OFFSET
1,2
COMMENTS
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10001
FORMULA
If n = p_a * p_b * ... * p_h * p_i * p_j * p_k, where p_a <= p_b <= ... <= p_k are (not necessarily distinct) primes (sorted into nondescending order) in the prime factorization of n, then a(n) = p_{k-j} * p_{k-i} * p_{k-h} * ... * p_{k-a} * p_k, where p_{0} = 1 and for k>=1, p_{k} = A000040(k).
a(1)=1, and for n>1, a(n) = p_{A243056(n)} * a(A032742(n)). Here p_{k} stands for 1 when k=0, and otherwise for the k-th prime, A000040(k).
For all n, a(n) = a(A243074(n)).
EXAMPLE
For n = 9 = 3*3 = p_2 * p_2, we have a(n) = p_{3-3} * p_3 = 1*3 = 3. [Like all terms in A070003 this is an example of "degenerate case", where some p's in the product get index 0, and thus are set to 1 by convention used here.]
For n = 10 = 2*5 = p_1 * p_3, we have a(n) = p_{3-1} * p_3 = 3*5 = 15.
For n = 12 = 2*2*3 = p_1 * p_1 * p_2, we have a(n) = p_{2-1} * p{2-1} * p_2 = p_1^2 * p_2 = 12.
For n = 15 = 3*5 = p_2 * p_3, we have a(n) = p_{3-2} * p_3 = 2*5 = 10.
For n = 2200 = 2*2*2*5*5*11 = p_1 * p_1 * p_1 * p_3 * p_3 * p_5, we have a(n) = p_{5-3} * p_{5-3} * p_{5-1} * p_{5-1} * p_{5-1} * p_5 = 3*3*7*7*7*11 = 33957.
For n = 33957 = 3*3*7*7*7*11 = p_2 * p_2 * p_4 * p_4 * p_4 * p_5, we have a(n) = p_{5-4} * p_{5-4} * p_{5-4} * p_{5-2} * p_{5-2} * p_5 = 2*2*2*5*5*11 = 2200.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 31 2014
STATUS
approved