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%I #21 Mar 30 2012 18:35:54
%S 8,548,21248,561248,18381248,212781248,5315781248,70925781248,
%T 9912425781248,364206425781248,4497636425781248,465079836425781248,
%U 5779489836425781248,181155019836425781248,2572639519836425781248
%N a(n) = smallest number k such that k is divisible by 2^n, k+1 is divisible by 3^n and k+2 is divisible by 5^n.
%C The sequence a(n) shows an interesting property : a(n) - a(n-1) == 0 mod 10^(n-1), for example a(7) - a(6) = 5103000000 == 0 mod 10^6.
%e a(2) = 548 = 2 ^ 2 * 137 ; 549 = 3 ^ 2 * 61 ; 550 = 2 * 5 ^ 2 * 11.
%p with(numtheory): for k from 1 to 4 do: kk:=0:for n from 1 to 600000 do: xx:=2^k:yy:=3^k:
%p zz:=5^k:if irem(n,xx) =0 and irem(n+1,yy) =0 and irem(n+2,zz) =0 and kk=0 then
%p kk:=1:print(k):print(n):else fi:od:od:
%t Table[ChineseRemainder[{0, -1, -2}, {2^n, 3^n, 5^n}], {n, 15}]
%Y Cf. A063528.
%K nonn
%O 1,1
%A _Michel Lagneau_, Feb 03 2011
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