 48 and 51. 4 divides 48 and 52. 5 divides 50. 6 divides 48. 7 divides 49. 8 divides 48. But 9 doesn't divide any integer in the span of consecutive integers 47 to 53. So 8 is the largest integer m such that 1,2,3,4,...m each divide at least one integer in the span 47 to 53. 48 is the multiple of 8 among the integers in the span. So a(15) = 48.
%Y Cf. A142972.
%K nonn
%O 1,1
%A _Leroy Quet_, Jul 21 2008
%E Extended by _Ray Chandler_, Jun 21 2009
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