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%I #21 Dec 02 2018 02:58:58
%S 2,20,220,5980,173420,8150740,431989220,25487363980,1707653386660,
%T 742559959944940,79453915714108580,10090647295691789660,
%U 1757620212758033277820,286492094679559424284660,47844179811486423855538220
%N Smallest number k such that among its first 2n divisors nonprimes and primes alternate, as {1, a prime, a nonprime, a prime, ...} up to the 2n-th smallest divisor of k.
%C The first 2n divisors of a(n) = k are of the form {1, 2, c(1), p(1), c(2), p(2), ..., c(n), p(n)} where c(i) are composite divisors and p(i) are prime divisors with k = 4*p(1)*p(2)*...*p(n). Suppose the (2*n+1)-th and the (2n+2)-th divisor of k are composite numbers c(n+1) and c(n+2), so we have two cases:
%C Case 1: if there exists a prime number p(n+1) such that c(n+1) < p(n+1) < c(n+2), then a(n+1) = a(n)*p(n+1);
%C Case 2: if no such prime exists, it is necessary to consider a'(n) = 4*p'(1)*p'(2)*...*p'(n) such that there exist two composites c'(n+1) and c'(n+2) where c'(n+1) < p'(n+1) < c'(n+2), and a(n+1) = a'(n)* p'(n+1).
%C Example 1: a(4) = 5980 = 4*5*13*23 has the divisors {1, 2, 4, 5, 10, 13, 20, 23, 26, 46, 52, 65, 92, 115, 130, 230, 260, 299, 460, 598, 1196, 1495, 2990, 5980} with 4 couples (1, 2), (4, 5), (10, 13), (20, 23). The smallest prime between the next two divisors, 26 and 46, is 29. So a(5) = 4*5*13*23*29.
%C Example 2: a(3) = 220 = 4*5*11 has the divisors {1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220} with 3 couples (nonprime, prime) = (1, 2), (4, 5), (10, 11). But no prime exists between the next two divisors, 20 and 22. We consider a'(3) = 4*5*13 with the divisors {1, 2, 4, 5, 10, 13, 20, 26, 52, 65, 130, 260} with 3 couples (nonprime, prime) = (1, 2), (4, 5), (10, 13) => there exists the prime 23 between the next two divisors, 20 and 26. So a(4) = 4*5*13*23 = 5980.
%e In the following examples, the { } contains the sequence of n couples (nonprime, prime divisors).
%e a(1) = 2 => {1, 2}
%e a(2) = 20 = 4*5 => {1, 2, 4, 5}
%e a(3) = 220 = 4*5*11 => {1, 2, 4, 5, 10, 11}
%e a(4) = 5980 = 4*5*13*23 => {1, 2, 4, 5, 10, 13, 20, 23}
%e a(5) = 173420 = 4*5*13*23*29 => {1, 2, 4, 5, 10, 13, 20, 23, 26, 29}
%e a(6) = 8150740 = 4*5*13*23*29*47 => {1, 2, 4, 5, 10, 13, 20, 23, 26, 29, 46, 47}
%p with(numtheory):for n from 2 by 2 to 30 do:ii:=0:for k from 2 to 10^8 while(ii=0) do: x:=divisors(k):n1:=nops(x):it:=0:for i from 1 by 2 to n do:if n1>=n and type(x[i],prime)=false then it:=it+1:else fi:od:for j from 2 by 2 to n do:if n1>=n and type(x[j],prime)=true then it:=it+1:else fi:od:if it=n then ii:=1: printf ( "%d %d \n",n/2,k):else fi:od:od:
%K nonn
%O 1,1
%A _Michel Lagneau_, Jul 07 2013
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