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%I #11 Sep 17 2017 22:24:39
%S 2,6,30,1155,15015,969969,37182145,1078282205,33426748355,
%T 7244053893505,311494317420715,14640232918773605,775932344695001065,
%U 64092011671807087969,3909612711980232366109,261944051702675568529303,18598027670889965365580513,1357656019974967471687377449
%N Smallest number k such that the n prime distinct divisors of k are also n consecutive divisors of k.
%C Property of the prime divisors: Let p(1), p(2), ..., p(n) the n prime divisors of k. Necessarily, p(1)*p(2) > p(n) for n > 2. Hence another definition of this sequence: smallest number k such that the product of n distinct primes p(1), p(2), ..., p(n) is minimized with the condition p(1)*p(2) > p(n).
%e a(4) = 1155 because the 4 prime divisors are {3, 5, 7, 11}and the divisors of 1155 are {1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155} with 4 prime consecutive divisors and the property 3*5 > 11.
%e Factorizations of the first several terms are as follows:
%e n=1 => 2;
%e n=2 => 2 * 3 = 6;
%e n=3 => 2 * 3 * 5 = 30;
%e n=4 => 3 * 5 * 7 * 11 = 1155;
%e n=5 => 3 * 5 * 7 * 11 * 13 = 15015;
%e n=6 => 3 * 7 * 11 * 13 * 17 * 19 = 969969;
%e n=7 => 5 * 7 * 11 * 13 * 17 * 19 * 23 = 37182145;
%e ...
%p with(numtheory): printf ( "%d %d \n",2,6):T:=array(1..2):i:=1:for n from 3 to 20 do:for it from 1 to 2 do:ii:=0:for j from i+1 to i+3 while(ii=0) do:if ithprime(i)*ithprime(j)> ithprime(j+n-2) then pp:=product('ithprime(a)', 'a'=j..j+n-2):ii:=1:pp:=pp*ithprime(i):T[it]:=pp:i:=i+1:else fi:od:od: ppp:=min(T[1],T[2]): printf(`%d, `,ppp):if T[2]> T[1] then i:=i-2:else i:=i-1:fi:od:
%K nonn
%O 1,1
%A _Michel Lagneau_, Jul 01 2013
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