OFFSET
1,1
COMMENTS
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:
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);
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).
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.
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.
EXAMPLE
In the following examples, the { } contains the sequence of n couples (nonprime, prime divisors).
a(1) = 2 => {1, 2}
a(2) = 20 = 4*5 => {1, 2, 4, 5}
a(3) = 220 = 4*5*11 => {1, 2, 4, 5, 10, 11}
a(4) = 5980 = 4*5*13*23 => {1, 2, 4, 5, 10, 13, 20, 23}
a(5) = 173420 = 4*5*13*23*29 => {1, 2, 4, 5, 10, 13, 20, 23, 26, 29}
a(6) = 8150740 = 4*5*13*23*29*47 => {1, 2, 4, 5, 10, 13, 20, 23, 26, 29, 46, 47}
MAPLE
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:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jul 07 2013
STATUS
approved